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php-db | 2001071

RE: [PHP-DB] Dynamic SQL + result resource error From: Ben Bleything (bleythbe <email protected>)
Date: 07/08/01

Sure he is. Right here:

$queryResult = mysql_query($sql);

what exact error is occurring?

-----Original Message-----
From: Matthew Loff [mailto:matt <email protected>]
Sent: Sunday, July 08, 2001 5:00 PM
To: 'Mark Gordon'; php-db <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:eyesyte <email protected>]
Sent: Sunday, July 08, 2001 7:54 PM
To: php-db <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
eyesyte <email protected>

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