CODEWITHSUNDEEP
phpmysqltry-catch
Nave Tseva
Nave Tseva

Reputation: 878

Try and Catch error / PHP

I have the following PHP code:

try{ 
    $article_ID =$_GET["articleID"];  
    if(!$article_ID) {
        throw new Exception("Invalid query: ". mysql_error());
    }
    else {
        $select_query = mysql_query("SELECT articleContent, articleTitle From articles WHERE articleID=$article_ID AND typeID=$type_ID");
    }
}
catch(Exception $e) { 
    //echo $e->getMessage();
    $select_query = mysql_query("SELECT articleContent, articleTitle From articles       WHERE typeID=$type_ID");
}
$row = mysql_fetch_assoc($select_query); 
echo '<h1>'.$row['articleTitle'].'</h1>';
echo  $row['articleContent'];

The condition in the if statment (if(!$article_ID) ) should try to get the value in get method, and if it can't so it will throw exception and pass to the catch part, it works fine, but I see error message in my webpage any time it comes to the catch (Notice: Undefined index: articleID on line 6) why? and how can I hide this message?

Upvotes: 0

Views: 157

Answers (1)

Voitcus
Voitcus

Reputation: 4446

Notices do not throw exceptions. The notice is in the first line, because there is no "ArticleID" key in $_GET array variable.

I think you could do something like this

$articleID = isset($_GET["articleID"]) ? $_GET["articleID"] : '';

Upvotes: 1

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