DJSnels Posted September 30, 2010 Share Posted September 30, 2010 I'm getting the following notice: Notice: Undefined variable: TotaalBerichten in C:\Users\Sander\Documents\**********************\index.php on line 60 Is it because it is in the IF? The code: //Aantal berichten totaal if (!$ResultTotaalBerichten = @mysql_query("SELECT COUNT(1) FROM " . $db_prefix . "berichten")) {$foutmelding .= $Taal['errorConfiguratieFromDB'];} else {$TotaalBerichten = mysql_result($ResultTotaalBerichten, 0); $TotaalPaginas = ceil($TotaalBerichten / $aantalBerichtenPerPagina);} @mysql_free_result($ResultTotaalBerichten); //Var eersteBericht is het eerste bericht if (isset($_GET['eersteBericht'])) {if (!is_numeric($_GET['eersteBericht'])) {$eersteBericht = '0';} else{$eersteBericht = $_GET['eersteBericht'];}} else{$eersteBericht = '0';} // Als er iets mis is met var eersteBericht, dan hem op 0 zetten if (($eersteBericht >= $TotaalBerichten) || ($eersteBericht < 0)) ///////////DIT IS LIJN 60 {$eersteBericht = 0;} EDIT: translated to english If i modify it to: $TotaalBerichten=0; /////////////ADDED //Aantal berichten totaal if (!$ResultTotaalBerichten = @mysql_query("SELECT COUNT(1) FROM " . $db_prefix . "berichten")) {$foutmelding .= $Taal['errorConfiguratieFromDB'];} else {$TotaalBerichten = mysql_result($ResultTotaalBerichten, 0); $TotaalPaginas = ceil($TotaalBerichten / $aantalBerichtenPerPagina);} @mysql_free_result($ResultTotaalBerichten); echo $TotaalBerichten; //////////////////////////////////////////////////////////////////////////ADDED //Var eersteBericht is het eerste bericht if (isset($_GET['eersteBericht'])) {if (!is_numeric($_GET['eersteBericht'])) {$eersteBericht = '0';} else{$eersteBericht = $_GET['eersteBericht'];}} else{$eersteBericht = '0';} // Als er iets mis is met var eersteBericht, dan hem op 0 zetten if (($eersteBericht >= $TotaalBerichten) || ($eersteBericht < 0)) ///////////DIT IS LIJN 60 {$eersteBericht = 0;} The $TotaalBerichten is '0' while there IS a message in the database??? Quote Link to comment https://forums.phpfreaks.com/topic/214812-notice-undefined-variable/ Share on other sites More sharing options...
DJSnels Posted September 30, 2010 Author Share Posted September 30, 2010 SOLVED after using echo mysql_error(); i found that the table i used was incorrect (or actually it was $prefix in stead of &db_prefix) Quote Link to comment https://forums.phpfreaks.com/topic/214812-notice-undefined-variable/#findComment-1117520 Share on other sites More sharing options...
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