PHP unable to display a form inside a table

Question

I must be missing something incredibly simple, but I can't see it. I have a function that generates a simple form that should be displayed several times within a table. Whenever I try to call the function with in a table cell, the form is displayed above the table for some reason.

Here's a simplified version of the code:

Php function that generates a form to be displayed multiple times

private function addTaskForm()
{
   // display add task form
   echo "<form method='post' action='#'>
           <select name='task'>
             <option value='1'>Clean</option>
             <option value='1'>Shop</option>
           </select>
           <input type='submit' name='btnSubmit'>
         </form>";
}

Table

echo "<table>
        <tr>
          <td>".addTaskForm()."</td> 
          <td>".addTaskForm()."</td>
          <td>".addTaskForm()."</td> 
          <td>".addTaskForm()."</td>
        </tr>
      </table>";

score 2 · Accepted Answer · edited May 23 '17 at 11:48

2

you dont need to echo on function just return string and i am sure you know about public/private/protected function

codepad

Good Read

edited May 23 '17 at 11:48

Community

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answered Dec 21 '12 at 05:21

NullPoiиteя

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PHP unable to display a form inside a table

1 Answers1