[SOLVED] Problem with serialize function

[Suchy]

 

 

....
<label><input type="checkbox" name="type[]" value="13">People</label><br>
<label><input type="checkbox" name="type[]" value="14">Erotica</label><br>
<label><input type="checkbox" name="type[]" value="16">Nature</label><br>
....

 

<?php
.....
$type1 = ($_POST['type']);
if ($type1 == 14) /
$type = serialize(14); // field should have: a:1:{i:0;s:2:"14";}
else
$type = serialize($type1); // field should have a:3:{i:0;s:2:"01";i:1;s:2:"02";i:2;s:2:"07";} or other
.....
?>

The problem is that if I select erotica (type 14) other types are also selected and stored in the db, how can I fix this?

 

 

 

[cooldude832]

issue in your if statement

<?php
$type1 = ($_POST['type']);
if ($type1 == 14){
   $type = serialize(14); // field should have: a:1:{i:0;s:2:"14";
}
else{
    $type = serialize($type1); // field should have a:3:{i:0;s:2:"01";i:1;s:2:"02";i:2;s:2:"07";}
}
?>

 

[teng84]

i think your condition always falls under your else because $type1 is an array and obviously array is not equal to any numbers or string...

i think you should not use if since your using array

 

just use foreach instead then serialize it one by one since this is checbox not radio button you're expecting the form to be selected on the same time

[Suchy]

did something else that works for me:

 

....
<label><input type="checkbox" name="type[]" value="13">People</label><br>
<label><input type="checkbox" name="type[]" value="x">Erotica</label><br>
<label><input type="checkbox" name="type[]" value="14">Nature</label><br>
....

 

if (isset($_POST['type']))	
  $type1 = join(',',($_POST['type']) );

if (strstr ($type1, "x"))
  $type= "x";
else
  $type= $type1;