PHP updating database

Question

I'm having problems with the following code to update a database. There are no errors, and I know it's not the database because I have another 'GET' script which uses the same &con information and returns data from the database just fine. However, the database just never updates.

 <?php
    define("DB_DSN","xxx");
    define("DB_HOST","xxx");
    define("DB_USER","xxx");
    define("DB_PASS","xxx");

    $con = mysql_connect(DB_HOST, DB_USER, DB_PASS) or die('Could not connect: ' .mysql_error());
    mysql_select_db(DB_DSN) or die('Could not select database');    
    $id = base64_decode($_POST["id"]);
    $deaths = base64_decode($_POST["deaths"]);
    $sql = "UPDATE Level01 SET Deaths =' . $id .' WHERE DeathID= ' . $deaths . '" ;     
    $uresult = mysql_query($sql,$con);
    if(! $uresult )
    {
      die('Could not update data: ' . mysql_error());
    }
    echo "Updated data successfully\n";
    mysql_close($con);
    exit;

Can anyone help??

Are you sure your $_POST variables have in tem what you expect? Are you sure they're Base64 encoded? — user1864610
– user1864610, Commented Sep 13, 2013 at 9:33
change your UPDATE statement to "UPDATE Level01 SET Deaths ='" . $id . "' WHERE DeathID= '" . $deaths . "'"; — wns
– wns, Commented Sep 13, 2013 at 9:34

anaconda · Accepted Answer · 2013-09-13 09:33:07Z

2

This

"UPDATE Level01 SET Deaths =' . $id .' WHERE DeathID= ' . $deaths . '" ;

should be

"UPDATE Level01 SET Deaths ='" . $deaths . "' WHERE DeathID= " . $id . " ;

I think.

answered Sep 13, 2013 at 9:33

anaconda

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Comments

user3200248 · Accepted Answer · 2014-01-16 12:23:10Z

1

Basicly insert this code without using variables:

mysql_query("UPDATE `Level1` SET `Deaths`= '".$deaths."' WHERE DeathID= ".$id.");

answered Jan 16, 2014 at 12:23

user3200248

1 Comment

SLaks Over a year ago

You have a SQL injection vulnerability.

Ben Fortune · Accepted Answer · 2013-09-13 09:33:01Z

0

$sql = "UPDATE Level01 SET Deaths =' . $id .' WHERE DeathID= ' . $deaths . '" ;

Should be

$sql = "UPDATE Level01 SET Deaths ='" . $deaths . "' WHERE DeathID= '" . $id . "'";

answered Sep 13, 2013 at 9:33

Ben Fortune

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1 Comment

Ben Fortune Over a year ago

Considering he's using double quotes and trying to concatenate using single quotes, I'd say it does.

Wikunia · Accepted Answer · 2013-09-13 09:35:34Z

0

Try to echo your variables at first and you don't neeed '.$id.' (actually it should be '".$id."') you can write Deaths = '$id' or Deaths = '$deaths' should be the right one.

answered Sep 13, 2013 at 9:35

Wikunia

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Comments

Charaf JRA · Accepted Answer · 2013-09-13 09:39:32Z

0

I suggest to do echo $sql ; and then you will know your error :)

Points are here interpreted as caracter not as concatenation operator,so update will update every Death that has id ='.value.' ,it not exists in database and then nothing happens.

Replace this :

$sql = "UPDATE Level01 SET Deaths =' . $id .' WHERE DeathID= ' . $deaths . '" ;

By this :

$sql = "UPDATE Level01 SET Deaths ='$id' WHERE DeathID= '$deaths'" ;

edited Sep 13, 2013 at 9:39

answered Sep 13, 2013 at 9:33

Charaf JRA

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1 Comment

Charaf JRA Over a year ago

@user2775839 please validate this answer so that it can be seen in the top and helps others

rohitr · Accepted Answer · 2013-09-13 09:40:45Z

0

Hi Please remove the dots from query as shown below it will solve problem.

$sql = "UPDATE Level01 SET Deaths ='$id' WHERE DeathID= '$deaths'" ;

answered Sep 13, 2013 at 9:40

rohitr

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