This my query:
PHP Code:
// Query for standort:
$result3 = mysql_query
(" SELECT standort.name from standort," .
"WHERE titles.standortID = standort.standortID");
;
Then when using it in:
PHP Code:
$standortcount=0;
L.299: while ($row = mysql_fetch_object($result3)){
$standortcount++;
echo $row->standort ? "<br>Standort: <i>" . htmlentities($row->standort) ."</i>" :"",
"</li>\n";
}
I get the message:Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in d:\programme\apache group\apache\htdocs\bibli\find.php on line 299
I ve changed everything 100 times but without succes.
So help woul be great.
Matthias
01-Jul-2003, 05:27
Toyota - 2008 July Promotion by Nihal
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