return json from php and handle - php

Question

I get this from ajax call
{"id":"381120774951940096","time":"Posted 0 minutes and 51 seconds ago.",....
How can I add each of these into variable, id, time etc? data.id won't work, it says undef.
<script>
$(function() {
$.ajax({
type: "POST",
url: "start.php",
cache: false,
success: function(data) {
console.log(data.name);
console.log(data);
}
})
});
</script>
This is what I return from start.php
$return = array('id' => $info['id_str'] ,'time' => timeDiff(new DateTime($info['created_at'])), 'name' => $info['user']['name'], 'text' => $info['text'], 'image' => $picture, 'url' => "http://twitter.com/{$info['user']['screen_name']}");
print_r(json_encode($return));
EDIT
I had that print_r inside foreach loop and that was the issue. So I added another array and used echo json_decode($array,true) at the end of file.Writing this just in case, might help someone.
Cheers

Answer 1

First, in your PHP, change your line to:
echo json_encode($return, true);
That second parameter ensures that the JSON is interpretable as an associative array. Also, you should use echo instead of print_r when returning JSON; print_r can alter the formatting.
Next, use the following AJAX call:
$.ajax({
type: "POST",
url: "start.php",
dataType: "json", // Add this option.
cache: false,
success: function(data) {
console.log(data.name);
console.log(data);
}
})
The dataType: "json" option ensures that data will be parsed as a JSON object as soon as it's retrieved. So, data.id should be immediately available in your success callback function.
Hope this helps!

Answer 2

Firstly, you should use echo in the server script. print_r is mainly for debugging arrays.
Secondly, you should declare a dataType option for the ajax call:
$.ajax({
dataType: "json", // <-- here
type: "POST",
url: "start.php",
cache: false,
success: function(data) {
console.log(data.name);
console.log(data);
}
});
The way you have it now, I think you are getting a string response as data.
You can verify that with console.log(JSON.stringify(data));

Answer 3

You have to set dataType: 'json' in your ajax. This means that jQuery will parse the result as JSON for you.
Then parse the data,
data = $.parseJSON(data);
Then read var id = data.id;
Also, in your PHP, No need to use print_r(). Just use echo instead print_r().like:
echo json_encode($return);

Answer 4

You'll need to parse the JSON string into an object. You can use JSON.parse to do this.
// Your JSON string
var json_str = "{'id': 'foo', 'time': 'bar'}";
// We need to parse it, converting it into a JS object
var json_obj = JSON.parse(json_str);
// We can now interact with it
console.log(json_obj.id);
console.log(json_obj.time);
Alternatively, you can parse JSON using a built in jQuery function parseJSON().
jQuery also has a built in function for fetching JSON called getJSON(). If memory serves, this is just a shorthand for doing an .ajax() call and specifying a datatype of json. This will handle the above (parsing JSON) for you, and is the recommended solution.