RE: [PHP-DB] Dynamic SQL + result resource error
| From: | Michael Rudel | Date: | Mon, 09 Jul 2001 08:44:17 +0000 |
| Subject: | RE: [PHP-DB] Dynamic SQL + result resource error | ||
| References: | 1 | Groups: | php.db |
| Request: | Send a blank email to php-db+get-10173@lists.php.net to get a copy of this message | ||
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
D-78467 Konstanz
Germany
fon: +49-(0)7531-89207-17
fax: +49-(0)7531-89207-13
e-mail: mailto:mru@suchtreffer.de
internet: http://www.suchtreffer.de
_______________________________________________________________
> -----Original Message-----
> From: Mark Gordon [mailto:eyesyte@yahoo.com]
> Sent: Monday, July 09, 2001 1:54 AM
> 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
>
> __________________________________________________
> Do You Yahoo!?
> Get personalized email addresses from Yahoo! Mail
> http://personal.mail.yahoo.com/
>
> --
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, e-mail: php-db-unsubscribe@lists.php.net
> For additional commands, e-mail: php-db-help@lists.php.net
> To contact the list administrators, e-mail:
> php-list-admin@lists.php.net
>