jquery, post multiple variables

Question

I am brand new to jquery and trying to modify a basic script from php academy.

the jquery script is:

<script type="text/javascript">
function get() {
    $.post('getpeopleinjobs.php', {
        postvarfname: form.firstname.value,
        postvarlname: form.d = surname.value
    }, function(output) {
        $('#age').html(output).show();
    });
}
</script>

and my form code is:

    <form name="form">
    <input type="text" id="firstname">
    <input type="text" id="surname">
    <input type="button" value="get" onclick="get();">
    </form>

This worked perfectly when I was passing just one variable and my code snippet was:

$.post('getpeopleinjobs.php', {
    postvarfname: form.firstname.value
}

I then tried to add a second variable with

$.post('getpeopleinjobs.php', {
    postvarfname: form.firstname.value,
    postvarlname: form.d = surname.value
}

It does not work now in Internet Explorer but interestingly enough it does work in Chrome?

Any advice for the beginner?=

Are you sure you mean to have that d= sitting in the second value? So, it should read $.post ('getpeopleinjobs.php',{postvarfname: form.firstname.value,postvarlname: form.surname.value} — Jim O'Brien
– Jim O'Brien, Commented Jul 11, 2012 at 16:26

Josh Mein · Accepted Answer · 2012-07-11 16:26:57Z

8

You have the concept down correctly but there appears to be a minor mistake

//Posted Code
$.post ('getpeopleinjobs.php',{postvarfname: form.firstname.value,postvarlname: form.d=surname.value}
// Fixed Code
$.post ('getpeopleinjobs.php',{postvarfname: form.firstname.value, postvarlname: form.surname.value}

answered Jul 11, 2012 at 16:26

Josh Mein

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Comments

Andrew Whitaker · Accepted Answer · 2012-07-11 16:26:29Z

5

You have a typo in your post code:

$.post('getpeopleinjobs.php',{
    postvarfname: form.firstname.value,
    postvarlname: form.surname.value
}, function () { /* ... */ });

Should work fine.

answered Jul 11, 2012 at 16:26

Andrew Whitaker

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Comments

John Kalberer · Accepted Answer · 2012-07-11 16:26:46Z

5

It looks like you have an issue in your script. Remove the = sign:

$.post ('getpeopleinjobs.php',{postvarfname: form.firstname.value,postvarlname: form.surname.value});

You may also simply want to write the script as:

$.post ('getpeopleinjobs.php',$(form).serialize());

This way if you add additional fields to the form you don't have to modify the script.

answered Jul 11, 2012 at 16:26

John Kalberer

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2 Comments

Smudger Over a year ago

Hi John, can you elaborate on the serialze function? will this send all form data to the next page? Thanks

John Kalberer Over a year ago

Yeah, the serialize function serializes all children input elements into the url-encoded text string api.jquery.com/serialize

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