Re: [PHP] XML Question
Sorry, but I am unaware of this function. It is not a native function to php, maybe it is a custom function that is part of a library that you are using. You would have to contact them for documentation on thier functions. - Original Message - From: Diana Castillo [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Monday, March 03, 2003 10:06 AM Subject: [PHP] XML Question what does harvest mean? what would the function harvest do? -- 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] Post method
Why not use sessions? - Original Message - From: Alex Shi [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Friday, February 28, 2003 3:09 PM Subject: [PHP] Post method Hi, Any one know in a php script, if it is possible to simulate a post method? I mean I want to header() to an url but don't like to embed the parameters into that url. Thanks in advance! Alex Shi -- == Cell Phone Batteries at 30-50%+ off retail prices! http://www.pocellular.com == TrafficBuilder Network: http://www.bestadv.net/index.cfm?ref=7029 == -- 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] Why PHP doesn't work with Apache2?
Well, I don't know about trying to uninstall it, but RH8.0 does come with Apache 2 It works fine and I have not ran into a reason why I should use the older version for this. Granted I only use it for testing, my live server is RH7.2 which uses Apache 1 - Original Message - From: Richard Baskett [EMAIL PROTECTED] To: PHP General [EMAIL PROTECTED] Sent: Thursday, January 23, 2003 2:28 PM Subject: Re: [PHP] Why PHP doesn't work with Apache2? Ok I am hearing a bad rumor that Red Hat 8.0 and the Mac xserve both come with Apache 2.. now is this a rumor or is this true? If it's true.. why? And does anybody have any experience in uninstalling Apache 2 on the xserve and does it break anything? Cheers! Rick Don't walk in front of me, I may not follow, Don't walk behind me, I may not lead, Just walk beside me, and be my friend. - Unknown From: Tim Thorburn [EMAIL PROTECTED] Date: Thu, 23 Jan 2003 05:03:00 -0500 To: [EMAIL PROTECTED] Subject: Re: [PHP] Why PHP doesn't work with Apache2? Just to add a little fuel to the fire, the only way I've seen or been able to get PHP and Apache 2 working together is by installing each from my Red Hat 8.0 CD's. When I tried to compile each together, I got many an error. When I start doing some actually web work on my testing server, I imagine I'll uninstall Apache 2, and go with some version of 1.3 -- 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] Form Validating Class (OOP misunderstandings...)
For the most part I found two errors
The first was a scope problem. $HTTP_*_VARS hashes are not accessible
inside functions unless you source them in. i.e. global
The other was an extra line in an if statement that did not contain braces.
I did a little formating while I checked the code. Try this to see if it
works as you expected it to.
?php
class FormV {
var $errorList;
function FormV() {
$this-reset_errorList();
}
function getValue($field) {
global $HTTP_POST_VARS;
return $HTTP_POST_VARS[$field];
}
function isEmpty($field, $msg) {
$value = $this-getValue($field);
if(!trim($value)) {
$this-errorList[] = array('field'=$field,
'value'=$value, 'msg'=$msg);
return true;
} else
return false;
}
function isString($field, $msg) {
$value = $this-getValue($field);
if(is_string($value))
return true;
else {
$this-errorList[] = array('field'=$field,
'value'=$value,'msg'=$msg);
return false;
}
}
function isEmail($field, $msg) {
$value = $this-getValue($field);
$pattern = '/^([a-zA-Z0-9])*@([a-zA-Z0-9_-])+(\.[a-zA-Z0-9_-]+)+/';
if(preg_match($pattern, $value))
return true;
else {
$this-errorList[] = array('field'=$field,
'value'=$value, 'msg'=$msg);
return false;
}
}
function isError() {
if(sizeof($this-errorList) 0)
return true;
else
return false;
}
function get_errorList() {
return $this-errorList;
}
function reset_errorList() {
$this-errorList = array();
}
function stringConvert($field) {
$value = $this-getValue($field);
$value = html_special_chars($value);
$value = stripslashes($value);
$value = strtolower($value);
return $value;
}
}
?
- Original Message -
From: Nicholas Wieland [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, January 21, 2003 8:12 AM
Subject: [PHP] Form Validating Class (OOP misunderstandings...)
Hello everybody,
I'm working on a class but having a lot of problems... probably my
understanding of PhP-OOP is not so good ...
Here's the class: I found it on the web and I've tried to personalize
it to fit my needs...
?php
class FormV
{
var $errorList;
function FormV()
{
$this-reset_errorList();
}
function getValue( $field )
{
$value = $HTTP_POST_VARS[ $field ];
return $value;
}
function isEmpty( $field, $msg )
{
$value = $this-getValue( $field );
if( trim( $value ) == )
{
$this-errorList[] = array( field = $field,
value = $value, msg = $msg );
return true;
}
else
return false;
}
function isString( $field, $msg )
{
$value = $this-getValue( $field );
if( is_string( $value ) )
return true;
else
{
$this-errorList[] = array( field = $field,
value = $value, msg = $msg );
return false;
}
}
function isEmail( $field, $msg )
{
$value = $this-getValue( $field );
$pattern =
/^([a-zA-Z0-9])*@([a-zA-Z0-9_-])+(\.[a-zA-Z0-9_-]+)+/;
if( preg_match( $pattern, $value ) )
return true;
else
{
$this-errorList[] = array( field = $field,
value = $value, msg = $msg );
return false;
}
}
function isError()
{
if( sizeof( $this-errorList ) 0 )
return true;
else
return false;
}
function get_errorList()
{
return $this-errorList;
}
function reset_errorList()
{
$this-errorList = array();
}
function stringConvert( $field )
{
$value = $this-getValue( $field );
$value = html_special_chars( $value );
$value = stripslashes( $value );
$value = strtolower( $value );
return $value;
}
};
?
Ok.
I think that the getter is *totally* wrong but I'm not able to debug
it, because it's private (I think...). Also the stringConvert() method
sucks, maybe I'll split it in some others accessor methods, for a
better design.
Now I'm totally lost, I don't have a clue how to continue, I've tried
for hours and hours to make it work, but didn't succeed :(
Any suggestion on improving and other enhancment for a good form
validating class is really really appreciated.
TIA,
Nicholas
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Re: [PHP] Parse error
Try this...
$sql = INSERT INTO personnel (firstname, lastname, nick, email, salary)
VALUES ('$first','$last','$nickname',);
- Original Message -
From: Brent Lee [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, January 14, 2003 10:02 AM
Subject: [PHP] Parse error
I'm new to PHP and am slowly learning it. I have the following code that
is
giving me errors could someone take a look and suggest a fix?
Parse error: parse error in /home//public_html/Info/datain.php on
line 7
HTML
?php
$db = mysql_connect(localhost, ,xx);
mysql_select_db(learndb,$db);
$sql = INSERT INTO personnel (firstname, lastname, nick, email, salary)
VALUES ('$first','$last','$nickname',)
$result = mysql_query'($sql);
echo Thank you! Information entered.\n;
?
/HTML
--
Thanks
__
Brent Lee
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Re: [PHP] MySQL problem with RedHat 8
Make sure that the shared module is in the correct directory. Check your php.ini file to make sure but it is most likely at /usr/lib/php4 make sure that you have mysql.so - Original Message - From: Daniel Elenius [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Tuesday, January 14, 2003 3:58 PM Subject: [PHP] MySQL problem with RedHat 8 Hi! I'm trying to connect to my mysql database using something like mysql_connect( 'localhost', 'root', 'thepassword' ) or die ( 'Unable to connect to server.' ); But I get the error message: Fatal error: Call to undefined function: mysql_connect() in /home/daniel/public_html/index.php on line 21 I have: [root@p85 /]# rpm -qa |grep sql php-mysql-4.2.2-8.0.5 mysql-3.23.52-3 mysql-server-3.23.52-3 mysql-devel-3.23.52-3 and: [root@p85 /]# rpm -q php php-4.2.2-8.0.5 Someone mentioned these two settings in php.ini, which I tried with no success: register_globals = On short_open_tag = On phpinfo() says that php was compiled with '--with-mysql=shared,/usr' Can someone help me please? regards, -- Daniel Elenius [EMAIL PROTECTED] -- 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] question on Header(location
No, I am afraid not. It will only send a redirection header to the same page that you are on. - Original Message - From: Don [EMAIL PROTECTED] To: php list [EMAIL PROTECTED] Sent: Monday, January 13, 2003 1:33 PM Subject: [PHP] question on Header(location When I use --- header(Location: http://www.somepage.com), it redirects to my page no problem. My question is, is there a PHP trick I can use to make it open in a new browser window? I'm adverse to using JavaScript as some users may not have it turned on (odd I know but a reality still). Thanks, Don --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.443 / Virus Database: 248 - Release Date: 1/10/2003 -- 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] undefined variable notice - how to furn of
In the php.ini file set error reporting to E_ALL ~E_NOTICE - Original Message - From: Borut Kovacec [EMAIL PROTECTED] To: PHP Mailing List [EMAIL PROTECTED] Sent: Thursday, January 09, 2003 9:57 AM Subject: [PHP] undefined variable notice - how to furn of Hi I just installed new php 4.2.3 on Win XP, Apache 1.3.24.. Everything works fine, just now I'm getting Notice messages for every undefined variable or undefined index in arrays.. So now I have to use issset() everytime to avoid this messages... ..is there any way to turn this messages off, because if prior versions I never got them..?? Borut -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] newbee
This is because of the default error reporting setting in the newer version of php. This was not the case when your book was written. In your php.ini file set error reporting = E_ALL ~E_NOTICE and you will quit getting the 'Undefined Variable' notices. - Original Message - From: Hendrik van Niekerk [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Thursday, January 09, 2003 10:50 AM Subject: [PHP] newbee Hi I'm very new to PHP. (have just bought the book PHP and MySQL Web Deelopment) SAMS authors Luke Welling and Laura Thompson I working with the examples in the book. I have loaded MySQL, Apache and PHP onto a NT server. All the test passed as per the examples in the book. Working with the the first example in chapter one. I have the HTML document that when the quantities are entered the processorder.php should be called to show the results and do some calcs. what happens is that the php file is called but I get an error message: Warning: Undefined variable typeqty in C:\apache group\apache\htdocs\processorder.php on line 23 this varialbe is defined in the html document exactly as in the php document. now this warning is repeated for all the variables that should have been passed to the php for processing. What am I missing ? thank you in advance Hendrik van Niekerk -- 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] newbee
Without seeing some code I would guess that it is register globals issue. It falls into the same explanation as the previous problem. You can fix this in one of two ways. 1) use the super globals instead of $typeqty use $_POST['typeqty'] or $_GET['typeqty'] depending on how it is passed. 2) turn register globals on in the php.ini file. - Original Message - From: Hendrik van Niekerk [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Thursday, January 09, 2003 11:58 AM Subject: Re: [PHP] newbee Joseph I can get the errors to go away but then my example just shows a blank page! The variable are not being passed through to the .php doc hendrik Joseph W. Goff [EMAIL PROTECTED] wrote in message 003201c2b803$ada54f60$bdcaa8c0@jg42000">news:003201c2b803$ada54f60$bdcaa8c0@jg42000... This is because of the default error reporting setting in the newer version of php. This was not the case when your book was written. In your php.ini file set error reporting = E_ALL ~E_NOTICE and you will quit getting the 'Undefined Variable' notices. - Original Message - From: Hendrik van Niekerk [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Thursday, January 09, 2003 10:50 AM Subject: [PHP] newbee Hi I'm very new to PHP. (have just bought the book PHP and MySQL Web Deelopment) SAMS authors Luke Welling and Laura Thompson I working with the examples in the book. I have loaded MySQL, Apache and PHP onto a NT server. All the test passed as per the examples in the book. Working with the the first example in chapter one. I have the HTML document that when the quantities are entered the processorder.php should be called to show the results and do some calcs. what happens is that the php file is called but I get an error message: Warning: Undefined variable typeqty in C:\apache group\apache\htdocs\processorder.php on line 23 this varialbe is defined in the html document exactly as in the php document. now this warning is repeated for all the variables that should have been passed to the php for processing. What am I missing ? thank you in advance Hendrik van Niekerk -- 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] Variables that persist through reload?
Your local version does not have register_globals turned on in php.ini.
You can either set the value = on or access posted values by using $_POST or
$HTTP_POST_VARS depending on your version of php.
- Original Message -
From: David Chamberlin [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Friday, January 03, 2003 1:58 PM
Subject: [PHP] Variables that persist through reload?
Hey,
I'm somewhat new to PHP and I'm doing a lot of forms stuff. Looking at
some of the examples at:
http://www.linuxguruz.org/z.php?id=33
The variables seem to persist through a reload. For example, the first
demo has:
?
if (!isset($pick)) {
echo Fill out and submit the form below.; }
else {
$j = count($pick);
for($i=0; $i$j; $i++) {
echo Pick b$pick[$i]/b is Checkedbr /; }
}
?
form action=? echo $PHP_SELF; ?
ol
liPaintinginput name=pick[] type=checkbox value=Painting //li
liPlumbinginput name=pick[] type=checkbox value=Plumbing //li
liElectricinput name=pick[] type=checkbox value=Electric //li
/ol
input type=submit value=Submit! /
input type=reset value=Reset /
/form
So the $pick variable is maitained through the submit. When I demo the
script on that site, it works as advertised. When I use the same code
on my web server, it doesn't work ($pick is never set after submitting).
Is there some configuration option that dictates whether variables
persist in this manner?
Thanks,
Dave
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Re: [PHP] Hello and help
You are on the right track, but missed one major thing. Anytime you want to retrieve information from a database result set you have to follow 3 basic steps. 1) Connect to the database *you got that one 2) Execute a query *you got that one 3) Retrieve the resultset ***missed this one See http://www.php.net/mysql_fetch_assoc for the appropriate syntax. - Original Message - From: Todd Barr [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Friday, January 03, 2003 6:32 PM Subject: [PHP] Hello and help I am a PHP newbie, and I am having difficulty in getting my records to display here is my code ?php $Host=localhost; $User=us34992a; $Password=*; $DBname=db34992a; $Tablename=calendar; $Link=mysql_pconnect($Host, $User, $Password) or die (YOu suck1); $db=mysql_select_db($DBname, $Link) or die (ISUCKEGGS); $result=mysql_query(SELECT * from calendar, $Link) or die (YOu suck3); print(table width=53% border=0 cellpadding=0 cellspacing=0\n); print (tr\n); print (td background=topbig.gif\n); print (div align=centerfont face=Arial, Helvetica, sans-serifbfont color=#FFCalender/font/b/font/div /td /tr\n); print (tr bgcolor=#009ACE\n); print(td$result[meeting_name]/td\n)or die (you suck); mysql_close ($Link); ? any help? -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Baffling output
Try viewing the source that is generated in html and I bet you will find
that it is broken html code.
- Original Message -
From: Lightfirst [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Thursday, January 02, 2003 4:25 PM
Subject: [PHP] Baffling output
Can someone explain to me why the loop that counts to 99 appears before
the
5 by 5 grid in the following php code?
Thanks!
?php
echo table border=\1\ align=\center\ width=\100%\
bgcolor=\#FF\ bordercolor=\#FF\tr;
for ($r=0; $r5; $r++){
for ($c=0; $c7; $c++){
if ($c==0 || $c%7==0)
echo td align=\center\ valign=\middle\ width=\15%\
height=\77\ bordercolor=\#00\/td; else if ($c%6==0)
echo td align=\center\ valign=\middle\ width=\14%\
height=\77\ bordercolor=\#00\/td/trtr;
else {
echo td align=\center\ valign=\middle\ width=\15%\
height=\77\ border=\1\ bordercolor=\#00\;
echo div align=\center\font size=\1\/font/div;
echo Hello ; $i++;
} //for else
}// for loop c
} //for loop r
for ($i=1; $i100; $i++)
echo $ibr\n;
?
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Re: [PHP] $_SERVER['DOCUMENT_ROOT'] on localhost
Is there a difference between PHP versions on your machine and your site? Have you looked at phpinfo() to see what variables have been set on your system? - Original Message - From: rolf vreijdenberger [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Wednesday, December 18, 2002 3:38 PM Subject: [PHP] $_SERVER['DOCUMENT_ROOT'] on localhost hey there, I include files outside the document root in my site. include $_SERVER['DOCUMENT_ROOT'].../inc.php ; when I use this on the localhost it doesn't find this variable: $_SERVER['DOCUMENT_ROOT'] which makes testing more annoying since I have to upload to my site all the time. any ideas on the how and why??? thanks a lot -- 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] $_SERVER['DOCUMENT_ROOT'] on localhost
Both sites are running winXP and IIS? Both same version of IIS? If the variable is on one and not on the other it could be either a version conflict, or a configuration problem. - Original Message - From: Rolf Vreijdenberger [EMAIL PROTECTED] To: Joseph W. Goff [EMAIL PROTECTED]; php-general [EMAIL PROTECTED] Sent: Wednesday, December 18, 2002 3:50 PM Subject: Re: [PHP] $_SERVER['DOCUMENT_ROOT'] on localhost yes, the server var doesn't exist in phpifno(); no difference winxp pro with iss -- 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] $_SERVER['DOCUMENT_ROOT'] on localhost
IIS may not have that variable. You might want to install apache on your local machine so that you are using the same tools. - Original Message - From: rolf vreijdenberger [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Wednesday, December 18, 2002 4:00 PM Subject: Re: [PHP] $_SERVER['DOCUMENT_ROOT'] on localhost the document_root just doesn't show up in my phpinfo(); do windows machines have one anyway? I have used this var before, just not testing it locally. It normally shows up when I use phpifno on the server. -- 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] Submitting a Form
Your error reporting is set to E_ALL. If you change it to E_ALL ~E_NOTICE you will not get the notice messages anymore. Also, if you are not getting the value, try $_POST['Test'] instead. - Original Message - From: Jay Thayer [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Tuesday, December 17, 2002 11:38 AM Subject: [PHP] Submitting a Form I have a php page, with a form I am attempting to submit. form method=Post action=submitTest.php input type=hidden name=Test value=Test Value input type=submit value=Submit name=SubmitTest form In submitTest.php, I simply put ?php print $Test; ? Yet, I continually get the browser message: Notice: Undefined variable: Test in ...\submittest.php on Line 2 1. Any reason that I can not pass the variable to the submit page? I am using an Apache Web Server on Windows XP. I am simplying attempting to pass a variable from one page to another, so I can update my database. This is just a simple test that is failing. 2. If I have a select name=testmultiple multiple HTML tag, how can I reference the x# of variables that are selected in that form value in the submittest page? I would want to get every option that is selected in the page. 3. Is there a variable so that I can see everything that is passed from the form, some debug variable I could print() to see everything that was passed from the form? Anyone have an idea? -- 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] PHP question
You can set this up in the php.ini file in the /etc directory. You can set php to run an any user and any group that you have set up on your system. Just make sure that you set permissions properly so that it can function in the directories that it is accessed. - Original Message - From: Tom Ray [EMAIL PROTECTED] To: PHP List [EMAIL PROTECTED] Sent: Tuesday, December 17, 2002 1:11 PM Subject: [PHP] PHP question Are PHP script supposed to run as the user or as the web server? Currently I'm running Red Hat 7.3 with apache 1.3.x and all my PHP scripts run as apache, not as the user. I'm wonder if I can run the scripts as the user and how do I fix this? Any help would be great! -- 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] PHP question
Actually, I was mistaken. It isn't php.ini that has to be changed. I believe it is httpd.conf. php runs as what ever the webserver is set up to run as. - Original Message - From: Joseph W. Goff [EMAIL PROTECTED] To: php-general [EMAIL PROTECTED]; Tom Ray [EMAIL PROTECTED] Sent: Tuesday, December 17, 2002 1:18 PM Subject: Re: [PHP] PHP question You can set this up in the php.ini file in the /etc directory. You can set php to run an any user and any group that you have set up on your system. Just make sure that you set permissions properly so that it can function in the directories that it is accessed. - Original Message - From: Tom Ray [EMAIL PROTECTED] To: PHP List [EMAIL PROTECTED] Sent: Tuesday, December 17, 2002 1:11 PM Subject: [PHP] PHP question Are PHP script supposed to run as the user or as the web server? Currently I'm running Red Hat 7.3 with apache 1.3.x and all my PHP scripts run as apache, not as the user. I'm wonder if I can run the scripts as the user and how do I fix this? Any help would be great! -- 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] $HTTP_POST_VARS problem
It looks to me like you are trying to get an uploaded file? If so, it isn't $HTTP_POST_VARS, it is $HTTP_POST_FILES or $_FILES if you are using a version of PHP that has super globals. See the PHP manual for more info: http://www.php.net/manual/en/language.variables.predefined.php - Original Message - From: Chris Shiflett [EMAIL PROTECTED] To: Lee P. Reilly [EMAIL PROTECTED]; PHP [EMAIL PROTECTED] Sent: Monday, December 16, 2002 10:04 AM Subject: Re: [PHP] $HTTP_POST_VARS problem --- Lee P. Reilly [EMAIL PROTECTED] wrote: The following statements have the following return values: echo $HTTP_POST_VARS['userfile']; = C:\\Documents and Settings\\Administrator\\Desktop\\IR Files\\gmp1.ir echo $userfile; = C:\\Documents and Settings\\Administrator\\Desktop\\IR Files\\gmp1.ir echo $HTTP_POST_VARS['userfile']['name']; = NOTHING RETURNED echo $HTTP_POST_VARS['userfile']['size']; = NOTHING RETURNED echo $userfile_size; = NOTHING RETURNED echo $userfile_name; = NOTHING RETURNED Does anyone know what the problem is? What do you think the problem is? I don't see anything unexpected, unless I'm missing something. Chris -- 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] FTP Can't create temp file?
What happens when you remove the suppresion operator (@) from the
move_upload_file and why not just use ftp_put()?
That has always worked fine for me.
- Original Message -
From: Marios Adamantopoulos [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Monday, December 16, 2002 11:06 AM
Subject: [PHP] FTP Can't create temp file?
Hi all again
I'm trying to upload to a unix server. The program I use works fine on my
2000 IIS machine but it has a problem on the online one. Permittions have
been checked and I've been looking all day to find something, but natha...
I would appreciate it if someone can have a look and suggest something.
Here
is the script:
$conn_id = ftp_connect($ftpServer);
$login_result = ftp_login($conn_id, $ftpUser, $ftpPass);
if ((!$conn_id) || (!$login_result)) {
echo FTP connection has failed!;
echo Attempted to connect to $ftpServer for user $ftpUser;
exit;
} else {
//echo Connected to $ftpServer, for user $ftpUser;
}
$phpftp_dir = ;
echo br-- . ftp_pwd($conn_id) . br;
$phpftp_tmpdir=/tmp;
srand((double)microtime()*100);
$randval = rand();
$tmpfile=$phpftp_tmpdir . / . $thefile_name . . . $randv;
echo Temp file: . $tmpfile . br;
if (!@move_uploaded_file($thefile,$tmpfile)) {
echo Upload failed! Can't create temp file;
} else {
ftp_chdir($conn_id,$phpftp_dir);
ftp_put($conn_id,$thefile_name,$tmpfile,FTP_BINARY);
//ftp_quit($conn_id);
unlink($tmpfile);
}
---
I always get: Upload failed! Can't create temp file
Thank you guys in advance
Mario
_
Marios Adamantopoulos
Senior Developer
Tonic
+44 (0)20 7691 2227
+44 (0)7970 428 372
www.tonic.co.uk
Recent projects
www.polydor.co.uk
www.adcecreative.org
www.sony-europe.com/pocketlife
Opinions, conclusions and other information in this message that do not
relate to the official business of Tonic Design Limited shall be
understood as neither given nor endorsed by them.
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Re: [PHP] URL field receiving Array for others
Whatever variable that is suppose to contain the URL is an array. If you
try to print a variable that is an array the value you get is 'Array'. You
will either have to step through it, or implode it, or something of that
nature to get what the array contains.
- Original Message -
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Monday, December 16, 2002 11:50 AM
Subject: [PHP] URL field receiving Array for others
Hello All!
I have a script which inserts a users info into mysql via PHP.
Here is the code which does so:
$sql = (INSERT INTO `business` (`id`, `bt_id`, `bus_name`, `bcity`,
`phone`,
`cell`, `email`, `url`, `details`, `duration`, `s_id`, `license`,
`datime`,
`comments`,`ip`,`user_id`,`user_pass`)
VALUES
('', '$bus_type_id', '$bus_name', '$city_id', '$phone', '$cell',
'$email', '$url', '$details', '$duration', '$state_id', '$license',
'$date','$comments', '$REMOTE_ADDR', '$user_id',
password('$userpass2')));
and it sends a mail to me, however crude:
100% ASAP Plumbing and Rooter Specialist, City ID: 2647
1-866-FIX- GUARD,
l
Array (this is supposed to be a url)
PlumberASAP.com Specializing in Drains, Faucets, Filters, Repairs,
Sprinkers,
Installations, Water Heaters, Disposals, Copper Plumbing, Leak Detection.
Fast,
Clean and Courteous Service Danny McGowan Owner/Operator
,
12 months,
I can't duplicate the problem.
Any idea what is happening here?
Thanks
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Re: [PHP] URL field receiving Array for others
I don't know. You need to show your code.
- Original Message -
From: [EMAIL PROTECTED]
To: Joseph W. Goff [EMAIL PROTECTED]
Cc: php-general [EMAIL PROTECTED]
Sent: Monday, December 16, 2002 12:01 PM
Subject: Re: [PHP] URL field receiving Array for others
But what could the user be typing into the field to return an array?
I'm completely at a loss.
Quoting Joseph W. Goff [EMAIL PROTECTED]:
### Whatever variable that is suppose to contain the URL is an array. If
you
### try to print a variable that is an array the value you get is 'Array'.
### You
### will either have to step through it, or implode it, or something of
that
### nature to get what the array contains.
### - Original Message -
### From: [EMAIL PROTECTED]
### To: [EMAIL PROTECTED]
### Sent: Monday, December 16, 2002 11:50 AM
### Subject: [PHP] URL field receiving Array for others
###
###
###
### Hello All!
###
### I have a script which inserts a users info into mysql via PHP.
###
### Here is the code which does so:
###
### $sql = (INSERT INTO `business` (`id`, `bt_id`, `bus_name`, `bcity`,
### `phone`,
### `cell`, `email`, `url`, `details`, `duration`, `s_id`, `license`,
### `datime`,
### `comments`,`ip`,`user_id`,`user_pass`)
### VALUES
### ('', '$bus_type_id', '$bus_name', '$city_id', '$phone',
### '$cell',
### '$email', '$url', '$details', '$duration', '$state_id', '$license',
### '$date','$comments', '$REMOTE_ADDR', '$user_id',
### password('$userpass2')));
###
### and it sends a mail to me, however crude:
###
###
### 100% ASAP Plumbing and Rooter Specialist, City ID: 2647
### 1-866-FIX- GUARD,
###l
###
### Array (this is supposed to be a url)
###
### PlumberASAP.com Specializing in Drains, Faucets, Filters, Repairs,
### Sprinkers,
### Installations, Water Heaters, Disposals, Copper Plumbing, Leak
### Detection.
### Fast,
### Clean and Courteous Service Danny McGowan Owner/Operator
### ,
### 12 months,
###
###
### I can't duplicate the problem.
###
### Any idea what is happening here?
###
### Thanks
###
###
### --
### PHP General Mailing List (http://www.php.net/)
### To unsubscribe, visit: http://www.php.net/unsub.php
###
###
###
### --
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### To unsubscribe, visit: http://www.php.net/unsub.php
###
###
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Re: [PHP] Re: Stumped!
You don't have a closing french brace for your while loop.
- Original Message -
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Sent: Monday, December 16, 2002 12:57 PM
Subject: Re: [PHP] Re: Stumped!
Hmm, I am still getting a parse error on the last line of code...
In a message dated 12/16/2002 1:49:26 PM Eastern Standard Time,
[EMAIL PROTECTED] writes:
$sql = SELECT .;
$sql_e = mysql_query($sql);
while ($result = mysql_fetch_array($query_e)) {
.
}
You were missing the mysql_query
[EMAIL PROTECTED] escribió en el mensaje
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
I am trying to display a column from my database as a list. Each listing
needs to be a URL that links to another script that brings up all of the
data
in the row to edit. I keep getting a parser error and I can't figure it
out.
Here is the code and any help is greatly appreciated.
?php
$db = mysql_connect(localhost, Uname, PW);
$select_db = mysql_select_db(machmedi_meetingRequest,$db);
$sql = SELECT * FROM requests;
while ($result = mysql_fetch_array($query)) {
$id= $result[id];
$meetingName= $result[meetingName];
echo (A HREF=/edit.php?id='$id'\$meetingName/A );
?
Christopher Parker
Senior Corporate Technical Specialist, Corporate Events
America Online Inc.
Phone: 703.265.5553
Fax: 703.265.2007
Cell: 703.593.3199
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Re: [PHP] php setup
You have the error_reporting set at E_ALL in the ini file. Set it to E_ALL ~E_NOTICE - Original Message - From: Edward Peloke [EMAIL PROTECTED] To: php [EMAIL PROTECTED] Sent: Monday, December 16, 2002 1:29 PM Subject: [PHP] php setup I recently got a laptop and was in the process this weekend of installing, php, apache, mysql etc. One thing I noticed is when I ran the code (that works fine everywhere else) on the laptop, I got errors of 'Variable not defined' and I have pictures that just won't show. I am not sure if this is the problem but most of the scripts giving the 'Variable not defined problem and the images where in other folders and referenced in the php page like /images/picture.gif and /includes/template1.inc. Is there something I need to look at in my config file? Thanks, Eddie -- 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] Date Formatting
Just a small question, but why not just do this through the SQL statement
instead of using php?
http://www.mysql.com/doc/en/Date_and_time_functions.html#IDX1295
[from the mysql manual]
mysql SELECT DATE_FORMAT('1997-10-04 22:23:00', '%W %M %Y');
- 'Saturday October 1997'
mysql SELECT DATE_FORMAT('1997-10-04 22:23:00', '%H:%i:%s');
- '22:23:00'
mysql SELECT DATE_FORMAT('1997-10-04 22:23:00',
'%D %y %a %d %m %b %j');
- '4th 97 Sat 04 10 Oct 277'
- Original Message -
From: Clint Tredway [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Friday, December 13, 2002 11:55 AM
Subject: [PHP] Date Formatting
How can I format a date coming out of a MySQL? I know how to format today's
date but not a date coming out of MySQL. I have looked through the manual,
but I must be blind because I cannot figure it out.
Thanks,
Clint
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Re: [PHP] newbie - decimal places in arthimetic functions
You can use the format() function for this. http://www.php.net/format - Original Message - From: Max Clark [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Friday, December 13, 2002 1:10 PM Subject: [PHP] newbie - decimal places in arthimetic functions Hi- How can I control the decimal places returned by an arthimetic function? I would like this divsion function to only return two decimal places (rounded up). $calcsize = $value / $oneGig; Thanks in advance, Max -- 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] FTP Listings for HTML page
As long as the script has reading capabilities for the directory then yes. - Original Message - From: Randum Ian [EMAIL PROTECTED] To: 'Kevin Stone' [EMAIL PROTECTED]; [EMAIL PROTECTED] Sent: Friday, December 13, 2002 1:25 PM Subject: RE: [PHP] FTP Listings for HTML page Would I be able to do this for an ftp server where I have to log in? -Original Message- From: Kevin Stone [mailto:[EMAIL PROTECTED]] Sent: 13 December 2002 19:07 To: Randum Ian; [EMAIL PROTECTED] Subject: Re: [PHP] FTP Listings for HTML page Yes it is doable. see. open_dir(); @ www.php.net You can also find a number of premade scripts on HotScripts.com or one of the class stores. Do a search on Google. -Kevin - Original Message - From: Randum Ian [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Friday, December 13, 2002 11:38 AM Subject: [PHP] FTP Listings for HTML page Hi guys, I am maintaining a list of files on an FTP server and it would be great if I could get a very simple list of all the files and their directory names so I can generate a HTML page with the information. Is this doable? Randum Ian [EMAIL PROTECTED] DJ / Reviewer / Webmaster, DancePortal (UK) Limited DancePortal.co.uk - Global dance music media -- 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] newbie - decimal places in arthimetic functions
I'm sorry, posted the wrong link. It is the number_format() function. http://www.php.net/number_format - Original Message - From: Joseph W. Goff [EMAIL PROTECTED] To: php-general [EMAIL PROTECTED]; Max Clark [EMAIL PROTECTED] Sent: Friday, December 13, 2002 1:31 PM Subject: Re: [PHP] newbie - decimal places in arthimetic functions You can use the format() function for this. http://www.php.net/format - Original Message - From: Max Clark [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Friday, December 13, 2002 1:10 PM Subject: [PHP] newbie - decimal places in arthimetic functions Hi- How can I control the decimal places returned by an arthimetic function? I would like this divsion function to only return two decimal places (rounded up). $calcsize = $value / $oneGig; Thanks in advance, Max -- 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] Progress bar for uploading files
Actually, I don't think you can make a progress bar function correctly for a file upload. The file is uploaded before control is ever given to the script. - Original Message - From: Oscar F [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Wednesday, October 16, 2002 1:17 PM Subject: [PHP] Progress bar for uploading files Hello, I'm working on a file uploading script, but I need to know if there is a way to know what's the progress of the file, so I can make la a progress bar?.. Any ideas? Please?. Thanks in advance. Oscar F.- -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Variables and Forms
This should open up a new window for the new page form action=your_new_page.php method=post target=_new As far as passing variables from page to page, have you ever considered sessions? - Original Message - From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Tuesday, September 24, 2002 2:46 PM Subject: [PHP] Variables and Forms I'm trying to pass variables from one page to the next but don't want to use a hidden input type=text box. I need to break from frames and I don't know a way to pass that to the action tag in the form. If there's a way to do that then great but I haven't found it. I've also tried the meta http-equiv=window-target content tag on the action page from the form and it didn't work either. TIA Ed Curtis -- 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] PHP source code
PHP is open source so anyone can get the source code to it. You can find it at http://www.php.net/downloads.php - Original Message - From: Oliver Witt [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Thursday, September 19, 2002 9:50 AM Subject: [PHP] PHP source code Hi, Is there any way to read php source code? I didn't think so until I heard about people you have done that... Kind regards, Oliver -- 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] Building a query on multiple variables, how to test for which variable A SOLUTION
I have done this two different ways. The first involves a function that creates a dropdown from the database. What is special about this function is that I assign it's own condition as part of the value for that option. Take this sql statement: select distinct column_name from the_table and it would generate this for a select statement: select name="column_name" option value=" "All/option option value=" and column_name='some_value'"some_value/option option value=" and column_name='another_value'"another_value/option option value=" and column_name='a_different_value'"a_different_value/option /select Then, when the form is processed, the sql statement would be as follows: $sql="select whatever_columns from the_table where 1=1$column_name"; The second method is somewhat similar in that the sql statement would remain the same, but you would have to check each select value to see if you needed to add a "and column_name ='$column_value'" to it. i.e. select statement would be like: select name="column_name" option value=" "All/option option value="some_value"some_value/option option value="another_value"another_value/option option value="a_different_value"a_different_value/option /select processing would be like: if (trim($column_name)) $column_name=" and column_name='$column_name'"; $sql="select whatever_columns from the_table where 1=1$column_name"; The important part on this is that you don't have to be concerned with whether or not to use and or where on the statement, always use and, and in your sql statement put a where clause that is always true. i.e. where 1=1. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] command line argument vs. urlencode ?
php is treating command line arguments as GET argument. So that means a + is a space. Try rawurlencode. - Original Message - From: Renato Lins [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Wednesday, August 28, 2002 1:37 PM Subject: [PHP] command line argument vs. urlencode ? Hi, Any one know why the + (plus sign) is not passed as argument to $argv when calling a php script from linux shell ? if this is not a bug, how do I pass a + as argument ? should I use urlencode/urldecode ? is that any php.ini variable to turn off this behavior ? try this and you will see : teste.php-- ?php foreach( $argv as $x ) printf(p: %s\n,$x); ? --- call with : php test.php 123 xxx+ -- 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] Exiting from an include or required file
You will have to enclose the code is some form of a container.
i.e. a function, while loop...
I would probably use the function, then you could exit with return
The other types of container would use break.
- Original Message -
From: Henry [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Wednesday, August 21, 2002 9:59 AM
Subject: Re: [PHP] Exiting from an include or required file
Thanks, but my situation is slight more convoluted than I describe.
I want to cascade down through a include chain and want to just stop
processing the current include and return to the one which called it.
Similar to exit() but only for the local scope.
TIA
Henry
Mark Roedel [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Seems like the easy way, at least in this case, would be to make the
second part of your File A an else to your if.
?php
if (empty($_POST['search_criteria'])) {
echo You must provide search criteria;
} else {
echo You provided search criteria;
}
---
Mark Roedel | Blessed is he who has learned to laugh
Systems Programmer| at himself, for he shall never cease
LeTourneau University | to be entertained.
Longview, Texas, USA | -- John Powell
-Original Message-
From: Henry [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, August 21, 2002 9:40 AM
To: [EMAIL PROTECTED]
Subject: [PHP] Exiting from an include or required file
Hi All,
I would like to exit from an include file and continue in the calling
script!
As an example
FILE A:
?php
if (empty($_POST['search_criteria'))
{
echo You must provide search criteria;
// exit the include file here But how?
}
echo You have provided serach criteria;
?
FILE B
#include A;
form method=post
input name=search_criteria type=text
/form
TIA
Henry
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Re: [PHP] Exiting from an include or required file
Actually, M.A.Bond is right, you don't even have to enclose it, just use the
return construct.
- Original Message -
From: Joseph W. Goff [EMAIL PROTECTED]
To: php-general [EMAIL PROTECTED]; Henry
[EMAIL PROTECTED]
Sent: Wednesday, August 21, 2002 10:03 AM
Subject: Re: [PHP] Exiting from an include or required file
You will have to enclose the code is some form of a container.
i.e. a function, while loop...
I would probably use the function, then you could exit with return
The other types of container would use break.
- Original Message -
From: Henry [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Wednesday, August 21, 2002 9:59 AM
Subject: Re: [PHP] Exiting from an include or required file
Thanks, but my situation is slight more convoluted than I describe.
I want to cascade down through a include chain and want to just stop
processing the current include and return to the one which called it.
Similar to exit() but only for the local scope.
TIA
Henry
Mark Roedel [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Seems like the easy way, at least in this case, would be to make the
second part of your File A an else to your if.
?php
if (empty($_POST['search_criteria'])) {
echo You must provide search criteria;
} else {
echo You provided search criteria;
}
---
Mark Roedel | Blessed is he who has learned to laugh
Systems Programmer| at himself, for he shall never cease
LeTourneau University | to be entertained.
Longview, Texas, USA | -- John Powell
-Original Message-
From: Henry [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, August 21, 2002 9:40 AM
To: [EMAIL PROTECTED]
Subject: [PHP] Exiting from an include or required file
Hi All,
I would like to exit from an include file and continue in the calling
script!
As an example
FILE A:
?php
if (empty($_POST['search_criteria'))
{
echo You must provide search criteria;
// exit the include file here But how?
}
echo You have provided serach criteria;
?
FILE B
#include A;
form method=post
input name=search_criteria type=text
/form
TIA
Henry
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Re: [PHP] variables not behaving as expected
You don't actually have a variable named $do at this point.
You can either have php set this variable automatically by turning on
register globals in the php.ini file or you can extract the variable or do
something like this:
foreach ($_POST as $key = $post)
{
echo $key = $postBr;
$$key = $post;
}
- Original Message -
From: ROBERT MCPEAK [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Monday, August 19, 2002 12:58 PM
Subject: [PHP] variables not behaving as expected
We've recently upgraded to PHP version 4.2.2 running on SuSE 7.2 with
the 2.4.4 kernel and I'm seeing variables behave in a way I didn't
expect. With the prior configuration http post variables were freely
available on the receiving page, but now, for some reason, they aren't.
This code:
foreach ($_POST as $key = $post)
{
echo $key = $postBr;
$key = $post;
}
echo hr;
echo do = $do;
renders this result:
do = addart
display_date = date
art_time = time
art_url = url
art_link = link
src_link = link
src_url = url
--
--
do =
Why is the $do variable null outside of the for loop? Shouldn't it
equal addart? Shouldn't it be available anywhere on the page? With
the previous configuration the http post variables were available
anywhere on the page. Is this a php configuration issue?
Thanks!
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Re: [PHP] OO code and private functions
Not currently, but this will be added in the Zend2 engine. You can find out about this on the Zend website. - Original Message - From: Richard Black [EMAIL PROTECTED] To: Php-General [EMAIL PROTECTED] Sent: Friday, August 16, 2002 9:24 AM Subject: [PHP] OO code and private functions Can I make a function private in PHP, so that it can only be called from within an object of that class??? Just discovering the wonders of OO PHP... :-) == Richard Black Systems Programmer, DataVisibility Ltd - http://www.datavisibility.com Tel: 0141 435 3504 Email: [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] so nobody knows how do get rid of everything outside or certain tags?
preg_match('!div[^]+(.*)/div!Uis',$info,$regs);
$everything_between_divs=$regs[1];
- Original Message -
From: Jay Blanchard [EMAIL PROTECTED]
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Thursday, August 15, 2002 11:44 AM
Subject: RE: [PHP] so nobody knows how do get rid of everything outside or
certain tags?
[snip]
I asked this question a while ago, and I am still seeking clarification,
I fooled with ereg_replace and they only do a piece of what I want, I
would like to know how to keep only a chunk of text defined between 2
tags let's say, div and /div and ignore everything else, I also have
to search and replace certain tags between, but I know how to do that
with ereg_replace. So can anyone help me echo only what is between the
div tags? Please?
[/snip]
Why not take the the beginning div, everything in the middle, and the
end
/div tag and explode() them into an array and then get what you want and
write it out? I didn't see your originl post and don't know if there was
more detail available, but I hope this helps.
Jay
I'm not tense.just terribly, terribly alert
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