How to pass multiple variables between jQuery and PHP files using JSON?

Question

How can I pass multiple variables between PHP files using jquery with json? Ex:

$.post("example.php", {asdf:asdf, sdfg:sdfg}, function(data){

        $('section').html(data);
});

But instead of just using the .html function I want to retrieve multiple variables from the PHP file and then use them for .html on my page. I can't just use data because that only outputs whatever PHP returns so to my understanding php can only return one thing in a form of an echo(), but how can I make it retrieve more than just that one variable? I think I have to use JSON but I have never used JSON before so I would appreciate it if someone could help me out. Thanks.

Community · Accepted Answer · 2017-05-23 12:11:57Z

1

You can return a JSON object from php ( Returning JSON from PHP to JavaScript? ). And then in your jquery you can acees it like this,

$.post("example.php", {asdf:asdf, sdfg:sdfg}, function(data){

 $('section1').html(data.Variable1);
 $('section2').html(data.Variable2);

});

edited May 23, 2017 at 12:11

CommunityBot

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answered Nov 21, 2011 at 3:21

Jayantha Lal Sirisena

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3 Comments

Dany Khalife Over a year ago

i was gonna say embed them in an object but you raced me :D +1

randomphp Over a year ago

Alright after I tried using this example I had my php file like this by the end: echo json_encode(array('page' => 'Working!', 'options' => 'yes', 'session' => 'active')); and my jquery was $('section1').html(data.page); $('section2').html(data.options); but that didn't work, any ideas on why it didn't work?

randomphp Over a year ago

Oh never mind, I got it working it was just a little problem in another part of my jquery, thank you very much!

Syntax Error · Accepted Answer · 2011-11-21 03:23:22Z

0

You could have the php script output javascript code creating your variables and then just eval() it.

answered Nov 21, 2011 at 3:23

Syntax Error

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3 Comments

Christopher Over a year ago

You could - however, eval() comes with significant performance hits, and generally can be avoided altogether. Some background on "eval is evil" can be found at jslint.com/lint.html. Particularly in the case of this specific question: there are well-understood and widely-accepted ways to do what's needed without eval().

Syntax Error Over a year ago

I was worried this was a little stinky, js is not my forte. Is this bad enough that it should not even used as a "quick and dirty" solution? Should I remove?

Christopher Over a year ago

No; for quick-and-dirty, for environments where you control everything and can guarantee (or at least assume you can guarantee :) ) the cleanliness of the data, it can be okay. While not necessarily the One True Answer to this question, it's not always wrong. Putting aside security concerns, eval() always carries with it a hit, and instead of using programming constructs you fire up a whole new interpreter and effectively make debugging much more difficult.

Christopher · Accepted Answer · 2011-11-21 03:23:28Z

0

See the jQuery post() documentation here. Check out the dataType parameter, which tells jQuery what kind of data to expect back from the server.

See the json_encode() and json_decode() PHP functions here. Use json_encode() to prepare your populated data structure for echo()'ing back to the client.

answered Nov 21, 2011 at 3:23

Christopher

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1 Comment

randomphp Over a year ago

Thanks very much! Those documentations really helped out as well!

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