send radiobutton's value to different fields at DB

Question

i have a submit form that consist of 1 group radio button.

<div id="defectclass">
        <input id="def1" type="radio" class="defect" name="defect" value="1"/>S
        <input id="def2" type="radio" class="defect" name="defect" value="1" />A
        <input id="def3" type="radio" class="defect" name="defect" value="1" />B
        <input id="def4" type="radio" class="defect" name="defect" value="1" />C
</div>

beside that i have 4 fields at DB table, that is :

1. S
2. A
3. B
4. C

i want after submit:

    - if def1 are checked send value to field "S"
     - if def2 are checked send value to field "A"
     - if def3 are checked send value to field "B"
     - if def4 are checked send value to field "C"
     - if all not checked or null send to all fields value="0"

how do i do that bcoz i've never try this?

Answer 1

By looking at the possible values of defect , in your table definition,

change defect VARCHAR(1) to defect VARCHAR(1) DEFAULT "0"

This simplifies your code to insert row/update columns

in your php script, after parsing the form, add a switch statement to decide which column to update.

// I assumed the row already exists

mysql_connect (localhost, $username, $password);
@mysql_select_db ($database) or die("Error");

$defect = mysql_real_escape_string($_POST["defect"]); //parse the defect field
$id = mysql_real_escape_string$_POST["id"]); //primary key of the table
$flag = True;

if(!empty($_POST['defect']))
{

   switch($defect)
   {
       case "S" : $var = "S";
                  break;
       case "A" : $var = "A";
                  break;
       case "B" : $var = "B";
                  break;
       case "C" : $var = "C";
                  break;
       default : $flag = False;
   }

   if($flag)
   {
      $query = "UPDATE TABLE_NAME SET" + $var + " = '1' WHERE Id = '$id'";
      mysql_query($query);
   }

   mysql_close(); 
}

Answer 2

<div id="defectclass">
        <input id="def1" type="radio" class="defect" name="defect['s']" value="1"/>S
        <input id="def2" type="radio" class="defect" name="defect['a']" value="1" />A
        <input id="def3" type="radio" class="defect" name="defect['b']" value="1" />B
        <input id="def4" type="radio" class="defect" name="defect['c']" value="1" />C
</div>

This will give you $_POST['defect']['s'], &tc. so you'll know which update path to take. See http://php.net/faq.html for more about this syntax.

Answer 3

1. You have to make just 1 field.
2. If you need numbers to represent these letters make an array.

like this

$defects_arr = array(0,"S","A","B","C");

this array can be used in many tasks:

• to represent these letters with numbers
• to validate user rinput
• to build this input dynamically

So, upon receiving your POST data just compare it against this array:

$defect = '0';
if(!empty($_POST['defect'])) {
  $key = array_search($_POST['defect'],$defect_arr));
}

and you will have a number in the $key variable. This number should be stored in the database.