Sure he is. Right here:
$queryResult = mysql_query($sql);
what exact error is occurring?
-----Original Message-----
From: Matthew Loff [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 08, 2001 5:00 PM
To: 'Mark Gordon'; [EMAIL PROTECTED]
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 "$myrow[0],<p>";
}
-----Original Message-----
From: Mark Gordon [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 08, 2001 7:54 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Dynamic SQL + result resource error
Why is this code generating an error when it outputs a
valid SQL statement? (there are no parse errors)
//$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
[EMAIL PROTECTED]
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