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up vote 0 down vote favorite

So I want to send a PHP array to jQuery but it's not working and I don't know why. Can someone help me out?

Here is the PHP:

$var1 = "hello";
$var2 = "world";
$values = array('value1'=>$var1,'value2'=>$var2);
echo (json_encode($values));

And the jQuery:

$.get("test.php", function(values){
var jsonValues = $.parseJSON(values);
$("showValue1").html(jsonValues.value1);
});
share|improve this question
    
Are you sure that you don't get any error from your php script or javascript? – marian0 Jun 9 '15 at 16:57
    
why are you not using .getJSON() it eliminates your parsing step. api.jquery.com/jquery.getjson – cmorrissey Jun 9 '15 at 17:00
    
Maybe use the dev-tools of your browser to inspect the requests and what's actually going on? – Jörn Hees Jun 9 '15 at 17:06
    
I have tried all suggestions but no luck. With some of the suggestions I am getting SyntaxError: JSON.parse: unexpected character at line 1 column 1 of the JSON data – user4958875 Jun 9 '15 at 19:19
    
Please look at my answer and remove what success method has inside and just write console.log(data, typeof data); and post what you get here. Also in php update echo json_encode($values); exit(); - remove brackets for echo and add exit – stefanz Jun 9 '15 at 22:02
up vote 1 down vote

I tested your php code and it checks out. I think you actually have some syntax errors with your JQuery code.

$("showValue1").html(jsonValues.value1);

most likely should have been one of

$("#showValue1")[0].html(jsonValues.value1);
$(".showValue1")[0].html(jsonValues.value1);

JQuery uses # and . to indicate an element's id and it's class repectively. the [0] is because it returns an array of every element that matches on the page and I assume you only want to match the first one, maybe not though. Check out this overview of jquery selectors: http://www.w3schools.com/jquery/jquery_ref_selectors.asp

share|improve this answer
    
Still not working when using # or . And [0] is not making any difference either. – user4958875 Jun 9 '15 at 19:21
    
Can you edit your post with these logs? And an output of console.log(values) – nchinda2 Jun 9 '15 at 19:31
up vote 1 down vote

I think you didn't include the index of the json object. Try to used this :

$.get("test.php", function(values){
var jsonValues = $.parseJSON(values);
$("showValue1").html(jsonValues[0].value1);
});

Update this answer as of 06/10/2015 09:42AM GMT+8.

If this code also has an error, return the code to this jsonValues.value1. I think this will be the solution. You haven't an error parsing the JSON. I think your error is on this part $("showValue1"). JQuery doesn't know what element is that. You can try this if your container is an ID : $("#showValue1") and else if it is class $(".showValue1")

For example your element is:

<div id="showValue1"></div>

Then your jquery is :

$.get("test.php", function(values){
    var jsonValues = $.parseJSON(values);
    $("#showValue1").html(jsonValues.value1);
});
share|improve this answer
    
He didn't use a json array, just a dictionary. The php array gets covered into a dictionary. At least that's what happened when I ran his code. – nchinda2 Jun 9 '15 at 17:22
    
Still not working but am getting SyntaxError: JSON.parse: unexpected character at line 1 column 1 of the JSON data – user4958875 Jun 9 '15 at 19:23
    
Please see my updates. I hope it will help now. – Aaron Hernandez Jun 10 '15 at 1:58
up vote 0 down vote

You also can try doing stuff like : 

$.ajax({
    url : 'test.php',
    method : 'GET',
    dataType : 'json',
    success : function( data ){
        var data = JSON.parse( data );
        $('#showValue1').html( data.value1 );
    }
});

Get more documentation here, it should not be a hard task. Just make sure your path to test.php is the right one and also is a good practice to exit() your php file

share|improve this answer
up vote 0 down vote accepted

I figured it out on my own! It's probably not the best way (I am just a beginner in this stuff) but it works.

So I was trying to put the data from my database into an array. The code in my original post was just a sample version because I wanted to try the code first.

This is what I came up with. The PHP:

$array = array();

while($row = mysql_fetch_assoc($rs)){     //$rs is mySQL query
    $array[] = $row;
}

echo $array[0]['post']. '&';       //post is the column name from my database
echo $array[1]['post']. '&';
echo $array[2]['post'];

With Ajax I did this (thanks to stefanz!):

 $.ajax({
    method:    'GET',
    url:       'test.php',
    cache:     false,
    success:   function(data){
       var tmp = data.split("&");
       $('#showValue1').html(tmp[0]);
       $('#showValue2').html(tmp[1]);
       $('#showValue3').html(tmp[2]);
    }
});

Thanks everybody for helping me out! It lead me to the code I am using now.

share|improve this answer
    
Ok, glad that all of us could help you, but what you did there is a totally begginer way ( don't say it s wrong since it works ). It s up to you if you want to go with this solution or you want to improve your skills. – stefanz Jun 10 '15 at 17:51
    
I definitely want to improve my skills, so any suggestions on how to do this thing the right way are welcome. But it worked for now and I had to fix this quickly but it's always good to know the right way ofcourse. And maybe it helps someone else out with the same problem. – user4958875 Jun 12 '15 at 8:01

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