Re: [PHP] Passing variable to a page in a frameset
I apologize for my ignorance, I don't really know much about
javascript. When I add all that into my form page, when I submit the
form, it just replaces the page I was on with the form results, rather
than in the new frame page. I'm assuming I need to put the url for the
frameset page in that code somewhere. Where does it go? And, which
part do I replace with the frame name on that frameset page?
Thank you for taking the time to help me with this, I really appreciate it!
Let's look at the code:
//var1 and var2 are the search criteria the user entered in your search
form(You may have more or less)
function submitForm(var1,var2) {
//top.leftFrame is how you set the focus to the frame you want. leftFrame
could be different on your system...whatever you named the frame.
//top.leftFrame.document.my_search.text1.value=var1 assumes the form name in
the frame you want data in is named
//my_search and it givs the text object text1 the value of var1 from your
search form
top.leftFrame.document.my_search.text1.value = var1;
//top.leftFrame.document.my_search.text2.value=var2 assumes the form name in
the frame you want data in is named
//my_search and it givs the text object text1 the value of var2 from your
search form
top.leftFrame.document.my_search.text2.value = var2;
//top.leftFrame.document.my_search.submit() will submit the form in your
target frame then you can use the $_POST values throughout the frame
top.leftFrame.document.my_search.submit();
//document.search_form.submit() will submit your search page. I use this so
I can reuse the $_POST values to display the search criteria after submit.
document.search_form.submit();
}
Re: [PHP] Passing variable to a page in a frameset
On Aug 15, 2008, at 7:33 AM, Dan Shirah wrote:
I apologize for my ignorance, I don't really know much about
javascript. When I add all that into my form page, when I submit the
form, it just replaces the page I was on with the form results, rather
than in the new frame page. I'm assuming I need to put the url for the
frameset page in that code somewhere. Where does it go? And, which
part do I replace with the frame name on that frameset page?
Thank you for taking the time to help me with this, I really
appreciate it!
Let's look at the code:
//var1 and var2 are the search criteria the user entered in your
search form(You may have more or less)
function submitForm(var1,var2) {
//top.leftFrame is how you set the focus to the frame you want.
leftFrame could be different on your system...whatever you named the
frame.
//top.leftFrame.document.my_search.text1.value=var1 assumes the form
name in the frame you want data in is named
//my_search and it givs the text object text1 the value of var1 from
your search form
top.leftFrame.document.my_search.text1.value = var1;
//top.leftFrame.document.my_search.text2.value=var2 assumes the form
name in the frame you want data in is named
//my_search and it givs the text object text1 the value of var2 from
your search form
top.leftFrame.document.my_search.text2.value = var2;
//top.leftFrame.document.my_search.submit() will submit the form in
your target frame then you can use the $_POST values throughout the
frame
top.leftFrame.document.my_search.submit();
//document.search_form.submit() will submit your search page. I use
this so I can reuse the $_POST values to display the search criteria
after submit.
document.search_form.submit();
}
In the head of my page, I have this:
script type=text/javascript
function submitForm(var1,var2) {
top.mainFrame.document.my_search.text1.value = var1;
top.mainFrame.document.my_search.text2.value = var2;
top.mainFrame.document.my_search.submit();
document.search_form.submit();
}
/script
Then, for the form code I have this:
form name=search_form method=post action=http://webcat.winnefox.org/web2/tramp2.exe/do_keyword_search/guest
id=qpl
input type=hidden name=SETTING_KEY value=Menasha /
input type=hidden name=index value=default /
input TYPE=hidden NAME=hitlist_screen VALUE=hitlist.html /
input TYPE=hidden NAME=record_screen VALUE=Record.html /
input type=hidden name=query_screen value=home.html /
input type=hidden name=servers value=1home /
Find library books, music, movies and more...input
class=searchbartextfield id=query name=query type=text a
href
=
javascript:submitForm
(document
.search_form.var1.value,document.search_form.var2.value)SEARCH/
anbsp;a id=topNav_advanced_search href=/sites/
beta.menashalibrary.org/themes/salamander/searchframe.htmlAdvanced
Search/a
/form
If I type in a word in the search box http://beta.menashalibrary.org/about
and hit enter, it searches the catalog just fine, but then just
replaces the current page with the search results. If I click on the
SEARCH link with the javascript, it does nothing.
I want the results to go to this url:
http://beta.menashalibrary.org/sites/beta.menashalibrary.org/themes/salamander/searchframe.html
in the frame name of mainFrame. I'm thinking that address needs to go
in the javascript somewhere so it knows where to go, but where? Does
the rest of the code look ok?
Thanks again for taking the time to help me with this.
- jody
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Re: [PHP] Passing variable to a page in a frameset
In the head of my page, I have this:
script type=text/javascript
function submitForm(var1,var2) {
top.mainFrame.document.my_search.text1.value = var1;
top.mainFrame.document.my_search.text2.value = var2;
top.mainFrame.document.my_search.submit();
document.search_form.submit();
}
/script
If I type in a word in the search box http://beta.menashalibrary.org/about and
hit enter, it searches the catalog just fine, but then just replaces the
current page with the search results. If I click on the SEARCH link with the
javascript, it does nothing.
It looks like you haven't adapted the example I gave you to conform to your
page/form structure.
top.mainFrame.document.my_search.text1.value = var1;
The frame you are populating may not be named mainFrame
The form in your destination frame may not be named my_search
The input in you destination frame may not be text1
You have to modify these areas to suit your application and add or remove
more as needed.
Re: [PHP] Passing variable to a page in a frameset
On Aug 15, 2008, at 11:32 AM, Dan Shirah wrote:
In the head of my page, I have this:
script type=text/javascript
function submitForm(var1,var2) {
top.mainFrame.document.my_search.text1.value = var1;
top.mainFrame.document.my_search.text2.value = var2;
top.mainFrame.document.my_search.submit();
document.search_form.submit();
}
/script
If I type in a word in the search box http://beta.menashalibrary.org/about
and hit enter, it searches the catalog just fine, but then just
replaces the current page with the search results. If I click on the
SEARCH link with the javascript, it does nothing.
It looks like you haven't adapted the example I gave you to conform
to your page/form structure.
top.mainFrame.document.my_search.text1.value = var1;
The frame you are populating may not be named mainFrame
The form in your destination frame may not be named my_search
The input in you destination frame may not be text1
You have to modify these areas to suit your application and add or
remove more as needed.
So, I had this all wrong before. Basically, I need two forms, right?
One on my originating page, and one on the page within the frameset I
want to pass the values to. Correct?
My form, which I named my_search has one input field, which I named
query. In the javascript, I put in this:
script type=text/javascript
function submitForm(var1,var2) {
top.mainFrame.document.my_search.query.value = var1;
top.mainFrame.document.my_search.submit();
document.search_form.submit();
}
/script
In the framset named mainFrame, I have a page which matches the form
on the originating page.
Am I on the right track? When I click on the search link, nothing
happens.
- jody
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Re: [PHP] Passing variable to a page in a frameset
So, I had this all wrong before. Basically, I need two forms, right? One
on my originating page, and one on the page within the frameset I want to
pass the values to. Correct?
My form, which I named my_search has one input field, which I named query.
In the javascript, I put in this:
script type=text/javascript
function submitForm(var1,var2) {
top.mainFrame.document.my_search.query.value = var1;
top.mainFrame.document.my_search.submit();
document.search_form.submit();
}
/script
In the framset named mainFrame, I have a page which matches the form on
the originating page.
Am I on the right track? When I click on the search link, nothing happens.
mainFrame should not be the name of your frameset. Say I have a frameset
with three sections. Top, Left, and Right...you will actually have 4 pages.
Frameset.php //The top level of the frameset that doesn't do much more than
assign the Frame names and pages associated with them
TopSection.php //Anything on this page will be displayed in your top frame
LeftSection.php // Anything on this page will be displayed in your left
frame
RightSection.php //Anything on this page will be displayed in your right
frame
In order to see the names assigned to your different frames you need to open
Frameset.php and look for them.
It will look something like this:
frameset rows=200,* cols=* framespacing=0 frameborder=NO
border=1
frame src=TopSection.php name=topFrame scrolling=NO noresize
frame src=LeftSection.php name=leftFrame
frame src=RightSection.php name=rightFrame
/frameset
You should never need to reference Framset.php directly.
Say I wanted to pass a value from TopSection.php to a text field in
LeftSection.php I would do this:
TopSection.php
form name=search_form method=post action=?php echo
$_SERVER['PHP_SELF']; ?
input type=text name=search_name value= //The default value will be
blank but will change if someone enters in text.
/form
//Now the persons submits the form and you want the search name to populate
to your other frame.
a
href=javascript:submitForm(document.search_form.search_name.value)SEARCH/a
//The Javascript function this calls will pass the
document.search_form.search_name.value to the new form
function submitForm(name) {
top.leftFrame.document.my_search.search_name.value = name;
document.search_form.submit();
}
//top goes to the highest level of your frameset, leftFrame is the frame
name specified for LeftSection.php on your Frameset.php page,
document.my_search is the name of your form in
//LeftSection.php, search_name is the name of your text field within the
form in LeftSection.php
//Your TopSection.php form has now passed the search_name value to the
LeftSection.php page
LeftSection.php
form name=my_search method=post action=
input type=text name=search_name value=
/form
That should be enough to get you going. If you have any further questions
lets take this Off List since it does not relate to PHP.
Dan
[PHP] Passing variable to a page in a frameset
Hello, I've got a website here: http://beta.menashalibrary.org/about On every page, i've got a search box at the top. This search box searches the library's web catalog. The problem is, when someone searches, it takes them away from the site. What I'd like to do is take what a person searches for, and load it into the bottom frame of this page: http://beta.menashalibrary.org/sites/beta.menashalibrary.org/themes/salamander/searchframe.html Is there a way, with php, to take what someone puts in the search box and put the results into the bottom frame of a frameset when the originating page does not contain frames? - jody -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Passing variable to a page in a frameset
On Thu, Aug 14, 2008 at 4:34 PM, Jody Cleveland [EMAIL PROTECTED]wrote: Hello, I've got a website here: http://beta.menashalibrary.org/about On every page, i've got a search box at the top. This search box searches the library's web catalog. The problem is, when someone searches, it takes them away from the site. What I'd like to do is take what a person searches for, and load it into the bottom frame of this page: http://beta.menashalibrary.org/sites/beta.menashalibrary.org/themes/salamander/searchframe.html Is there a way, with php, to take what someone puts in the search box and put the results into the bottom frame of a frameset when the originating page does not contain frames? - jody -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php This is more a javascript / html -- Bastien Cat, the other other white meat
Re: [PHP] Passing variable to a page in a frameset
Hello,
I've got a website here: http://beta.menashalibrary.org/about
On every page, i've got a search box at the top. This search box searches
the library's web catalog. The problem is, when someone searches, it
takes
them away from the site. What I'd like to do is take what a person
searches
for, and load it into the bottom frame of this page:
http://beta.menashalibrary.org/sites/beta.menashalibrary.org/themes/salamander/searchframe.html
Is there a way, with php, to take what someone puts in the search box and
put the results into the bottom frame of a frameset when the originating
page does not contain frames?
- jody
Bastien is right, this is more of a Javascript issue.
It can be accomplished very easily by doing the following.
Once the user enters the search criteria have the submit button/link call a
javascript function like the following:
function submitForm(var1,var2) {
top.leftFrame.document.my_search.text1.value = var1;
top.leftFrame.document.my_search.text2.value = var2;
top.leftFrame.document.my_search.submit();
document.search_form.submit();
}
And this would be your submit link:
a
href=javascript:submitForm(document.search_form.var1.value,document.search_form.var2.value)SEARCH/a
Re: [PHP] Passing variable to a page in a frameset
On 14 Aug 2008, at 21:34, Jody Cleveland wrote: I've got a website here: http://beta.menashalibrary.org/about On every page, i've got a search box at the top. This search box searches the library's web catalog. The problem is, when someone searches, it takes them away from the site. What I'd like to do is take what a person searches for, and load it into the bottom frame of this page: http://beta.menashalibrary.org/sites/beta.menashalibrary.org/themes/salamander/searchframe.html Is there a way, with php, to take what someone puts in the search box and put the results into the bottom frame of a frameset when the originating page does not contain frames? Set the target attribute of the search form to the name of the bottom frame. The form will then be POSTed in that frame. -Stut -- http://stut.net/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Passing variable to a page in a frameset
On Thu, Aug 14, 2008 at 3:43 PM, Dan Shirah [EMAIL PROTECTED] wrote:
Hello,
I've got a website here: http://beta.menashalibrary.org/about
On every page, i've got a search box at the top. This search box
searches
the library's web catalog. The problem is, when someone searches, it
takes
them away from the site. What I'd like to do is take what a person
searches
for, and load it into the bottom frame of this page:
http://beta.menashalibrary.org/sites/beta.menashalibrary.org/themes/salamander/searchframe.html
Is there a way, with php, to take what someone puts in the search box
and
put the results into the bottom frame of a frameset when the originating
page does not contain frames?
- jody
Bastien is right, this is more of a Javascript issue.
It can be accomplished very easily by doing the following.
Once the user enters the search criteria have the submit button/link call a
javascript function like the following:
function submitForm(var1,var2) {
top.leftFrame.document.my_search.text1.value = var1;
top.leftFrame.document.my_search.text2.value = var2;
top.leftFrame.document.my_search.submit();
document.search_form.submit();
}
And this would be your submit link:
a
href=javascript:submitForm(document.search_form.var1.value,document.search_form.var2.value)SEARCH/a
I apologize for my ignorance, I don't really know much about
javascript. When I add all that into my form page, when I submit the
form, it just replaces the page I was on with the form results, rather
than in the new frame page. I'm assuming I need to put the url for the
frameset page in that code somewhere. Where does it go? And, which
part do I replace with the frame name on that frameset page?
Thank you for taking the time to help me with this, I really appreciate it!
- jody
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Re: [PHP] Passing variable to a page in a frameset
At 4:06 PM -0500 8/14/08, george wrote: Thank you for taking the time to help me with this, I really appreciate it! - jody Make it easy on yourself and try this: http://sperling.com/examples/search/ Cheers, tedd -- --- http://sperling.com http://ancientstones.com http://earthstones.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Passing variable number of parameters by reference
Hello,
Using: PHP Version 5.0.3
I build the following function:
function Set4HTMLOut() {
$C = func_num_args();
for ($I = 0; $I $C; $I++) {
$A = func_get_arg($I);
echo 'I['.$I.']('.strlen($A).'): '.$A.'br /';
$A = htmlspecialchars($A);
echo 'I['.$I.']('.strlen($A).'): '.$A.'br /';
}
}
And called it like this:
Set4HTMLOut($LName, $LOrg, $LWebSite, $LPhones, $LComments);
echo 'br /After call:br /'.
'LName: '.$LName.'br /'.
'LOrg: '.$LOrg.'br /'.
'LWebSite: '.$LWebSite.'br /'.
'LPhones: '.$LPhones.'br /'.
'LComments: '.$LComments.'br /';
The echo statements in the Set4HTMLOut() function echos all the correct
data and proves the data was received correctly then altered by the
htmlspecialchars() function appropriately. Notice that the caller passes
each variable as a reference. Yet, the echo statement after the call
displays the data without any changes, like the first echo statement in
the Set$HTMLOut() function does.
Is it not possible to pass variables by reference to a variable-length
parameter list? Or am I doing something wrong? If so, what?
I have already built a work around. I do not need any work arounds. I just
think one should be able to pass variables by reference to a function that
accepts a variable-length parameter list.
Regards
Robert
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Re: [PHP] Passing variable number of parameters by reference
I believe it's mentioned somewhere in the manual you can't do that.
func_get_arg() always return a copy. You can workaround this using an array.
Robert Meyer wrote:
Hello,
Using: PHP Version 5.0.3
I build the following function:
function Set4HTMLOut() {
$C = func_num_args();
for ($I = 0; $I $C; $I++) {
$A = func_get_arg($I);
echo 'I['.$I.']('.strlen($A).'): '.$A.'br /';
$A = htmlspecialchars($A);
echo 'I['.$I.']('.strlen($A).'): '.$A.'br /';
}
}
And called it like this:
Set4HTMLOut($LName, $LOrg, $LWebSite, $LPhones, $LComments);
echo 'br /After call:br /'.
'LName: '.$LName.'br /'.
'LOrg: '.$LOrg.'br /'.
'LWebSite: '.$LWebSite.'br /'.
'LPhones: '.$LPhones.'br /'.
'LComments: '.$LComments.'br /';
The echo statements in the Set4HTMLOut() function echos all the correct
data and proves the data was received correctly then altered by the
htmlspecialchars() function appropriately. Notice that the caller passes
each variable as a reference. Yet, the echo statement after the call
displays the data without any changes, like the first echo statement in
the Set$HTMLOut() function does.
Is it not possible to pass variables by reference to a variable-length
parameter list? Or am I doing something wrong? If so, what?
I have already built a work around. I do not need any work arounds. I just
think one should be able to pass variables by reference to a function that
accepts a variable-length parameter list.
Regards
Robert
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[PHP] Passing variable on include
Hi, I have to pass variable on include include(includes/include.php); include(includes/include.php?name=$name); wouldn't work. How do I do that? Cokies are not an option, btw. -- Tadas Talaikis [EMAIL PROTECTED] http://www.xongoo.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Passing variable on include
Xongoo!com: Central unit wrote: Hi, I have to pass variable on include include(includes/include.php); include(includes/include.php?name=$name); wouldn't work. How do I do that? Cokies are not an option, btw. Code in includes/include.php is executed in the scope of the including file, so all variables available at the include() will be available in the included file. Thus the solution is: $name = 'Your Name'; include(includes/include.php); -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Passing variable on include
Duh, really, and just read on varianle scopes on php.net. Gotta to sleep for some time :) -- Tadas Talaikis [EMAIL PROTECTED] http://www.xongoo.com - Original Message - From: Marek Kilimajer [EMAIL PROTECTED] To: Xongoo!com: Central unit [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Sent: Sunday, August 29, 2004 11:41 AM Subject: Re: [PHP] Passing variable on include Xongoo!com: Central unit wrote: Hi, I have to pass variable on include include(includes/include.php); include(includes/include.php?name=$name); wouldn't work. How do I do that? Cokies are not an option, btw. Code in includes/include.php is executed in the scope of the including file, so all variables available at the include() will be available in the included file. Thus the solution is: $name = 'Your Name'; include(includes/include.php); -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Passing variable from webpage to php (newbie?)
Hiya It could need setting register_globals =on in your php.ini if after that still problems then you may need to look into sessions and in particular session_start() and $_SESSION['varname'] and make sure the variables are global so that more than one script can use them. Hope this steers you in right direction *warning im a newbie too so you may wait for some more replies to confirm what im saying* Bobster From: Joe Kupiszewski [EMAIL PROTECTED] Reply-To: Joe Kupiszewski [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: [PHP] Passing variable from webpage to php (newbie?) Date: Wed, 19 Mar 2003 10:57:11 -0500 I think I already tried to post once, so sorry if this is a duplicate, but I don't see my first attempt. I am trying to do what should be a relatively simple and basic task. I've got a php script/page that has a switch/case selection statement in it. Obviously, depending on what value a particular variable takes when passed to the script, the script SHOULD :) do different things. However, when I invoke the script using www.somedomain.com/somephpscript.php?action=1 (substitute one with, 2, 3, 4 or whatever) and then do a check whether any value is passed to my script, it always tells me the value is empty ( if (empty ($action)) - it just always thinks its empty. I'm copying this script from a book, so I do not have any reason to believe there is an error in the code, but obviously something is not happening properly. My thought is that perhaps something needs to be turned on in the php.ini or in the apache httpd.conf file to allow this variable passing to work. Is there some other way to do this? Sorry for the long paragraph sentence. I'll be happy to post the code if needed or provide any additional information or give the actual URL so you can see what is happening. Thanks for any thoughts -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php _ Overloaded with spam? With MSN 8, you can filter it out http://join.msn.com/?page=features/junkmailpgmarket=en-gbXAPID=32DI=1059 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Passing variable from webpage to php (newbie?)
I think I already tried to post once, so sorry if this is a duplicate, but I don't see my first attempt. I am trying to do what should be a relatively simple and basic task. I've got a php script/page that has a switch/case selection statement in it. Obviously, depending on what value a particular variable takes when passed to the script, the script SHOULD :) do different things. However, when I invoke the script using www.somedomain.com/somephpscript.php?action=1 (substitute one with, 2, 3, 4 or whatever) and then do a check whether any value is passed to my script, it always tells me the value is empty ( if (empty ($action)) - it just always thinks its empty. I'm copying this script from a book, so I do not have any reason to believe there is an error in the code, but obviously something is not happening properly. My thought is that perhaps something needs to be turned on in the php.ini or in the apache httpd.conf file to allow this variable passing to work. Is there some other way to do this? Sorry for the long paragraph sentence. I'll be happy to post the code if needed or provide any additional information or give the actual URL so you can see what is happening. Thanks for any thoughts -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] passing variable question
Hello all, I have a php script that reads all the files in a directory (pictures), the outputs each picture as a link to be displayed in the main frame of the html page. All of that works fine. I want to be able to pretty up the picture out put by placing the picture into a html table so it is centered, add color, etc. But I cannot seem to figure out how to pass the variable from the menu page to the main frame page without using a goofy gray button form. Anyone know how to pass a variable on just a link alone? Thanks, Ronnie -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] passing variable question
You just need to add a question mark, the name of the variable, an equal sign and its value (properly encoded): echo 'a href=myurl?myvalue=' . urlencode ($value) . ''; In the receiving script, you can read the value using $_GET['myvalue']; Hope this helps! Marco -- php|architect - The magazine for PHP Professionals The first monthly worldwide magazine dedicated to PHP programmers Come visit us at http://www.phparch.com! ---BeginMessage--- Hello all, I have a php script that reads all the files in a directory (pictures), the outputs each picture as a link to be displayed in the main frame of the html page. All of that works fine. I want to be able to pretty up the picture out put by placing the picture into a html table so it is centered, add color, etc. But I cannot seem to figure out how to pass the variable from the menu page to the main frame page without using a goofy gray button form. Anyone know how to pass a variable on just a link alone? Thanks, Ronnie -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php ---End Message--- -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] passing variable question
Pass if via the URL a href=yourlink.php?picture_data=whateverLink/a On the display page, access the data you passed by... $_GET['picture_data'] Do a google search for get query. Ron Clark wrote: Hello all, I have a php script that reads all the files in a directory (pictures), the outputs each picture as a link to be displayed in the main frame of the html page. All of that works fine. I want to be able to pretty up the picture out put by placing the picture into a html table so it is centered, add color, etc. But I cannot seem to figure out how to pass the variable from the menu page to the main frame page without using a goofy gray button form. Anyone know how to pass a variable on just a link alone? Thanks, Ronnie -- By-Tor.com It's all about the Rush http://www.by-tor.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] passing variable via url ( newbye question)
I can pass one variable using the url on this way http://myadress/php/mypage.php?modo=123 and to read i have the code echo $_GET['modo'].BR; That's work fine on 4.21 How can I pass and read 3 variables instead of one ? I nedd to pass mode = 123 color= red size=3 using the url I tried http://myadress/php/mypage.php?modo=123color=redsize=3 and to get echo $_GET['modo'].BR; echo $_GET['color'].BR; echo $_GET['size'].BR; but does not work How can i do that -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] passing variable via url ( newbye question)
http://myaddress/php/mypage.php?modo=123color=redsize=3 Just separate name/value pairs with ampersand (). HTH Martin Clifford Homepage: http://www.completesource.net Developer's Forums: http://www.completesource.net/forums/ Saci [EMAIL PROTECTED] 07/30/02 04:09PM I can pass one variable using the url on this way http://myadress/php/mypage.php?modo=123 and to read i have the code echo $_GET['modo'].BR; That's work fine on 4.21 How can I pass and read 3 variables instead of one ? I nedd to pass mode = 123 color= red size=3 using the url I tried http://myadress/php/mypage.php?modo=123color=redsize=3 and to get echo $_GET['modo'].BR; echo $_GET['color'].BR; echo $_GET['size'].BR; but does not work How can i do that -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] passing variable via url ( newbye question)
You are passing the variables correctly. If that's your actual code echo $_GET['modo'].BR; echo $_GET['color'].BR; echo $_GET['size'].BR; then you need br (note the closing ) - PHP is doing what you told it, and your browser is confused by the output. If PHP were the problem you'd probably see an error message. Also, upgrade to 4.2.2 please :) Regards, Colin Teubner -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] passing variable via url ( newbye question)
Is this a direct copy from your code? echo $_GET['modo'].BR; echo $_GET['color'].BR; echo $_GET['size'].BR; if so, you'll need to close the br tags see if that helps Martin -Original Message- From: Saci [mailto:[EMAIL PROTECTED]] Sent: Wednesday, July 31, 2002 6:09 AM To: [EMAIL PROTECTED] Subject: [PHP] passing variable via url ( newbye question) I can pass one variable using the url on this way http://myadress/php/mypage.php?modo=123 and to read i have the code echo $_GET['modo'].BR; That's work fine on 4.21 How can I pass and read 3 variables instead of one ? I nedd to pass mode = 123 color= red size=3 using the url I tried http://myadress/php/mypage.php?modo=123color=redsize=3 and to get echo $_GET['modo'].BR; echo $_GET['color'].BR; echo $_GET['size'].BR; but does not work How can i do that -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] passing variable arguments from select
I still cant get it right,
in query.php i have this piece of code;
echo form action='test2.php' action='post'
select name='language'
option value='0'selected='yes'one/option
option value='1'two/option
/select
input type=text name=query
input type=submit name=submit
/form;
and in test2.php i have this;
if ($_POST['language'] = 1)
{$lang = EngP=1;}
else {$lang = EngP=0;}
echo $lang ;
But it always echoes EngP=1 no matter what I choose, what is wrong here??
Erik Price [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
On Wednesday, June 26, 2002, at 02:51 PM, Haddad Said wrote:
Hi, i am having a problem passing variables from a drop down menu in a
form;
from action=test.php action=post
select name=example
option selected one/option
option two/option
/form
You should do it like this (be sure to properly quote your attributes,
and use the 'value' attribute of the 'option' tag):
print form action='test.php' method='post'
select name='example'
option selected='yes' value='1'One/option
option value='2'Two/option
/select
/form\n;
In the test.php page, the selected value will be found in the
$_POST['example'] variable.
Erik
Erik Price
Web Developer Temp
Media Lab, H.H. Brown
[EMAIL PROTECTED]
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Re: [PHP] passing variable arguments from select
On Thu, 27 Jun 2002, Haddad Said wrote:
in query.php i have this piece of code;
echo form action='test2.php' action='post'
select name='language'
option value='0'selected='yes'one/option
option value='1'two/option
/select
input type=text name=query
input type=submit name=submit
/form;
and in test2.php i have this;
if ($_POST['language'] = 1)
{$lang = EngP=1;}
else {$lang = EngP=0;}
echo $lang ;
But it always echoes EngP=1 no matter what I choose, what is wrong here??
if ($_POST['language'] == 1)
You have an assignment operator instead of a comparison in your if
statement.
mh.
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Re: [PHP] passing variable arguments from select
Yes I changed that, but now it always echoes EngP=0
Mark Heintz Php Mailing Lists [EMAIL PROTECTED] wrote in
message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
On Thu, 27 Jun 2002, Haddad Said wrote:
in query.php i have this piece of code;
echo form action='test2.php' action='post'
select name='language'
option value='0'selected='yes'one/option
option value='1'two/option
/select
input type=text name=query
input type=submit name=submit
/form;
and in test2.php i have this;
if ($_POST['language'] = 1)
{$lang = EngP=1;}
else {$lang = EngP=0;}
echo $lang ;
But it always echoes EngP=1 no matter what I choose, what is wrong
here??
if ($_POST['language'] == 1)
You have an assignment operator instead of a comparison in your if
statement.
mh.
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Re: [PHP] passing variable arguments from select
I just found out the $_POST['language'] is always assigned the value Array.
Since a user must choose either of the options and not both, I would rather
do this without arrays, is there another way?
Haddad Said [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Yes I changed that, but now it always echoes EngP=0
Mark Heintz Php Mailing Lists [EMAIL PROTECTED] wrote in
message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
On Thu, 27 Jun 2002, Haddad Said wrote:
in query.php i have this piece of code;
echo form action='test2.php' action='post'
select name='language'
option value='0'selected='yes'one/option
option value='1'two/option
/select
input type=text name=query
input type=submit name=submit
/form;
and in test2.php i have this;
if ($_POST['language'] = 1)
{$lang = EngP=1;}
else {$lang = EngP=0;}
echo $lang ;
But it always echoes EngP=1 no matter what I choose, what is wrong
here??
if ($_POST['language'] == 1)
You have an assignment operator instead of a comparison in your if
statement.
mh.
--
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[PHP] passing variable arguments from select
Hi, i am having a problem passing variables from a drop down menu in a form; from action=test.php action=post select name=example option selected one/option option two/option /form So how will the values of $example be assigned in test.php? THanks -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] passing variable arguments from select
On Wednesday, June 26, 2002, at 02:51 PM, Haddad Said wrote: Hi, i am having a problem passing variables from a drop down menu in a form; from action=test.php action=post select name=example option selected one/option option two/option /form You should do it like this (be sure to properly quote your attributes, and use the 'value' attribute of the 'option' tag): print form action='test.php' method='post' select name='example' option selected='yes' value='1'One/option option value='2'Two/option /select /form\n; In the test.php page, the selected value will be found in the $_POST['example'] variable. Erik Erik Price Web Developer Temp Media Lab, H.H. Brown [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Passing variable to new page and pulling the rest of info from database
Hello, I am passing a variable to the new page, when user clicks on the link. Something like that: a href=showimage.php?ID=38img src=/some/image.jpg/a How can I extract all other information out of my database for that ID in the next page (showimage.php)? Thanks
Re: [PHP] Passing variable to new page and pulling the rest of info from database
SELECT * FROM table WHERE ID = $_GET['ID'] Then create a page to display all of that information. Look at the mysql functions and learn some PHP. We can't help you without knowing what's in your table and how you want it displayed, etc... So keep learning and reading and you'll figure out how to do it. Your question is way to broad for any help... ---John Holmes... - Original Message - From: Igor Portnoy [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Wednesday, June 05, 2002 2:40 PM Subject: [PHP] Passing variable to new page and pulling the rest of info from database Hello, I am passing a variable to the new page, when user clicks on the link. Something like that: a href=showimage.php?ID=38img src=/some/image.jpg/a How can I extract all other information out of my database for that ID in the next page (showimage.php)? Thanks -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Passing variable to new page and pulling the rest of info from database
Ahhh! quote that ID number before using it in a query! :) // for mysql mysql_quote($_GET['ID']); --- Scott Hurring Systems Programmer EAC Corporation [EMAIL PROTECTED] Voice: 201-462-2149 Fax: 201-288-1515 -Original Message- From: 1LT John W. Holmes [mailto:[EMAIL PROTECTED]] Sent: Wednesday, June 05, 2002 4:29 PM To: Igor Portnoy; [EMAIL PROTECTED] Subject: Re: [PHP] Passing variable to new page and pulling the rest of info from database SELECT * FROM table WHERE ID = $_GET['ID'] Then create a page to display all of that information. Look at the mysql functions and learn some PHP. We can't help you without knowing what's in your table and how you want it displayed, etc... So keep learning and reading and you'll figure out how to do it. Your question is way to broad for any help... ---John Holmes... - Original Message - From: Igor Portnoy [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Wednesday, June 05, 2002 2:40 PM Subject: [PHP] Passing variable to new page and pulling the rest of info from database Hello, I am passing a variable to the new page, when user clicks on the link. Something like that: a href=showimage.php?ID=38img src=/some/image.jpg/a How can I extract all other information out of my database for that ID in the next page (showimage.php)? Thanks -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Passing variable to new page and pulling the rest of info from database
Or simply just validate it's an integer, like it should be, like you would do with any user input... There is no mysql_quote() function...or am I missing something? ---John Holmes... - Original Message - From: Scott Hurring [EMAIL PROTECTED] To: Php-General (E-mail) [EMAIL PROTECTED] Sent: Wednesday, June 05, 2002 4:30 PM Subject: RE: [PHP] Passing variable to new page and pulling the rest of info from database Ahhh! quote that ID number before using it in a query! :) // for mysql mysql_quote($_GET['ID']); --- Scott Hurring Systems Programmer EAC Corporation [EMAIL PROTECTED] Voice: 201-462-2149 Fax: 201-288-1515 -Original Message- From: 1LT John W. Holmes [mailto:[EMAIL PROTECTED]] Sent: Wednesday, June 05, 2002 4:29 PM To: Igor Portnoy; [EMAIL PROTECTED] Subject: Re: [PHP] Passing variable to new page and pulling the rest of info from database SELECT * FROM table WHERE ID = $_GET['ID'] Then create a page to display all of that information. Look at the mysql functions and learn some PHP. We can't help you without knowing what's in your table and how you want it displayed, etc... So keep learning and reading and you'll figure out how to do it. Your question is way to broad for any help... ---John Holmes... - Original Message - From: Igor Portnoy [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Wednesday, June 05, 2002 2:40 PM Subject: [PHP] Passing variable to new page and pulling the rest of info from database Hello, I am passing a variable to the new page, when user clicks on the link. Something like that: a href=showimage.php?ID=38img src=/some/image.jpg/a How can I extract all other information out of my database for that ID in the next page (showimage.php)? Thanks -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Passing variable to new page and pulling the rest of info from database
Sorry, i meant: mysql_escape_string( $value ) I use a class to handle accessing the database, and i named the function $db-quote(...); i keep forgetting that the actual function call that the class is making ;-) And yes, validating that it's an integer works also, but even after pregging vars to all hell and back to verify contents, i still quote. i'm paranoid. :) --- Scott Hurring Systems Programmer EAC Corporation [EMAIL PROTECTED] Voice: 201-462-2149 Fax: 201-288-1515 -Original Message- From: 1LT John W. Holmes [mailto:[EMAIL PROTECTED]] Sent: Wednesday, June 05, 2002 5:10 PM To: Scott Hurring; Php-General (E-mail) Subject: Re: [PHP] Passing variable to new page and pulling the rest of info from database Or simply just validate it's an integer, like it should be, like you would do with any user input... There is no mysql_quote() function...or am I missing something? ---John Holmes... - Original Message - From: Scott Hurring [EMAIL PROTECTED] To: Php-General (E-mail) [EMAIL PROTECTED] Sent: Wednesday, June 05, 2002 4:30 PM Subject: RE: [PHP] Passing variable to new page and pulling the rest of info from database Ahhh! quote that ID number before using it in a query! :) // for mysql mysql_quote($_GET['ID']); --- Scott Hurring Systems Programmer EAC Corporation [EMAIL PROTECTED] Voice: 201-462-2149 Fax: 201-288-1515 -Original Message- From: 1LT John W. Holmes [mailto:[EMAIL PROTECTED]] Sent: Wednesday, June 05, 2002 4:29 PM To: Igor Portnoy; [EMAIL PROTECTED] Subject: Re: [PHP] Passing variable to new page and pulling the rest of info from database SELECT * FROM table WHERE ID = $_GET['ID'] Then create a page to display all of that information. Look at the mysql functions and learn some PHP. We can't help you without knowing what's in your table and how you want it displayed, etc... So keep learning and reading and you'll figure out how to do it. Your question is way to broad for any help... ---John Holmes... - Original Message - From: Igor Portnoy [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Wednesday, June 05, 2002 2:40 PM Subject: [PHP] Passing variable to new page and pulling the rest of info from database Hello, I am passing a variable to the new page, when user clicks on the link. Something like that: a href=showimage.php?ID=38img src=/some/image.jpg/a How can I extract all other information out of my database for that ID in the next page (showimage.php)? Thanks -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Passing variable to the next page
Here is scenario that I am working on but have a problem. I have 10 thumbnail images on the page with information for them created from database. For example: + + + Some Image + + + Name of Image: Blah Author of Image: Blah (I am pulling more info about that image from database, but on this page I only display Name of Image and Author of image, the rest of the info is stored in variables.) I want my users to click on the thumbnail. When they do I want a new window to open and I want information for that particular image that they clicked (along with other info that stored in variables for the same image) to be passed to that new opened page. How can I pass information about that particular image to the new page? Any ideas? Please Help.
Re: [PHP] Passing variable to the next page
Tag the variables onto the url: http://mysite.com/newpage.php?firstvar=somevaluesecondvar=othervalue;. for arrays ?layer[]=alayer[]=blayer[]=c. or use sessions. Here is scenario that I am working on but have a problem. I have 10 thumbnail images on the page with information for them created from database. For example: + + + Some Image + + + Name of Image: Blah Author of Image: Blah (I am pulling more info about that image from database, but on this page I only display Name of Image and Author of image, the rest of the info is stored in variables.) I want my users to click on the thumbnail. When they do I want a new window to open and I want information for that particular image that they clicked (along with other info that stored in variables for the same image) to be passed to that new opened page. How can I pass information about that particular image to the new page? Any ideas? Please Help. -Pushkar S. Pradhan -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Passing variable to the next page
Here is scenario that I am working on but have a problem. I have 10 thumbnail images on the page with information for them created from database. (I am pulling more info about that image from database, but on this page I only display Name of Image and Author of image, the rest of the info is stored in variables.) I want my users to click on the thumbnail. When they do I want a new window to open and I want information for that particular image that they clicked (along with other info that stored in variables for the same image) to be passed to that new opened page. How can I pass information about that particular image to the new page? Any ideas? Speaking personally, I'd only extract the data you actually need to display - it will lower your overheads (even if just a little) and you won't be in a situation where you have to keep data around on the off-chance that they click the thumbnail for more info. Then if they do click the thumbnail get the data you need from the database. In this scenario about the only thing you'd need to pass your new page would be the ID number from the database - then just do a query for what you need on the new page. The more images you're dealing with, and the more information you extract per image, the more effective this will be at saving you overheads. CYA, Dave -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] passing variable to anti spam mailto script
Hello All, What is the best way (or opinions rather) to pass variables to a anti spam script. This is used in place of mailto links in webpages, but am looking for any simple way to pass information so more than one address can easily be used in the script, I could use mysql or other, however want to have something easier than this. Suggestions ? (please CC me if mailing to the list) , any examples ? ?PHP $mailto = 'mailto:[EMAIL PROTECTED]'; header(Location: $mailto); ? TIA -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Passing Variable
I know that people have asked about this before, so let me appologize ahead of time but I need quick help. I have a url which looks like http://www.domain.com/remove/index.php?id=test. I use the echo function ? echo $id; ? to show the word test. What I want to do is send this variable along with other information that a user inputs via a form to the next page so it can be e-mailed via the mail() function. Can someone please advice how to do so. Thanks, Ben -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Passing Variable
Hello, store value in hidden field in form, like this: input type=hidden name=id value=?php echo $_GET['id']; ? when form is submitted field id contains required value. or use cookie (but here is test if cookie enabled) Regards Michal Dvoracek [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Passing variable with slashes in URL
Currently I have a problem with IE. The IE browser will not bring up my encoded url. When clicking on the link I either receive a script error or nothing happens. The issue is with the double quotes. IE does not recognized the encoded character %22 for quotes. IE does seem to process single quotes (%27) OK. Here is my code to encode the string variable and create the link The string I am having problems with = All Inclusive Casa De Campo Golf = $headline $url_title = rawurlencode( $headline ); print trtd align = 'left'font face='arial' size='2'a href='javascript:openQuote(\$main_url/miniquote_form.inc.php3?site=$sitead =$adheader=$url_title\, \helpWin\);'Contact us for information/a/font; Here is what the url looks like in the Netscape browser window header=%22All%20Inclusive%22%20Casa%20De%20Campo%20Golf If I include this code: $headline = addslashes ($headline); before the: $url_title = rawurlencode( $headline ); IE now processes the url link and the string displays correctly for me in my popup window. Here is the IE url: header=All%20Inclusive%20Casa%20De%20Campo%20Golf However -- NS now displays the string with backslahes. Here is the comparable NS url: header=%5C%22All%20Inclusive%5C%22%20Casa%20De%20Campo%20Golf Here is the code in my form to display the header variable (which I thought would strip out the slashes) ?php $headertitle = stripslashes( $header ); print $headertitle; ? I have tried htmlentities as: $url_title = htmlentites( $headline ); and neither IE nor Netscape would work -- the url link would not work. perhaps my syntax is incorrect for this function? Claudia -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] Passing variable with slashes in URL
cant say for sure, but I dont play around with rawencode and addslasses for urls, just use urlencode() echo a href='index.php?test=. urlencode($text) .test2=. urlencode($test2) .'test/a ; -- Chris Lee [EMAIL PROTECTED] Claudia Smith [EMAIL PROTECTED] wrote in message 9duqke$9g1$[EMAIL PROTECTED]">news:9duqke$9g1$[EMAIL PROTECTED]... Currently I have a problem with IE. The IE browser will not bring up my encoded url. When clicking on the link I either receive a script error or nothing happens. The issue is with the double quotes. IE does not recognized the encoded character %22 for quotes. IE does seem to process single quotes (%27) OK. Here is my code to encode the string variable and create the link The string I am having problems with = All Inclusive Casa De Campo Golf = $headline $url_title = rawurlencode( $headline ); print trtd align = 'left'font face='arial' size='2'a href='javascript:openQuote(\$main_url/miniquote_form.inc.php3?site=$sitead =$adheader=$url_title\, \helpWin\);'Contact us for information/a/font; Here is what the url looks like in the Netscape browser window header=%22All%20Inclusive%22%20Casa%20De%20Campo%20Golf If I include this code: $headline = addslashes ($headline); before the: $url_title = rawurlencode( $headline ); IE now processes the url link and the string displays correctly for me in my popup window. Here is the IE url: header=All%20Inclusive%20Casa%20De%20Campo%20Golf However -- NS now displays the string with backslahes. Here is the comparable NS url: header=%5C%22All%20Inclusive%5C%22%20Casa%20De%20Campo%20Golf Here is the code in my form to display the header variable (which I thought would strip out the slashes) ?php $headertitle = stripslashes( $header ); print $headertitle; ? I have tried htmlentities as: $url_title = htmlentites( $headline ); and neither IE nor Netscape would work -- the url link would not work. perhaps my syntax is incorrect for this function? Claudia -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP] passing variable via url
using PHP 3.x on Linux, I've always been able to pass along a variable like http://www.domain.com/index.php3?foo=bar and then I could print out this with something like echo $foo; would print out "bar" I've recently installed the latest PHP4 onto a Windows2000 server, I'm trying the same thing here, a simple test, and echo $foo; gives me nothing however including a PHPINFO() on the page displays this: HTTP_GET_VARS["foo"] bar I'm I missing something here ? Kevin Kevin Heflin | ShreveNet, Inc. | Ph:318.222.2638 x103 VP/Production | 333 Texas St #175| FAX:318.221.6612 [EMAIL PROTECTED]| Shreveport, LA 71101 | http://www.shreve.net -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] passing variable via url
On Monday 05 March 2001 17:43, you wrote: using PHP 3.x on Linux, I've always been able to pass along a variable like http://www.domain.com/index.php3?foo=bar and then I could print out this with something like echo $foo; would print out "bar" I've recently installed the latest PHP4 onto a Windows2000 server, I'm trying the same thing here, a simple test, and echo $foo; gives me nothing however including a PHPINFO() on the page displays this: HTTP_GET_VARS["foo"] bar I'm I missing something here ? Check whether register_globals is set to "on" -- Christian Reiniger LGDC Webmaster (http://sunsite.dk/lgdc/) ...to paraphrase Churchill, while representative democracy may be terrible, it's still the best system that large corporations can buy. - David Weinberger JOHO January 25, 2000 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
