PHP checkbox with if statement issue

Question

i have this checkbox input in my form

<input name='new' type='checkbox' value='0' />

the column new in my database is TINYINT based on the value of the checkbox , i want to display a div with a class if the value is 1 so i implement this code

if(isset($_POST[$URL['new']]) 
   && isset($_POST[$URL['new']]) == 1)
   {
    echo '<div class="premiere">Premiere</div>';
    }

everything seems working fine , there's no error,warnings and notices.but the div didnot display in the page . what is wrong with the code and how i can fix it? thanks

the is the name i gave the array element like that i have it <?php foreach ($URLS as $URL).i can change it — Obed Lorisson
– Obed Lorisson, Commented Dec 24, 2012 at 6:23
@ObedLorisson if you are using foreach, $URL is the key. It is not an array. so you cannot call it as $_POST[$URL['new']] if you need more help let us know what you really going to do. — vinu
– vinu, Commented Dec 24, 2012 at 6:32
yeah true , it's the key, i change the syntax , still not displaying — Obed Lorisson
– Obed Lorisson, Commented Dec 24, 2012 at 6:36

Lock · Accepted Answer · 2012-12-24 06:21:10Z

2

if ( isset($_POST["new"]) && $_POST["new"] == 1 )
{
  echo '<div class="premiere">Premiere</div>';
}

answered Dec 24, 2012 at 6:21

Lock

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Comments

Rukmi Patel · Accepted Answer · 2012-12-24 06:18:22Z

0

i think, you have apply second condition to check value of "new" variable.

you should write condition like this

if(isset($_POST[$URL['new']]) && $_POST[$URL['new']] == 1){
}
else{
}

answered Dec 24, 2012 at 6:18

Rukmi Patel

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Comments

fmask · Accepted Answer · 2012-12-24 06:41:33Z

0

if ( isset($_POST["new"]) && $_POST["new"] == 1 )
{
  echo '<div class="premiere">Premiere</div>';
}

Make sure class "premiere" not hidden through css code.

answered Dec 24, 2012 at 6:41

fmask

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Comments

Mehul Gohil · Accepted Answer · 2012-12-24 07:11:00Z

0

You can use this query to display div on checking the checkbox and on unchecking checkbox you can hide the div displayed.

    $(document).ready(function(){
     $("#new").change(function(e){
     var addDiv = $("#new").val();
     if($("#new").attr('checked'))
      { 
       $("#form2").append("<div class="premiere">Premiere</div>");
      }
      else
      {
       $(".premiere").remove();
      }
     });
    });

HTML Code:

    <form method="post" id="form2">
    <input id="new" name='new' type='checkbox'/>
    </form>

Hoping that this post will be very much useful to you.

answered Dec 24, 2012 at 7:11

Mehul Gohil

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Comments

Obed Lorisson · Accepted Answer · 2012-12-24 15:59:42Z

0

The prob was , i was trying to access the key from an array in a foreach loops , so i just remove the $_POST from the key and reference like URL['new'] instead. and it works fine.

if(isset($URL['new']) 
   && isset($URL['new']) == 1)
      {
  echo '<div class="new">Premiere</div>';
   }

answered Dec 24, 2012 at 15:59

Obed Lorisson

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