returns I have a php script which is called by ajax and i do some operation on mysql in this file and now i want to return the result stored in php array to ajax call. So how it can be possible.
6 Answers
You should use it.
echo json_encode($array);
Comments
Try this:
Use json for that. Read this for more details.
JSON is used especially for this purpose.
--
Thanks
4 Comments
Jite
This answer only tells him how to convert data into
json (even though its a good idea to use json in a case like this) it don't really answer the question.Jite
Link in answer is about the php function
json_encode which, like i said, encodes data to json, it don't tell the OP how to access the data in the ajax callbacks or anything, just shows him how to formate the data into another type of object (which he might as well have done already for all we know). :PJite
I'm sorry, but i see nowhere in the question anything about how to properly format the data, he wants to return the array after script run.
Anand Solanki
If you want to pass PHP Array to an AJAX call convert it as JSON and pass it.
The below is an example
<?php
$arr = array('a' => 10, 'b' => 23, 'c' => 32, 'd' => 42, 'e' => 52);
echo json_encode($arr);
?>
2 Comments
Joel Youngberg
This is a great way to setup an object in JavaScript as the php associative array can then be parsed with JSON.parse:
Joel Youngberg
Example in a JQuery AJAX success:
success: function(arr) { var alphaNum = JSON.parse(arr); var a = alphaNum.a; //10 var b = alphaNum.b; //23 }Anything you print in the php script will be 'returned' to the ajax call and caught in the success callback.
$.ajax({
type: "POST",
url: url,
data: data,
success: function(response){
alert(response); // Will print whatever the php script 'echos' out.
// or with the php code below as the script you call, you can use:
if(response.result) {
alert(response.data); // For the data in 'data' to be printed.
// Where response.data is your array:
for(var i=0;i<response.data.length;i++){
console.log(response.data[i]);
}
} else {
console.log(response.error);
}
},
});
Edit: As other states in the other answers, json is perfect to return from a php script to an ajax callback, cause JavaScript and PHP both handles this very well.
<?php
// I like to return an associative array,
// which in the javascript part of the code can be used as a object.
// I always include a 'result' boolean value which will let me know on the js
// side if the request was successfully done or not, and if false,
// i usually include a 'error' property with a error message.
$myreturndata = array("result" => true, "data" => $yourArray, "error" => null);
header('Content-Type: application/json');
die(json_encode($myreturndata));
Comments
In your ajax call you can specify the return datatype like this
$.ajax({
url : url,
data : data,
dataType : 'json',
success : function(data){
console.log(data);
}
})
In you php script you can print the json encoded string like this
print_r(json_enocde(array()));
Comments
A useful example on top of the other answers. Splitting each array value into javascript variables.
Improved with useful advice :)
PHP
echo json_encode($array);
jQuery.ajax in your success function
// pass each value to variables (this example array looks like this: [100,200,300])
var variableOne = response[0]; // This variable will = 100
var variableTwo = response[1]; // This variable will = 200
var variableThree = response[2]; // This variable will = 300
// Now you can easily use each individual value.
alert(variableOne);
4 Comments
Jite
The result would look like:
[100,200,300] and there is no reason to parse/stringify it, if its JSON that is returned, javascript can use it like a standard javascript object/array: alert(response[0]) <- will be 100. But its always best to validate the response of course.
misterManSam
Hangon,
alert(response[0]) gives me: [Jite
First thing in the callback? Does your php script return the data as application/json mime type?
misterManSam
Come to think of it... no!
alert(response[0]) works fine with it. I will update my answer, it was off the top of my head anyway.