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POphp query execution error when used as method
 

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  1. POphp query execution error when used as method
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    <p>In my opinion i have a weird problem. The part below works fine</p> <pre><code> $sql = " UPDATE ".$table[0]." SET p_title = '".$_POST['p_title']."', p_date = '".$_POST['p_date']."' WHERE p_id = '".$_POST['p_id']."' "; if(!$db->exec($sql)){ echo($defaults->throwError('MySql error',$sql,implode(":",$db->errorInfo()))); }else{ $defaults->writeLog($table,$db->lastInsertId(),'update'); } </code></pre> <p>But when i try the code below i get an 'Fatal error: Call to a member function exec() on a non-object in'</p> <pre><code> class Defaults{ [..] public function query($sql){ if(!$db->exec($sql)){ echo($defaults->throwError('MySql error',$sql,implode(":",$db->errorInfo()))); }else{ $defaults->writeLog($table,$db->lastInsertId(),'update'); } } [..] } </code></pre> <p>and then on my page</p> <pre><code> $defaults = new Defaults(); $defaults->query(" UPDATE ".$table[0]." SET p_title = '".$_POST['p_title']."', p_date = '".$_POST['p_date']."' WHERE p_id = '".$_POST['p_id']."' "); </code> </pre> <p>How come?</p> <p> Ah, it was about the <code>$db</code> in the class. However, when i do something like</p> <pre><code> public function query($sql){ $db = new PDO($dbdata->hostname,$dbdata->username,$dbdata->password); if(!$db->exec($sql)){ echo($defaults->throwError('MySql error',$sql,implode(":",$db->errorInfo()))); }else{ $defaults->writeLog($table,$db->lastInsertId(),'update'); } } </code></pre> <p>I get a nasty</p> <pre> Fatal error: Uncaught exception 'PDOException' with message 'invalid data source name' in /class.defaults.php:8 Stack trace: #0 /class.defaults.php(8): PDO->__construct('', NULL, NULL) #1 /class.form.php(269): Defaults->query('?????????UPDATE...') #2 /module.projectbeheer.edit.php(25): Form->proceed('update', 'p_id', 'rows', Array, Array, '') #3 /class.content.php(16): include_once('/path/') #4 /administratie.php(72): Content->write('/BraamsArchief/...') #5 {main} thrown in /class.defaults.php on line 8 </pre>
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    1. USMaurice
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    1. CObad guy will kill your database ....your code is vulnerable
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      1. USNullPoiиteя
    2. COThe first snippet has `$db` variable in scope; the second does not. You didn't show the place where it's initialized, but it's actually the key here.
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      1. USraina77ow
    3. COit's only for backend use for maybe two users who, i assume, are not interested in destroying their own database. But it should be safe, what am i doing wrong?
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      1. USMaurice
 

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