Hi Mark,
... beside of what was mentioned be4, you have also
a logical error in the way you append to your $sql-
query in the while loop.
Look: letz say, $find is "The Fields of the Nephilim",
so your $wordsarray would be:
0 ==> The
1 ==> Fields
2 ==> of
3 ==> the
4 ==> Nephilim
in your while-loop, you append every array-entry to
your $sql with "$word)".
So your whole SQL-Statement would read:
"SELECT bandname FROM bands WHERE (bandname LIKE The)Fields)of)the)Nephilim)"
This statement will result in error !
Try this 1:
/////////////////////////////////////////////////////////////////
//$find is text box input
$wordsarray = explode(" ",$find);
$sql = "SELECT bandname FROM bands WHERE bandname <> ''";
while ( list( $key, $word ) = each( $wordsarray ) )
{
$sql .= " OR bandname like '%".$word."%'";
}
echo $sql."<hr>";
$queryResult = mysql_query($sql);
while ( $myrow = mysql_fetch_assoc( $queryResult ) )
{
echo "<p> ".$myrow["bandname"]." </p>"."\n";
}
/////////////////////////////////////////////////////////////////
Hope this helps ;)
Greetinx,
Mike
Michael Rudel
- Web-Development, Systemadministration -
Besuchen Sie uns am 20. und 21. August 2001 auf der
online-marketing-düsseldorf in Halle 1 Stand E 16
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_______________________________________________________________
> -----Original Message-----
> From: Mark Gordon [mailto:[EMAIL PROTECTED]]
> Sent: Monday, July 09, 2001 1:54 AM
> 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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