MySQL delete query in PHP doesn't work in IF statement

Question

I made a script which inserts a row from one table in another, but then it has to delete the record from one table. For some reason that isn't working. Could someone please help me out with this?

My code:

<html>
    <head>
        <link rel="stylesheet" type="text/css" href="css/layout.css"/>
        <?php session_start();
           if(!isset($_SESSION['login_id'])){
           $url = 'helpdesklogin.php';
           header("Location: $url");
            }
        ?>
    </head>
    <body>
    <?php
    $server="localhost";
    $username="root";
    $password="";
    $connect_mysql=mysql_connect($server,$username,$password) or die ("Connection Failed!");
    $mysql_db=mysql_select_db("helpdesk_middenpolder",$connect_mysql) or die ("Could not Connect to Database");
    $id=$_GET['id'];
    $query=mysql_query("INSERT INTO afgehandelden_incidenten SELECT * FROM incidenten WHERE incidentID='$id'");
    $result=mysql_query($query);
    if($result=mysql_query($query)){
        $query2=mysql_query("DELETE FROM incidenten WHERE incidentID=$id");
    }
    else {
    echo mysql_error();
    }

    ?>
    </body>
</html>

@gvgvgvijayan His question is still valid you cannot simply mark it down because you prefer pdo instead of mysql. I know its deprecated but he didn't ask advice on which method to use. — damien hawks
– damien hawks, Commented Jun 5, 2014 at 7:22

Rahul Goel · Accepted Answer · 2014-06-05 07:18:34Z

2

Wrong way:

$query=mysql_query("INSERT INTO afgehandelden_incidenten SELECT * FROM incidenten WHERE incidentID='$id'");
$result=mysql_query($query);

You are executing your insert query twice by using "mysql_query()" twice

you can do:

$query="INSERT INTO afgehandelden_incidenten SELECT * FROM incidenten WHERE incidentID='$id'";
$result=mysql_query($query);

answered Jun 5, 2014 at 7:18

Rahul Goel

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2 Comments

Rahul Goel Over a year ago

please tick it correct after five minutes. It will help me gain reputation

user3709993 Over a year ago

I'll do that in a few minutes

FelixHJ · Accepted Answer · 2014-06-05 07:16:35Z

1

if($result=mysql_query($query))

will always return true, since it is just an assignment

answered Jun 5, 2014 at 7:16

FelixHJ

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2 Comments

GreyRoofPigeon Over a year ago

Your answer tickled me. I guess you are right, but if I want to check if the function is called right, could I use if( ( $result=mysql_query($query) ) === true )? Would that work?

FelixHJ Over a year ago

No you would have to test against $result (the assingment will always return true AFAIK) but you would have to test if $result is false or something, $result will contain the result from the database query, or false on error (i.e. testing for true will generally not work). See php.net/mysql_query

damien hawks · Accepted Answer · 2014-06-05 07:17:31Z

0

You have forgotten the '' marks that inclose the id and also to check whether your query executed or not and whether you have any results use this

 row = mysql_fetch_row($result);
 if(!(is_empty($row)){
        $query2=mysql_query("DELETE FROM incidenten WHERE incidentID='$id'");
    }

answered Jun 5, 2014 at 7:17

damien hawks

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Comments

Jan · Accepted Answer · 2014-06-05 07:34:15Z

First, the lines

$result=mysql_query($query);
if($result=mysql_query($query)){
    $query2=mysql_query("DELETE FROM incidenten WHERE incidentID=$id");
}

will execute the $query twice. So, you should omit the first line, since if($result=mysql_query($query)) is enough.

Apart from that, checking $result on true or false will just tell you if an error occured or not. What you should do is to check if the INSERT statement affected any rows by using mysql_affected_rows:

if($result=mysql_query($query)){
    if(mysql_affected_rows($connect_mysql) > 0) {
        if($result2=mysql_query("DELETE FROM incidenten WHERE incidentID='".mysql_real_escape_string($id)."'") {
            /* The query did not return errors */
        }
        else { /* add error handling here */ }
    }
}
else { /* add error handling here */ }

Please note:

I renamed $query2 to $result2, because the variable does not contain a query, but a query result
Your code is very unsecure because of two reasons:
- You put a $_GET parameter into a query without escaping it. That makes SQL Injection as easy as possible! I added the mysql_real_escape_string function as an easy way to most likely avoid SQL injection. You should add this function in the INSERT statement as well.
- You use the mysql_* functions. These are deprecated and should not be used anymore. See here. Use mysqli functions or PDO istead.

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