RE: [PHP-DB] Dynamic SQL + result resource error
| From: | Ben Bleything | Date: | Mon, 09 Jul 2001 00:04:12 +0000 |
| Subject: | RE: [PHP-DB] Dynamic SQL + result resource error | ||
| References: | 1 | Groups: | php.db |
| Request: | Send a blank email to php-db+get-10158@lists.php.net to get a copy of this message | ||
Sure he is. Right here:
$queryResult = mysql_query($sql);
what exact error is occurring?
-----Original Message-----
From: Matthew Loff [mailto:matt@willetts.com]
Sent: Sunday, July 08, 2001 5:00 PM
To: 'Mark Gordon'; php-db@lists.php.net
Subject: RE: [PHP-DB] Dynamic SQL + result resource error
You aren't calling mysql_query() to execute the query.
//$find is text box input
$wordsarray = explode(" ",$find);
$sql = "SELECT bandname FROM bands WHERE (bandname
LIKE ";
$i = 0;
while ($i < count($wordsarray))
{
$word = current($wordsarray);
next($wordsarray);
$sql=$sql."$word)";
$i++;
}
print "$sql<hr>";
$queryResult = mysql_query($sql);
while ($myrow=mysql_fetch_row($queryResult))
{
print "$myrow[0],<p>";
}
-----Original Message-----
From: Mark Gordon [mailto:eyesyte@yahoo.com]
Sent: Sunday, July 08, 2001 7:54 PM
To: php-db@lists.php.net
Subject: [PHP-DB] Dynamic SQL + result resource error
Why is this code generating an error when it outputs a
valid SQL statement? (there are no parse errors)
//$find is text box input
$wordsarray = explode(" ",$find);
$sql = "SELECT bandname FROM bands WHERE (bandname
LIKE ";
$i = 0;
while ($i < count($wordsarray))
{
$word = current($wordsarray);
next($wordsarray);
$sql=$sql."$word)";
$i++;
}
print "$sql<hr>";
while ($myrow=mysql_fetch_row($sql))
{
print "$myrow[0],<p>";
}
=====
Mark
eyesyte@yahoo.com
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