RE: [PHP] Links with parameters in DB
[snip] $link=eval(mysql_result($result,$i,link)); [/snip] Try just eval on the field you pull from the database... echo eval($databaseItem) then work your processing. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Links with parameters in DB
[snip]
$i=0;
$q=select * from links';
while ($imysql_num_rows($result))
{
$link=eval(mysql_result($result,$i,link));
.
.
and then i put this:
[/snip]
What happens if you do this?
while($i mysql_num_rows($result)){
$link = eval($result);
echo $link;
}
P.S. Please reply to the list too, I am quite busy and may not be able
to follow up.
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Re: [PHP] Links with parameters in DB
It doesn´t work, I have other fields in the DB apart from the field 'Link',
so if I use
$link = eval($result);
I get a parse error. Apart from that, I have to write the name of the field
(link), if not the server won´t know the field I´m refering to.
Thanks
- Original Message -
From: Jay Blanchard [EMAIL PROTECTED]
To: WebMaster. Radio ECCA [EMAIL PROTECTED];
[EMAIL PROTECTED]
Sent: Thursday, August 19, 2004 1:28 PM
Subject: RE: [PHP] Links with parameters in DB
[snip]
$i=0;
$q=select * from links';
while ($imysql_num_rows($result))
{
$link=eval(mysql_result($result,$i,link));
.
.
and then i put this:
[/snip]
What happens if you do this?
while($i mysql_num_rows($result)){
$link = eval($result);
echo $link;
}
RE: [PHP] Links with parameters in DB
[snip]
It doesn´t work, I have other fields in the DB apart from the field 'Link',
so if I use
$link = eval($result);
I get a parse error. Apart from that, I have to write the name of the field
(link), if not the server won´t know the field I´m refering to.
[/snip]
Then did you eval that? I meant for you to use a proper rendering of the eval
statement by itself, i.e.
$sql = SELECT * FROM table ;
$result = mysql_query($sql, $connection);
while($row = mysql_fetch_array($result)){
$link = eval($row['Link']);
echo $link;
}
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Re: [PHP] Links with parameters in DB
I tried what you said but i get an eval error:
Parse error: parse error, unexpected '=' in /index.php(135) : eval()'d
code on line 1
;(
- Original Message -
From: Jay Blanchard [EMAIL PROTECTED]
To: WebMaster. Radio ECCA [EMAIL PROTECTED];
[EMAIL PROTECTED]
Sent: Thursday, August 19, 2004 1:38 PM
Subject: RE: [PHP] Links with parameters in DB
[snip]
It doesn´t work, I have other fields in the DB apart from the field 'Link',
so if I use
$link = eval($result);
I get a parse error. Apart from that, I have to write the name of the field
(link), if not the server won´t know the field I´m refering to.
[/snip]
Then did you eval that? I meant for you to use a proper rendering of the
eval statement by itself, i.e.
$sql = SELECT * FROM table ;
$result = mysql_query($sql, $connection);
while($row = mysql_fetch_array($result)){
$link = eval($row['Link']);
echo $link;
}
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RE: [PHP] Links with parameters in DB
[snip] I tried what you said but i get an eval error: Parse error: parse error, unexpected '=' in /index.php(135) : eval()'d code on line 1 [/snip] You will probably have to escape the equals sign -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Links with parameters in DB
[snip] [snip] I tried what you said but i get an eval error: Parse error: parse error, unexpected '=' in /index.php(135) : eval()'d code on line 1 [/snip] You will probably have to escape the equals sign [/snip] Have you RTFM on eval? http://www.php.net/eval You have to use valid PHP code. Your stored code is currently not valid if I read this correctly. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Links with parameters in DB
On Thu, 19 Aug 2004 14:01:45 +0100, WebMaster. Radio ECCA
[EMAIL PROTECTED] wrote:
I tried what you said but i get an eval error:
Parse error: parse error, unexpected '=' in /index.php(135) : eval()'d
code on line 1
;(
- Original Message -
From: Jay Blanchard [EMAIL PROTECTED]
To: WebMaster. Radio ECCA [EMAIL PROTECTED];
[EMAIL PROTECTED]
Sent: Thursday, August 19, 2004 1:38 PM
Subject: RE: [PHP] Links with parameters in DB
[snip]
It doesn´t work, I have other fields in the DB apart from the field 'Link',
so if I use
$link = eval($result);
I get a parse error. Apart from that, I have to write the name of the field
(link), if not the server won´t know the field I´m refering to.
[/snip]
Then did you eval that? I meant for you to use a proper rendering of the
eval statement by itself, i.e.
$sql = SELECT * FROM table ;
$result = mysql_query($sql, $connection);
while($row = mysql_fetch_array($result)){
$link = eval($row['Link']);
echo $link;
}
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try this
eval(\$link = \$row[Link]\;);
echo $link;
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