Re: [PHP-DB] How to add 2 years to todays date ?
In php $your_timestamp += gmmktime(0,0,0,1,1,1972); MySQL has INTERVALling options. Regards, Andrey - Original Message - From: Dave Carrera [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Monday, March 18, 2002 10:11 AM Subject: [PHP-DB] How to add 2 years to todays date ? Hi All I think the subject line says it all. How to add 2 years to today's date ? Any helps as always most appreciated. Dave Carrera Php Developer http://davecarrera.freelancers.net http://www.davecarrera.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] javascript problem
hi all,
i know this is not exactly the place to ask this, but i don't know where
else to ask. i'm doing a multiframe project and i'm trying to get one of
the pages to change the content of another frame when it loads.
in the main.htm i have:
html
head
titleA Global Village?/title
...
script src=../../media/scripts/rightside.js type=text/javascript
script type=text/javascript
!--
rightopen();
rightwrite('Hello', 'www.yahoo.com', 'Yahoo is really good.');
rightclose();
//--
/script
/script
/head
body
...
/body
/html
the included rightside.js file has the three functions:
function rightopen () {
parent.rightFrame.document.open();
parent.rightFrame.document.write(!DOCTYPE html PUBLIC \-\/\/W3C\/\/DTD
XHTML 1.0 Strict\/\/EN\\n);
parent.rightFrame.document.write(
\http:\/\/www.w3.org\/TR\/xhtml1\/DTD\/xhtml1-strict.dtd\\n);
parent.rightFrame.document.write(html\n);
parent.rightFrame.document.write(head\n);
parent.rightFrame.document.write(title\/title\n);
parent.rightFrame.document.write(meta http-equiv=\Content-Type\
content=\text\/html; charset=iso-8859-1\ \/\n);
parent.rightFrame.document.write(link rel=\stylesheet\
href=\media\/scripts\/sidelayout.css\ type=\text\/css\ \/\n);
parent.rightFrame.document.write(link rel=\stylesheet\
href=\media\/scripts\/sidetext.css\ type=\text\/css\ \/\n);
parent.rightFrame.document.write(script
src=\media\/scripts\/scripts.js\ type=\text\/javascript\\/script\n);
parent.rightFrame.document.write(\/head\n);
parent.rightFrame.document.write(body\n);
}
function rightwrite (name, url, desc) {
parent.rightFrame.document.write(p class=\rightlink\a href=\ + url
+ \ target=\_blank\ + name + \/abr \/);
parent.rightFrame.document.write(desc + \/p);
}
function rightclose () {
parent.rightFrame.document.write(\/body\n);
parent.rightFrame.document.write(\/html\n);
parent.rightFrame.document.close();
}
when i start the file i get a javascript error saying expecting object
i'm not sure what i'm doing wrong. any help is appreciated.
thanks
_
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[PHP-DB] Re: Conflicting results using PHP/Mysql
Doug, Sorry for leaving this so long but I've been tied up in business meetings. It may be that my query is restricting the data retrieval but I cannot see how. If I look at the data I am playing with, I have 3 tables and I want to show transactions from a certain course. I also want to show related information from the bib_extract table and the scanrates (prices) table. I've checked the data in these tables and nothing is odd. Maybe there is a different way to build my query so that it displays the full set of records. Any suggestions? George - Original Message - From: Doug Thompson [EMAIL PROTECTED] To: George Pitcher [EMAIL PROTECTED]; [EMAIL PROTECTED]; [EMAIL PROTECTED] Sent: Wednesday, March 13, 2002 5:59 PM Subject: Re: Conflicting results using PHP/Mysql I would guess it's because only three records match the ANDed tests in the WHERE clause (last 3 lines). Isn't that what you intended? Doug On Wed, 13 Mar 2002 16:08:43 -, George Pitcher wrote: Hi all, Posted this yesterday and got no response. Trying again today. I'm having a small problem with a biggish query. The query: $Itemlistquery= select [a whole load of fields from 3 tables] ending with ; $Itemlistquery.= transactions.Pdownload ; $Itemlistquery.= from bib_extract,scanrates,transactions where ; $Itemlistquery.= (transactions.CourseID = '$Course_ID' and ; $Itemlistquery.= bib_extract.E_ID=transactions.ExtractID and ; $Itemlistquery.= scanrates.finrate=transactions.finrate) ; The problem: If I do a simple count of transactions.CourseID='$Course_ID' I get 18 (for a particular course) and the above query only displays 3 results. Any suggestions? I didn't want to clog the list with the whole query but I can if it's necessary. George, in Edinburgh - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] increase date()
Try this: $six_months_from_now = date(Y-m-d, mktime(0,0,0, date(m)+6, date(1), date(Y))); echo Six Months From Now: $six_months_from_nowP; -Original Message- From: its me [mailto:[EMAIL PROTECTED]] Sent: Sunday, March 17, 2002 12:28 AM To: [EMAIL PROTECTED] Subject: [PHP-DB] increase date() i have: $nowdate=date(Y-m-d); and i want to increase it by 6 months,how can i do that?? Rehab M.Shouman - Express yourself with a super cool email address from BigMailBox.com. Hundreds of choices. It's free! http://www.bigmailbox.com - -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] javascript problem
I am not a JavaScript expert, but I had a similar problem. I THINK that
you need to reference your frames like this:
window.parent.frames.rightFrame.document.open();
Give it a try, hope that helps...
Josh
hi all,
i know this is not exactly the place to ask this, but i don't know
where else to ask. i'm doing a multiframe project and i'm trying to
get one of the pages to change the content of another frame when it
loads.
in the main.htm i have:
html
head
titleA Global Village?/title
...
script src=../../media/scripts/rightside.js type=text/javascript
script type=text/javascript
!--
rightopen();
rightwrite('Hello', 'www.yahoo.com', 'Yahoo is really good.');
rightclose();
//--
/script
/script
/head
body
...
/body
/html
the included rightside.js file has the three functions:
function rightopen () {
parent.rightFrame.document.open();
parent.rightFrame.document.write(!DOCTYPE html PUBLIC
\-\/\/W3C\/\/DTD
XHTML 1.0 Strict\/\/EN\\n);
parent.rightFrame.document.write(
\http:\/\/www.w3.org\/TR\/xhtml1\/DTD\/xhtml1-strict.dtd\\n);
parent.rightFrame.document.write(html\n);
parent.rightFrame.document.write(head\n);
parent.rightFrame.document.write(title\/title\n);
parent.rightFrame.document.write(meta http-equiv=\Content-Type\
content=\text\/html; charset=iso-8859-1\ \/\n);
parent.rightFrame.document.write(link rel=\stylesheet\
href=\media\/scripts\/sidelayout.css\ type=\text\/css\ \/\n);
parent.rightFrame.document.write(link rel=\stylesheet\
href=\media\/scripts\/sidetext.css\ type=\text\/css\ \/\n);
parent.rightFrame.document.write(script
src=\media\/scripts\/scripts.js\
type=\text\/javascript\\/script\n);
parent.rightFrame.document.write(\/head\n);
parent.rightFrame.document.write(body\n);
}
function rightwrite (name, url, desc) {
parent.rightFrame.document.write(p class=\rightlink\a href=\ +
url
+ \ target=\_blank\ + name + \/abr \/);
parent.rightFrame.document.write(desc + \/p);
}
function rightclose () {
parent.rightFrame.document.write(\/body\n);
parent.rightFrame.document.write(\/html\n);
parent.rightFrame.document.close();
}
when i start the file i get a javascript error saying expecting
object
i'm not sure what i'm doing wrong. any help is appreciated.
thanks
_
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Re: [PHP-DB] javascript problem
If your framecontainer is your top-document perhaps you can use:
top.rightFrame.document.open();
or maybe
top.rightFrame.location='';
Remco
- Original Message -
From: Josh Trutwin [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Sent: Monday, March 18, 2002 4:25 PM
Subject: Re: [PHP-DB] javascript problem
I am not a JavaScript expert, but I had a similar problem. I THINK that
you need to reference your frames like this:
window.parent.frames.rightFrame.document.open();
Give it a try, hope that helps...
Josh
hi all,
i know this is not exactly the place to ask this, but i don't know
where else to ask. i'm doing a multiframe project and i'm trying to
get one of the pages to change the content of another frame when it
loads.
in the main.htm i have:
html
head
titleA Global Village?/title
...
script src=../../media/scripts/rightside.js type=text/javascript
script type=text/javascript
!--
rightopen();
rightwrite('Hello', 'www.yahoo.com', 'Yahoo is really good.');
rightclose();
//--
/script
/script
/head
body
...
/body
/html
the included rightside.js file has the three functions:
function rightopen () {
parent.rightFrame.document.open();
parent.rightFrame.document.write(!DOCTYPE html PUBLIC
\-\/\/W3C\/\/DTD
XHTML 1.0 Strict\/\/EN\\n);
parent.rightFrame.document.write(
\http:\/\/www.w3.org\/TR\/xhtml1\/DTD\/xhtml1-strict.dtd\\n);
parent.rightFrame.document.write(html\n);
parent.rightFrame.document.write(head\n);
parent.rightFrame.document.write(title\/title\n);
parent.rightFrame.document.write(meta http-equiv=\Content-Type\
content=\text\/html; charset=iso-8859-1\ \/\n);
parent.rightFrame.document.write(link rel=\stylesheet\
href=\media\/scripts\/sidelayout.css\ type=\text\/css\ \/\n);
parent.rightFrame.document.write(link rel=\stylesheet\
href=\media\/scripts\/sidetext.css\ type=\text\/css\ \/\n);
parent.rightFrame.document.write(script
src=\media\/scripts\/scripts.js\
type=\text\/javascript\\/script\n);
parent.rightFrame.document.write(\/head\n);
parent.rightFrame.document.write(body\n);
}
function rightwrite (name, url, desc) {
parent.rightFrame.document.write(p class=\rightlink\a href=\ +
url
+ \ target=\_blank\ + name + \/abr \/);
parent.rightFrame.document.write(desc + \/p);
}
function rightclose () {
parent.rightFrame.document.write(\/body\n);
parent.rightFrame.document.write(\/html\n);
parent.rightFrame.document.close();
}
when i start the file i get a javascript error saying expecting
object
i'm not sure what i'm doing wrong. any help is appreciated.
thanks
_
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RE: [PHP-DB] I can't get Oracle record structure from PHP
what error does it give u. also check some of these examples on http://conf.php.net/pres/index.php?p=slides%2Fociid=oci -Original Message- From: Csillag Zsolt [mailto:[EMAIL PROTECTED]] Sent: 17 March 2002 18:40 To: [EMAIL PROTECTED] Subject: [PHP-DB] I can't get Oracle record structure from PHP Hi, I want to call a Pl/Sql function from Php ( I'm using oci8 functions). If the Pl/Sql function returns with VarChar2, number etc. it works perfectly. However when I use the following record for return it doesn't work: Type T_CegTabla IS TABLE OF VarChar2(80) INDEX BY BINARY_INTEGER; Type T_CimTabla IS TABLE OF VarChar2(110) INDEX BY BINARY_INTEGER; TYPE T_CegLista_REK IS RECORD ( Cegnev T_CegTabla , Cim T_CimTabla ); Function DoSomething (KeresoStr Varchar2,MinRow Integer, MaxRow Integer) Return T_CegLista_REK ; Is there a way in Php to workaround this? I noticed that there isn't record structure in Php. Thank you in advance Zsolt Csillag, Hungary -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Fill select boxes with dependent data from a database. How?
Hi all, i am searching for a solution for filling 2 (or 3) dependent select boxes with values from a database. That means, when you select a value in selectbox 1 the options for the second selectbox should be filled with values related to select box 1. Short example Value for box 1Value for box 2 1a 1b 1c 2d 2e 2f If you now select 1 in select box 1 then you should only see a,b and c in selectbox 2. Was that clear? I hope so. Does anybody has a solution for this situation? Or maybe sample code? Thank you in advance for any help. Kind regards and greetings from Cologne/Germany, Marco. [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] RE: [PHP-WIN] Stange 'page-loading' effect
You should check your error logs after getting the server not found error. There will frequently be PHP error messages stored in there that might help discern the problem. You can also set checkpoints in your code that write out Passed this part, passed this part, etc to a temporary debugging file (whatever you want to call it), so you can determine where the script failed if you don't get any help from the logfiles). - Jonathan -Original Message- From: alain samoun [mailto:[EMAIL PROTECTED]] Sent: Friday, March 15, 2002 9:19 AM To: George Pitcher; [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: [PHP-DB] RE: [PHP-WIN] Stange 'page-loading' effect You probably have an error in your code, difficult to help without at least a snippet of it. A+ Alain -Original Message- From: George Pitcher [mailto:[EMAIL PROTECTED]] Sent: Friday, March 15, 2002 4:14 AM To: [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: [PHP-WIN] Stange 'page-loading' effect Hi all, I have a part of my development site where the user hits a button, sending a stream of values to a response page. However, on testing, the response page comes up with 'server not found' but when I refresh that error page, it loads the correct page but without any of the transferred values. Any ideas as to where it going wrong? I'm using PHP-4.1.1, Apache (console) on Win 2000Pro. Regards George George Pitcher is Technical Manager for the HERON Project at Napier University, Edinburgh. -- PHP Windows Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] How to bring down the mysql server smoothly
Hi there, how is it possible to bring down the server after starting with safe_mysqld. I read, that killing the ps with -9 is demmaging the tables. Thanx, Andy -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] textarea and mysql query
I need some help with a query to place the results within a text area...
here is the code and so far every time I try it out all that happens is it
displays the textbox but not the results of the query.
form name=wel_area method=post action=index_confirm.php3 target=box
?php
$record = mysql_fetch_array(mysql_query(SELECT wel_area FROM
cm_index,$dbh));
echo textarea name='wel_area' value='{$record['wel_area']}' cols='25'
rows='8'/textareabrinput type='submit' name='save'
value='save'br\n;
?
/form
Any help would be great.
Jas
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RE: [PHP-DB] textarea and mysql query
Try this:
form name=wel_area method=post action=index_confirm.php3 target=box
?php
$record = mysql_fetch_array(@mysql_query(SELECT wel_area FROM
cm_index,$dbh));
echo textarea name='wel_area' value='.$record['wel_area'].' cols='25'
rows='8'/textareabrinput type='submit' name='save'
value='save'br\n;
?
/form
If not then let me know...
Thank you,
Ray Hunter
Firmware Engineer
ENTERASYS NETWORKS
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Sunday, March 17, 2002 11:54 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] textarea and mysql query
I need some help with a query to place the results within a text area...
here is the code and so far every time I try it out all that happens is it
displays the textbox but not the results of the query. form name=wel_area
method=post action=index_confirm.php3 target=box
?php
$record = mysql_fetch_array(@mysql_query(SELECT wel_area FROM
cm_index,$dbh)); echo textarea name='wel_area'
value='{$record['wel_area']}' cols='25' rows='8'/textareabrinput
type='submit' name='save' value='save'br\n; ?
/form
Any help would be great.
Jas
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Re: [PHP-DB] textarea and mysql query
On Monday 18 March 2002 14:53, jas wrote:
I need some help with a query to place the results within a text area...
here is the code and so far every time I try it out all that happens is it
displays the textbox but not the results of the query.
form name=wel_area method=post action=index_confirm.php3
target=box ?php
$record = mysql_fetch_array(mysql_query(SELECT wel_area FROM
cm_index,$dbh));
echo textarea name='wel_area' value='{$record['wel_area']}' cols='25'
rows='8'/textareabrinput type='submit' name='save'
value='save'br\n;
?
/form
If this is *all* your code then you're missing the important step of
connecting to the mysql server and selecting a database.
Have a look at the examples in the manual.
--
Jason Wong - Gremlins Associates - www.gremlins.com.hk
/*
Most people don't need a great deal of love nearly so much as they need
a steady supply.
*/
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Re: [PHP-DB] textarea and mysql query
Yeah that didn't work either... If I put the same value (
value=\{$record['wel_area']}\ ) on a text line it works just fine however
using the same value for a textarea doesnt give me any results nor does it
give me an error message to go on, if there is a way to output the code to
the screen like a dos echo command that would help out alot. Thanks in
advance,
Jas
Ray Hunter [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Try this:
form name=wel_area method=post action=index_confirm.php3
target=box
?php
$record = mysql_fetch_array(@mysql_query(SELECT wel_area FROM
cm_index,$dbh));
echo textarea name='wel_area' value='.$record['wel_area'].' cols='25'
rows='8'/textareabrinput type='submit' name='save'
value='save'br\n;
?
/form
If not then let me know...
Thank you,
Ray Hunter
Firmware Engineer
ENTERASYS NETWORKS
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Sunday, March 17, 2002 11:54 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] textarea and mysql query
I need some help with a query to place the results within a text area...
here is the code and so far every time I try it out all that happens is it
displays the textbox but not the results of the query. form
name=wel_area
method=post action=index_confirm.php3 target=box
?php
$record = mysql_fetch_array(@mysql_query(SELECT wel_area FROM
cm_index,$dbh)); echo textarea name='wel_area'
value='{$record['wel_area']}' cols='25' rows='8'/textareabrinput
type='submit' name='save' value='save'br\n; ?
/form
Any help would be great.
Jas
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Re: [PHP-DB] textarea and mysql query
Sorry, the connection is established already so that part works just fine.
It is getting the results of said connection \ query to appear in a textarea
vs. a text line.
Jason Wong [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]...
On Monday 18 March 2002 14:53, jas wrote:
I need some help with a query to place the results within a text area...
here is the code and so far every time I try it out all that happens is
it
displays the textbox but not the results of the query.
form name=wel_area method=post action=index_confirm.php3
target=box ?php
$record = mysql_fetch_array(@mysql_query(SELECT wel_area FROM
cm_index,$dbh));
echo textarea name='wel_area' value='{$record['wel_area']}' cols='25'
rows='8'/textareabrinput type='submit' name='save'
value='save'br\n;
?
/form
If this is *all* your code then you're missing the important step of
connecting to the mysql server and selecting a database.
Have a look at the examples in the manual.
--
Jason Wong - Gremlins Associates - www.gremlins.com.hk
/*
Most people don't need a great deal of love nearly so much as they need
a steady supply.
*/
--
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To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] textarea and mysql query
What is the result of $record['wel_area']. If this is a multidimensional
array then you will need to access it differently...
Example:
echo $record[0][0].\n;
Try this is your text area and see if that comes out...
Thank you,
Ray Hunter
Firmware Engineer
ENTERASYS NETWORKS
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Monday, March 18, 2002 12:07 AM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] textarea and mysql query
Yeah that didn't work either... If I put the same value (
value=\{$record['wel_area']}\ ) on a text line it works just fine however
using the same value for a textarea doesnt give me any results nor does it
give me an error message to go on, if there is a way to output the code to
the screen like a dos echo command that would help out alot. Thanks in
advance, Jas Ray Hunter [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Try this:
form name=wel_area method=post action=index_confirm.php3
target=box
?php
$record = mysql_fetch_array(@mysql_query(SELECT wel_area FROM
cm_index,$dbh)); echo textarea name='wel_area'
value='.$record['wel_area'].' cols='25'
rows='8'/textareabrinput type='submit' name='save'
value='save'br\n; ?
/form
If not then let me know...
Thank you,
Ray Hunter
Firmware Engineer
ENTERASYS NETWORKS
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Sunday, March 17, 2002 11:54 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] textarea and mysql query
I need some help with a query to place the results within a text
area... here is the code and so far every time I try it out all that
happens is it displays the textbox but not the results of the query.
form
name=wel_area
method=post action=index_confirm.php3 target=box
?php
$record = mysql_fetch_array(@mysql_query(SELECT wel_area FROM
cm_index,$dbh)); echo textarea name='wel_area'
value='{$record['wel_area']}' cols='25' rows='8'/textareabrinput
type='submit' name='save' value='save'br\n; ?
/form
Any help would be great.
Jas
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Re: [PHP-DB] textarea and mysql query
Yeah it is not multidimensional at all, a simple text blob is what should be
pulled into the textarea. I am not trying to query multipled record fields
at all. One field per textarea. Do you know of a good tutorial on pulling
sql queries into textareas? If you do please let me know as I have searched
php.net and other sites for a tutorial on textarea's but have not found any
nor is there any reference to such a problem in any of the php or mysql
books i have.
Ray Hunter [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
What is the result of $record['wel_area']. If this is a multidimensional
array then you will need to access it differently...
Example:
echo $record[0][0].\n;
Try this is your text area and see if that comes out...
Thank you,
Ray Hunter
Firmware Engineer
ENTERASYS NETWORKS
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Monday, March 18, 2002 12:07 AM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] textarea and mysql query
Yeah that didn't work either... If I put the same value (
value=\{$record['wel_area']}\ ) on a text line it works just fine
however
using the same value for a textarea doesnt give me any results nor does it
give me an error message to go on, if there is a way to output the code to
the screen like a dos echo command that would help out alot. Thanks in
advance, Jas Ray Hunter [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Try this:
form name=wel_area method=post action=index_confirm.php3
target=box
?php
$record = mysql_fetch_array(@mysql_query(SELECT wel_area FROM
cm_index,$dbh)); echo textarea name='wel_area'
value='.$record['wel_area'].' cols='25'
rows='8'/textareabrinput type='submit' name='save'
value='save'br\n; ?
/form
If not then let me know...
Thank you,
Ray Hunter
Firmware Engineer
ENTERASYS NETWORKS
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Sunday, March 17, 2002 11:54 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] textarea and mysql query
I need some help with a query to place the results within a text
area... here is the code and so far every time I try it out all that
happens is it displays the textbox but not the results of the query.
form
name=wel_area
method=post action=index_confirm.php3 target=box
?php
$record = mysql_fetch_array(@mysql_query(SELECT wel_area FROM
cm_index,$dbh)); echo textarea name='wel_area'
value='{$record['wel_area']}' cols='25' rows='8'/textareabrinput
type='submit' name='save' value='save'br\n; ?
/form
Any help would be great.
Jas
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Re: [PHP-DB] textarea and mysql query
jas [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
I need some help with a query to place the results within a text area...
here is the code and so far every time I try it out all that happens is it
displays the textbox but not the results of the query.
form name=wel_area method=post action=index_confirm.php3
target=box
?php
$record = mysql_fetch_array(@mysql_query(SELECT wel_area FROM
cm_index,$dbh));
echo textarea name='wel_area' value='{$record['wel_area']}' cols='25'
rows='8'/textareabrinput type='submit' name='save'
value='save'br\n;
?
/form
Any help would be great.
Jas
If the wel_area is a TEXT field in the database, why the array treatment.
I have a database that contains several TEXT fields that I can call up and
edit. To do that I bring the data into a textarea.
Here's part of my code:
The SQL query:
$sql = SELECT subj, body, info, fname, lname, title, city, other,
record_date, status, batch, modified_by
FROM $table_name
WHERE id = '$id'
;
The results are read into variables:
while ($row = mysql_fetch_array($result)) {
$subj = $row['subj'];
$body = $row['body'];
$info = $row['info'];
$fname = $row['fname'];
$lname = $row['lname'];
$title = $row['title'];
$city = $row['city'];
$other = $row['other'];
$record_date = $row['record_date'];
$status = $row['status'];
$batch = $row['batch'];
$modified_by = $row['modified_by'];
} #end while
In the body of the HTML section I have areas such as this:
td
textarea cols=60 rows=20 name=body wrap=soft ? echo $body;
?/textarea
/td
This creates an editable TEXTAREA that can then be UPDATEd in the database.
John Hughes
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Re: [PHP-DB] textarea and mysql query
On Monday 18 March 2002 15:09, jas wrote:
Sorry, the connection is established already so that part works just fine.
It is getting the results of said connection \ query to appear in a
textarea vs. a text line.
Jason Wong [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]...
On Monday 18 March 2002 14:53, jas wrote:
I need some help with a query to place the results within a text
area... here is the code and so far every time I try it out all that
happens is
it
displays the textbox but not the results of the query.
form name=wel_area method=post action=index_confirm.php3
target=box ?php
action=index_confirm.php3
Note the closing double-quote.
$record = mysql_fetch_array(@mysql_query(SELECT wel_area FROM
cm_index,$dbh));
echo textarea name='wel_area' value='{$record['wel_area']}' cols='25'
rows='8'/textareabrinput type='submit' name='save'
value='save'br\n;
?
/form
Text areas are specified as:
textarea name=abcStuff which goes into text area/textarea
--
Jason Wong - Gremlins Associates - www.gremlins.com.hk
/*
Let's not complicate our relationship by trying to communicate with each
other.
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RE: [PHP-DB] textarea and mysql query
You problem lies in the html that you are outputting.
For normal html input tags, you use the value attribute to initialize
the default display. For textarea tags, you do not use the value
attribute; instead, enclose what you want displayed between the openining
textarea tag and the closing /textarea tag. For example:
input type=text valuePlease edit this textbox
textareaPlease edit this textarea/textarea
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Sunday, March 17, 2002 10:54 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] textarea and mysql query
I need some help with a query to place the results within a
text area...
here is the code and so far every time I try it out all that
happens is it
displays the textbox but not the results of the query.
form name=wel_area method=post
action=index_confirm.php3 target=box
?php
$record = mysql_fetch_array(@mysql_query(SELECT wel_area FROM
cm_index,$dbh));
echo textarea name='wel_area' value='{$record['wel_area']}'
cols='25'
rows='8'/textareabrinput type='submit' name='save'
value='save'br\n;
?
/form
Any help would be great.
Jas
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[PHP-DB] SQL statement - PHP/mySQL - brain fart
For some reason I keep thinking that this should be simple - but I just can't seem to figure it out. Help??? Please??? [I've been working on it for two days now.] Overview: What I'm trying to do is query one table by passing it two different variables - and return only the results that are COMMON to both variables. [PHP 4.1.2/mySQL 3.23.44/FreeBSD] Assume I have a site_category table: --- site_category --- site_category_id site_id category_id --- Perhaps a dump of this looks something like this: --- --- site_category_idsite_id category_id --- --- 1 2 10 2 3 11 3 4 12 4 4 10 5 5 12 6 5 14 --- --- Using values for the varibles I'm passing to the query (see below) of ... $category_id_1 = 10 $category_id_2 = 12 ... the result I'm looking for is: site_id = 4 ... as this is the only site_id which is common to both ... category_id = 10 category_id = 12 I've tried a bazillion variations on the following query: SELECT sc.* FROMsite_category sc WHERE (sc.category_id = $category_id_1 OR sc.category_id = $category_id_2) Breaking out the parts ... So, if category_id_1 = 10, I'm returned: site_id = 2 site_id = 4 So, if category_id_2 = 12, I'm returned: site_id = 4 site_id = 5 How can I get that 4 which you can clearly see is common to both of the parts above? But just about no matter how I write my queries, I keep getting: site_id = 2 site_id = 4 site_id = 4 site_id = 5 Or if use SELECT DISTINCT we get: site_id = 2 site_id = 4 site_id = 5 [I want that extra 4 that the DISTINCT threw out!!!] I keep thinking that I can do this in a single query - but I don't know for sure. I've tried sub-selects with no luck [E.g. IN()]. Do I need to do something with arrays and array_intersect? [I've even tried messing with the PHP3 hacks for array_unique - trying to reverse them 'n stuff - but still no luck.] Does anyone have a simple solution? [I'll even take a hard solution - but I keep thinking that I'm just looking at the the wrong way.] TIA, Summit There is no such thing as a stupid person - there are only those who choose not to learn! Summit - [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] How display a BLOB field contents in a HTML file?
Hi. I work in a client project that need to show a images of his products in a HTML file. I work with MySQL 3.23.47 for windows 98, Apache 1.3.23 for windows and PHP4. I insert the images with EMS MySQL Manager (is too easy) My databese work ok wend i retrive text information like product_ID , Product_Name, Maximun amount of product, etc. but i can't show the image of the product. The PHP code is: html body ? $link = mysql_connect(localhost, username,password); mysql_select_db(MyDB, $link); $result = mysql_query(SELECT * FROM producto, $link); echo ID: .mysql_result($result, 0, PRO_ID).br; echo Nombre: .mysql_result($result, 0, PRO_NOMBRE).br; echo Cant max :.mysql_result($result, 0, PRO_CANTMAX).br; echo Cant min :.mysql_result($result, 0, PRO_CANTMIN).br; echo Foto :.mysql_result($result, 0, PRO_CANTMIN).br; ? /body /html I know that some is wrong but i don't know what its Sergio Cornejo -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] passing array through $PHP_SELF
Sorry to say but the inelegant solution is the only way I have found how to do it :(. I read it on php FAQTs on how to pass arrays on multi-step forms. This is a very similar thing. So yes you have to implode the array then pass it in a hidden field. Then explode the array after being passed to form again. Hope this helps. -Mike de Libero [EMAIL PROTECTED] http://www.soreye.com - Original Message - From: John Hughes, Jomari Works [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Monday, March 18, 2002 5:27 PM Subject: [PHP-DB] passing array through $PHP_SELF I have a page that selects students from a mysql table who meet a certain requirement. Those who meet the requirement have a check box variable that is checked. (The other students have the check box, but it is unchecked.) The checked student id numbers are sent to the next PHP page. That page allows the user to edit the text of a message that will be sent to each of the students checked on the previous page. Up to this point I'm fine. I can pass the array of student ids and read them on the other page. What I want to do is use $PHP_SELF to allow the user to preview the letter after editing. My initial tries haven't worked. If I just submit $PHP_SELF the student id array disappears. When I attempt to create a hidden form field I get the message: Parse error: parse error, expecting `T_STRING' or `T_VARIABLE' or `T_NUM_STRING' Right now I have the book open to implode and explode, figuring I can translate the array and then pass it back and forth, but that seems rather inelegant. Is there a simple way to pass an array when using $PHP_SELF to reload a page? TIA John Hughes -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] passing array through $PHP_SELF
Why not write hidden inputs and name them appropriately so that when the form gets submitted again, the array gets created automatically--no implode/explode. input name=input[var1] value=hello input name=input[var2] value=world then, upon submittal, the array, $_POST['input'], would look like this: Array ( var1 = hello var2 = world ) Please let me know if this is unclear Court -Original Message- From: Mike de Libero To: John Hughes, Jomari Works; [EMAIL PROTECTED] Sent: 3/18/02 7:22 PM Subject: Re: [PHP-DB] passing array through $PHP_SELF Sorry to say but the inelegant solution is the only way I have found how to do it :(. I read it on php FAQTs on how to pass arrays on multi-step forms. This is a very similar thing. So yes you have to implode the array then pass it in a hidden field. Then explode the array after being passed to form again. Hope this helps. -Mike de Libero [EMAIL PROTECTED] http://www.soreye.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] passing array through $PHP_SELF
On Tuesday 19 March 2002 09:27, John Hughes, Jomari Works wrote: I have a page that selects students from a mysql table who meet a certain requirement. Those who meet the requirement have a check box variable that is checked. (The other students have the check box, but it is unchecked.) The checked student id numbers are sent to the next PHP page. That page allows the user to edit the text of a message that will be sent to each of the students checked on the previous page. Up to this point I'm fine. I can pass the array of student ids and read them on the other page. What I want to do is use $PHP_SELF to allow the user to preview the letter after editing. My initial tries haven't worked. If I just submit $PHP_SELF the student id array disappears. When I attempt to create a hidden form field I get the message: Parse error: parse error, expecting `T_STRING' or `T_VARIABLE' or `T_NUM_STRING' Right now I have the book open to implode and explode, figuring I can translate the array and then pass it back and forth, but that seems rather inelegant. Is there a simple way to pass an array when using $PHP_SELF to reload a page? You can serialize() the array then pass it along in a hidden form element. Or you can use cookies (sessions). -- Jason Wong - Gremlins Associates - www.gremlins.com.hk /* Why must you tell me all your secrets when it's hard enough to love you knowing nothing? -- Lloyd Cole and the Commotions */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] SQL statement - PHP/mySQL - brain fart
On Tuesday 19 March 2002 10:22, Summit wrote: For some reason I keep thinking that this should be simple - but I just can't seem to figure it out. Help??? Please??? [I've been working on it for two days now.] Overview: What I'm trying to do is query one table by passing it two different variables - and return only the results that are COMMON to both variables. [PHP 4.1.2/mySQL 3.23.44/FreeBSD] Assume I have a site_category table: --- site_category --- site_category_id site_id category_id --- Perhaps a dump of this looks something like this: --- --- site_category_idsite_id category_id --- --- 1 2 10 2 3 11 3 4 12 4 4 10 5 5 12 6 5 14 --- --- Using values for the varibles I'm passing to the query (see below) of ... $category_id_1 = 10 $category_id_2 = 12 ... the result I'm looking for is: site_id = 4 ... as this is the only site_id which is common to both ... category_id = 10 category_id = 12 I've tried a bazillion variations on the following query: SELECT sc.* FROMsite_category sc WHERE (sc.category_id = $category_id_1 OR sc.category_id = $category_id_2) Breaking out the parts ... So, if category_id_1 = 10, I'm returned: site_id = 2 site_id = 4 So, if category_id_2 = 12, I'm returned: site_id = 4 site_id = 5 How can I get that 4 which you can clearly see is common to both of the parts above? But just about no matter how I write my queries, I keep getting: site_id = 2 site_id = 4 site_id = 4 site_id = 5 Or if use SELECT DISTINCT we get: site_id = 2 site_id = 4 site_id = 5 [I want that extra 4 that the DISTINCT threw out!!!] I keep thinking that I can do this in a single query - but I don't know for sure. I've tried sub-selects with no luck [E.g. IN()]. Do I need to do something with arrays and array_intersect? [I've even tried messing with the PHP3 hacks for array_unique - trying to reverse them 'n stuff - but still no luck.] Does anyone have a simple solution? [I'll even take a hard solution - but I keep thinking that I'm just looking at the the wrong way.] SELECT sc.* FROM site_category sc WHERE sc.category_id = $category_id_1 AND sc.category_id = $category_id_2 Or am I missing something? -- Jason Wong - Gremlins Associates - www.gremlins.com.hk /* Let's just be friends and make no special effort to ever see each other again. */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] post to a url and return to same page
Hello, I am trying to post to a URL and return to the same page with the result and or error messages. The shared server that is hosting this site was not compiled with CURL. Is there an alternative way to post a form to an alternate url and return to the same page without having to rely on the alternate URL's server? Regards, Adrian -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] login problem wasn't solved
i have sell.php page which requires loggin so i put in it:
?
session_start();
if(!(session_is_registered(valusername)))
{
$page=$PHP_SELF;
session_register(page);
header(Location:http://$HTTP_HOST/auction/login.php;);
exit();
}
?
then in login.php i have;
$sql = SELECT * FROM registration WHERE username = '$username' and
password='$password';
$result = mysql_query($sql,$conn);
if(mysql_num_rows($result) 0)
{
$row = mysql_fetch_array($result);
$valusername=$row[username];
session_register(valusername);
mysql_close($conn);
if(session_is_registered(page))
{
header(location:http://$HTTP_HOST/.$page);
session_unregister(page);
}
else
header(location:http://$HTTP_HOST/auction/index.php;);
exit();
}
else
{
echo p align='center'font face='Comic Sans MS' size='6'
color='#FF'Sorry!!/font/p
p align='center'font color='#80' size='4'You entered either a wrong username or
password.bra href='login.php'try again/a/font/p;
}
and yes it goes to sell.php page but the url is:
http://localhost//php/php.exe/auction/seller/sell.php
and i tried instead:
if(session_is_registered(page))
{
header(location:http://$HTTP_HOST/.$$HTTP_REFERRER);
session_unregister(page);
}
but it goes to
http://localhost
anyhelp?
Rehab M.Shouman
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RE: [PHP-DB] SQL statement - PHP/mySQL - brain fart
ok it is 1:37 am here and i have been up since 6:00 am (the day before) and have to be awake ths mornign again at 6:00 am so if I make any mistakes trying to answer your question, please take it easy:) I was wondering if you'd wanna use temporary tables to accomplish it.. You may wanna do: CREATE TEMPORARY TABLE tmp1 SELECT * FROM site_category WHERE category_id=$category_id_1; ok this will give us the records that matches $category_id_1 in category_id field put them into a temporary table called tmp1 Then do: CREATE TEMPORARY TABLE tmp2 SELECT * FROM site_category WHERE category_id=$category_id_2; and this will give us the records matching $category_id_2 in category_id field and put them into a temporary table called tmp2.. Now you have 2 temporary tables, tmp1 and tmp2 as shown below: tmp1: +--+--+--+ | sci | si | ci | +--+--+--+ |1 |2 | 10 | |4 |4 | 10 | +--+--+--+ tmp2: +--+--+--+ | sci | si | ci | +--+--+--+ |3 |4 | 12 | |5 |5 | 12 | +--+--+--+ Now you can use use join syntax to find the si value thats common to the both tables... select * from tmp1, tmp2 where tmp1.si=tmp2.si; Does this work for you?? Gurhan -Original Message- From: Summit [mailto:[EMAIL PROTECTED]] Sent: Monday, March 18, 2002 9:22 PM To: [EMAIL PROTECTED] Subject: [PHP-DB] SQL statement - PHP/mySQL - brain fart For some reason I keep thinking that this should be simple - but I just can't seem to figure it out. Help??? Please??? [I've been working on it for two days now.] Overview: What I'm trying to do is query one table by passing it two different variables - and return only the results that are COMMON to both variables. [PHP 4.1.2/mySQL 3.23.44/FreeBSD] Assume I have a site_category table: --- site_category --- site_category_id site_id category_id --- Perhaps a dump of this looks something like this: --- --- site_category_idsite_id category_id --- --- 1 2 10 2 3 11 3 4 12 4 4 10 5 5 12 6 5 14 --- --- Using values for the varibles I'm passing to the query (see below) of ... $category_id_1 = 10 $category_id_2 = 12 ... the result I'm looking for is: site_id = 4 ... as this is the only site_id which is common to both ... category_id = 10 category_id = 12 I've tried a bazillion variations on the following query: SELECT sc.* FROMsite_category sc WHERE (sc.category_id = $category_id_1 OR sc.category_id = $category_id_2) Breaking out the parts ... So, if category_id_1 = 10, I'm returned: site_id = 2 site_id = 4 So, if category_id_2 = 12, I'm returned: site_id = 4 site_id = 5 How can I get that 4 which you can clearly see is common to both of the parts above? But just about no matter how I write my queries, I keep getting: site_id = 2 site_id = 4 site_id = 4 site_id = 5 Or if use SELECT DISTINCT we get: site_id = 2 site_id = 4 site_id = 5 [I want that extra 4 that the DISTINCT threw out!!!] I keep thinking that I can do this in a single query - but I don't know for sure. I've tried sub-selects with no luck [E.g. IN()]. Do I need to do something with arrays and array_intersect? [I've even tried messing with the PHP3 hacks for array_unique - trying to reverse them 'n stuff - but still no luck.] Does anyone have a simple solution? [I'll even take a hard solution - but I keep thinking that I'm just looking at the the wrong way.] TIA, Summit There is no such thing as a stupid person - there are only those who choose not to learn! Summit - [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: php-db Digest 18 Mar 2002 18:40:25 -0000 Issue 1105
At 06:40 PM 3/18/2002 +, you wrote: Subject: How to add 2 years to todays date ? Hi All I think the subject line says it all. How to add 2 years to today's date ? Any helps as always most appreciated. Dave Carrera Php Developer Subject: Re: [PHP-DB] How to add 2 years to todays date ? In php $your_timestamp += gmmktime(0,0,0,1,1,1972); MySQL has INTERVALling options. Regards, Andrey Subject: RE: [PHP-DB] increase date() Try this: $six_months_from_now = date(Y-m-d, mktime(0,0,0, date(m)+6, date(1), date(Y))); echo Six Months From Now: $six_months_from_nowP; In my coding, I've used: $date_var=date(Y-m-d,strtotime(+0 days)); In your case, you could do it like: $date_var=date(Y-m-d, strtotime(+2 years)) or $date_var=date(Y-m-d H:i, strtotime(+2 years)) if you wanted hours/minutes either example formats in a way that MySQL wants to see a date or datetime field (respectively) hope it helps, ~ Rob Small c0demonkey Fresno, CA -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] SQL statement - PHP/mySQL - brain fart
Well it's sorta' simple once you get your mind around this - you need two tables but you've only got one. And there's no OR, you need both to be true for a site_id ... The query would be easy if you actually had two tables so we'll search off of the same table twice giving it different names each time (sc1 and sc2). SELECT sc1.* FROMsite_category sc1, site_category sc2 WHERE sc1.category_id = $category_id_1 AND sc2.category_id = $category_id_2 AND sc1.site_id = sc2.site_id DISTINCT and such are not necessary unless it's possible you'll have duplicate rows (and you'd mind getting them back multiple times). I know this will work in all the big ones (Oracle, MS SQL, IBM) but I'm less sure about MySQL as I don't use it all that much. This is standard SQL. Good Luck, Frank On 3/18/02 10:53 PM, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: From: Summit [EMAIL PROTECTED] Date: Mon, 18 Mar 2002 19:22:22 -0700 To: [EMAIL PROTECTED] Subject: SQL statement - PHP/mySQL - brain fart For some reason I keep thinking that this should be simple - but I just can't seem to figure it out. Help??? Please??? [I've been working on it for two days now.] Overview: What I'm trying to do is query one table by passing it two different variables - and return only the results that are COMMON to both variables. [PHP 4.1.2/mySQL 3.23.44/FreeBSD] Assume I have a site_category table: --- site_category --- site_category_id site_id category_id --- Perhaps a dump of this looks something like this: --- --- site_category_idsite_id category_id --- --- 1 2 10 2 3 11 3 4 12 4 4 10 5 5 12 6 5 14 --- --- Using values for the varibles I'm passing to the query (see below) of ... $category_id_1 = 10 $category_id_2 = 12 ... the result I'm looking for is: site_id = 4 ... as this is the only site_id which is common to both ... category_id = 10 category_id = 12 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Conflicting results using PHP/Mysql
Thanks to Kevin, who yesterday emailed me with a solution to my JOIN problem. Thanks to all who helped me see the wood for the trees. George - Original Message - From: Doug Thompson [EMAIL PROTECTED] To: George Pitcher [EMAIL PROTECTED]; [EMAIL PROTECTED]; [EMAIL PROTECTED] Sent: Monday, March 18, 2002 9:20 PM Subject: Re: Conflicting results using PHP/Mysql Your base table is transactions. I assume you want to do a LEFT JOIN. I think you might see an improvement if transactions were consistently named on the left side of the equalities as: $Itemlistquery.= transactions.Pdownload ; $Itemlistquery.= from bib_extract,scanrates,transactions where ; $Itemlistquery.= (transactions.CourseID = '$Course_ID' and ; $Itemlistquery.= transactions.ExtractID=bib_extract.E_ID and ; $Itemlistquery.= transactions.finrate=scanrates.finrate) ; On Mon, 18 Mar 2002 10:58:02 -, George Pitcher wrote: It may be that my query is restricting the data retrieval but I cannot see how. If I look at the data I am playing with, I have 3 tables and I want to show transactions from a certain course. I also want to show related information from the bib_extract table and the scanrates (prices) table. I've checked the data in these tables and nothing is odd. Maybe there is a different way to build my query so that it displays the full set of records. Any suggestions? George - Original Message - From: Doug Thompson [EMAIL PROTECTED] To: George Pitcher [EMAIL PROTECTED]; [EMAIL PROTECTED]; [EMAIL PROTECTED] Sent: Wednesday, March 13, 2002 5:59 PM Subject: Re: Conflicting results using PHP/Mysql I would guess it's because only three records match the ANDed tests in the WHERE clause (last 3 lines). Isn't that what you intended? Doug On Wed, 13 Mar 2002 16:08:43 -, George Pitcher wrote: Hi all, Posted this yesterday and got no response. Trying again today. I'm having a small problem with a biggish query. The query: $Itemlistquery= select [a whole load of fields from 3 tables] ending with ; $Itemlistquery.= transactions.Pdownload ; $Itemlistquery.= from bib_extract,scanrates,transactions where ; $Itemlistquery.= (transactions.CourseID = '$Course_ID' and ; $Itemlistquery.= bib_extract.E_ID=transactions.ExtractID and ; $Itemlistquery.= scanrates.finrate=transactions.finrate) ; The problem: If I do a simple count of transactions.CourseID='$Course_ID' I get 18 (for a particular course) and the above query only displays 3 results. Any suggestions? I didn't want to clog the list with the whole query but I can if it's necessary. George, in Edinburgh - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
