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php-db | 2001071

RE: [PHP-DB] Dynamic SQL + result resource error From: Michael Rudel (mru <email protected>)
Date: 07/09/01

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
_______________________________________________________________

Suchtreffer AG
Bleicherstraße 20
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Germany
fon: +49-(0)7531-89207-17
fax: +49-(0)7531-89207-13
e-mail: mailto:mru <email protected>
internet: http://www.suchtreffer.de
_______________________________________________________________

> -----Original Message-----
> From: Mark Gordon [mailto:eyesyte <email protected>]
> Sent: Monday, July 09, 2001 1:54 AM
> 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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