Send a form using jQuery and check if it was submitted

Question

I have 2 forms on my page and 1 of them has 2 different ways to be submited, a submit button and a jQuery on click event, it looks something like this:

<script>
    $(document).ready(function() {
        $('#img_send_form').click(function() {
            $('#form2').submit();
        });
    });
</script>

<form name="form1" action="#" method="post">
    <input type="text" name="field1"/>
    <input type="submit" name="send1"/>
</form>
<form name="form2" action="#" method="post">
    <input type="text" name="field1"/>
    <input type="text" name="field2"/>
    <input type="text" name="field3"/>
    <input type="text" name="field4"/>
    <input type="text" name="field5"/>
    <input type="text" name="field6"/>
    <input type="text" name="field7"/>
    <input type="submit" name="send2"/>
</form>
<img src="xxx" id="img_send_form"/>

What is the best way to check if form2 was submmitted on php? Do I need to use isset for every form field ?

if (isset($_POST['field1'])||isset($_POST['field2'])||isset($_POST['field3'])||isset($_POST['field4'])||isset($_POST['field5'])||isset($_POST['field6'])||isset($_POST['field7']))

or is there another "better" way to do it?

add hidden field in both form and check hidden field value on server — Girish
– Girish, Commented Feb 25, 2014 at 8:45

Sagar Mahajan · Accepted Answer · 2014-02-25 08:57:31Z

2

Take Hidden Field with Same Name in Both Forms (but differ Ids if you need)

Then you will only need to check the that hidden field

answered Feb 25, 2014 at 8:57

Sagar Mahajan

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Comments

Anas Bouhtouch · Accepted Answer · 2014-02-25 08:43:42Z

1

just add a hidden field to the second form, and in PHP check if it's set, in this case was used the second form

answered Feb 25, 2014 at 8:43

Anas Bouhtouch

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Comments

Rohan Kumar · Accepted Answer · 2014-07-16 13:07:30Z

1

Not necessary to take hidden fields,

PHP :

if(isset['send2'])) { echo "Form2 submitted !" ;?> }

edited Jul 16, 2014 at 13:07

Rohan Kumar

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answered Feb 25, 2014 at 9:32

Supriya Pansare

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1 Comment

gbestard Over a year ago

it doesn't work to me when I send it using the image

Clément Andraud · Accepted Answer · 2014-02-25 08:49:06Z

<script>
    $(document).ready(function() {
        $('#img_send_form').click(function() {
            $('#form2').submit();
        });
    });
</script>

<form name="form1" action="#" method="post">
    <input type="text" name="field1"/>
    <input type="submit" name="send1"/>
</form>
<form name="form2" action="#" method="post">
    <input type="hidden" name="form2_send"/>
    <input type="text" name="field1"/>
    <input type="text" name="field2"/>
    <input type="text" name="field3"/>
    <input type="text" name="field4"/>
    <input type="text" name="field5"/>
    <input type="text" name="field6"/>
    <input type="text" name="field7"/>
    <input type="submit" name="send2"/>
</form>
<img src="xxx" id="img_send_form"/>

And php :

if(isset['form2_send'])) { echo "Form2 submitted !" ;?> }

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