RE: [PHP] parse error driving me nuts ...
Instead of
if ((mysql_num_rows($result)) = 1)
try
if ((mysql_num_rows($result)) == 1)
HTH
Jon
-Original Message-
From: Andrew [mailto:[EMAIL PROTECTED]]
Sent: 15 February 2001 15:34
To: [EMAIL PROTECTED]
Subject: [PHP] parse error driving me nuts ...
hi, can anyone help me spot the parse error?
It's throwing up on the first "if" line, and for the life of me I cannot
find it! :)
tia,
andrew
?php
include("db_connect_params.inc");
$sql="select path from PHOTO where pid =1";
$link_id = mysql_connect($host, $usr, $pass);
$result = mysql_db_query($database, $sql, $link_id) or die("no result");
if ((mysql_num_rows($result)) = 1)
{
$row=mysql_fetch_array($result);
extract($row);
echo "img src =\"$path\"/imgbr";
} else {
if (!isset($ii)) $ii = 1;
$i = 1;
while($row=mysql_fetch_array($result))
{
extract($row);
$displayed[$i]=$path;
echo "a href=\"$PHP_SELF?ii=$i\"$i/anbspnbsp";
$i++;
}
echo "img src =\"$displayed[$ii]\"/imgbr";
}
?
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Re: [PHP] parse error driving me nuts ...
Try to use == instead of =
Sorin Ifrim
- Original Message -
From: Andrew [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Thursday, February 15, 2001 5:33 PM
Subject: [PHP] parse error driving me nuts ...
hi, can anyone help me spot the parse error?
It's throwing up on the first "if" line, and for the life of me I cannot
find it! :)
tia,
andrew
?php
include("db_connect_params.inc");
$sql="select path from PHOTO where pid =1";
$link_id = mysql_connect($host, $usr, $pass);
$result = mysql_db_query($database, $sql, $link_id) or die("no result");
if ((mysql_num_rows($result)) = 1)
{
$row=mysql_fetch_array($result);
extract($row);
echo "img src =\"$path\"/imgbr";
} else {
if (!isset($ii)) $ii = 1;
$i = 1;
while($row=mysql_fetch_array($result))
{
extract($row);
$displayed[$i]=$path;
echo "a href=\"$PHP_SELF?ii=$i\"$i/anbspnbsp";
$i++;
}
echo "img src =\"$displayed[$ii]\"/imgbr";
}
?
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Re: [PHP] parse error driving me nuts ...
hi, can anyone help me spot the parse error?
It's throwing up on the first "if" line, and for the life of me I cannot
find it! :)
tia,
andrew
?php
include("db_connect_params.inc");
$sql="select path from PHOTO where pid =1";
$link_id = mysql_connect($host, $usr, $pass);
$result = mysql_db_query($database, $sql, $link_id) or die("no result");
if ((mysql_num_rows($result)) = 1)
Well I don't know if it will give a parse error or not, but you have made
the oh way too common C error in your IF statement.
if ((mysql_num_rows($result)) = 1)
should be:
if ((mysql_num_rows($result)) == 1)
- John Vanderbeck
- Admin, GameDesign
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RE: [PHP] parse error driving me nuts ...
Jon, John, Ifrim, the == worked ... thank you!!
gotta read up on it now :)
Cameron, dropping the extra () did not work :(
thanks for the suggestions folks!
andrew
-Original Message-
From: Jon Haworth [mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 15, 2001 10:38 AM
To: '[EMAIL PROTECTED]'; '[EMAIL PROTECTED]'
Subject: RE: [PHP] parse error driving me nuts ...
Instead of
if ((mysql_num_rows($result)) = 1)
try
if ((mysql_num_rows($result)) == 1)
HTH
Jon
-Original Message-
From: Andrew [mailto:[EMAIL PROTECTED]]
Sent: 15 February 2001 15:34
To: [EMAIL PROTECTED]
Subject: [PHP] parse error driving me nuts ...
hi, can anyone help me spot the parse error?
It's throwing up on the first "if" line, and for the life of me I cannot
find it! :)
tia,
andrew
?php
include("db_connect_params.inc");
$sql="select path from PHOTO where pid =1";
$link_id = mysql_connect($host, $usr, $pass);
$result = mysql_db_query($database, $sql, $link_id) or die("no result");
if ((mysql_num_rows($result)) = 1)
{
$row=mysql_fetch_array($result);
extract($row);
echo "img src =\"$path\"/imgbr";
} else {
if (!isset($ii)) $ii = 1;
$i = 1;
while($row=mysql_fetch_array($result))
{
extract($row);
$displayed[$i]=$path;
echo "a href=\"$PHP_SELF?ii=$i\"$i/anbspnbsp";
$i++;
}
echo "img src =\"$displayed[$ii]\"/imgbr";
}
?
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To contact the list administrators, e-mail: [EMAIL PROTECTED]
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Re: [PHP] parse error driving me nuts ...
Jon, John, Ifrim, the == worked ... thank you!! gotta read up on it now :) Cameron, dropping the extra () did not work :( thanks for the suggestions folks! andrew Remember that '=' means assign what is on the right to what is on the left, whereby '==' means is what is on the left equal to what is on the right. C, and C based languages allow you to out an assignemtn statement even when you intend a relational expression. It is a very common pitfall. I know I have fallen into it in my C work many times :( - John Vanderbeck - Admin, GameDesign -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
