Send only one input value with form submit in php

Question

I have a condition where there are 5 different menu items and in a row and one form.

For each of the menu item there is a input value in the form, which looks like this:

<form action='' method='post'>
    <?php 
           foreach($menu_items as $item){?>
           <div id='' class=''>
               <?php echo ('menu item description here');?>
               <input type='text' name='something' value='something'>
           </div>
       <?php 
       } 
    ?>
   <button type='submit'>send</button>
</form>

Now i need only to send only one input value with form submit that has the menu class 'active'.

Is there a simple approach to this?

You need to process your value before submit using JQuery event handler $("form").submit(function(event) { /* ... */ });. — Anwar
– Anwar, Commented Jan 19, 2016 at 17:11
id's must be unique. Inserting a blank id for each item is essentially setting them all to the same id. you're not giving anything the 'active' class. — Jay Blanchard
– Jay Blanchard, Commented Jan 19, 2016 at 17:14
@JayBlanchard yes the ids are set unique, was just giving the overview of the problem. — Armed
– Armed, Commented Jan 19, 2016 at 17:21

user5811191 · Accepted Answer · 2016-01-19 17:17:18Z

1

This is a simpler way of doing it:

<form action='' method='post'>
   <select id="example" name="example">
     <option value="something">Option 1</option>
     <option value="something">Option 2</option>
     <option value="something">Option 3</option>
     <option value="something">Option 4</option>
     <option value="something">Option 5</option>
   </select>

   <button type='submit'>send</button>
</form>

now when you send the form using PHP it will only send the value of the selected option. <select> can be followed by as many <option> blocks as you want.

answered Jan 19, 2016 at 17:17

user5811191

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2 Comments

Armed Over a year ago

but i need the user to enter the input as text

user5811191 Over a year ago

You could create a single <input> where the user inputs the text and then selects for what option it is?

Anwar · Accepted Answer · 2016-01-19 17:19:25Z

You need to listen to the event thrown by your form, and process your form before sending it.

You can follow this example to perform this task :

$(function() {
    $("form").submit(function(event) {
        event.preventDefault(); // prevent sending the form before we processed the form

        var active_input        =   $(".active")[0];        
        var active_input_value  =   active_input.val();

        $.ajax({
                url     :   "php_file_to_call.php"
            ,   method  :   "post"
            ,   data    :   { something : active_input_value }
        });
    });
});

The Ajax call (which could have been a $.post()) is necessary since you cannot submit the actual form because it does not prevent un-active element to be sent.

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