CODEWITHSUNDEEP
php
Obed Lorisson
Obed Lorisson

Reputation: 449

PHP checkbox with if statement issue

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

Upvotes: 0

Views: 2174

Answers (5)

Obed Lorisson
Obed Lorisson

Reputation: 449

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>';
   }

Upvotes: 0

Mehul Gohil
Mehul Gohil

Reputation: 245

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.

Upvotes: 0

fmask
fmask

Reputation: 481

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

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

Upvotes: 0

Lock
Lock

Reputation: 5522

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

Upvotes: 2

Rukmi Patel
Rukmi Patel

Reputation: 2561

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{
}

Upvotes: 0

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