I am sending an ajax request with two post values, the first is "action" which defines what actions my php script has to parse, the other is "id" which is the id of the user it has to parse the script for.
The server returns 6 values inside an array() and is then encoded to JSON with the PHP function: json_encode();
Some of my responses are HTML, but when I encode it to JSON, it escapes "/" so it becomes "\/"
How do I disable that?
Also when I don't know how to display this in jQuery when I get the server response, I just thought that putting it all into a div would just display the numbers and HTML codes I had requested, but it displays the array as it is encoded in PHP.
PHP
$response = array();
$response[] = "<a href=''>link</a>";
$response[] = 1;
echo json_encode($response);
jQuery:
$.ajax({
type: "POST",
dataType: "json",
url: "main.php",
data: "action=loadall&id=" + id,
complete: function(data) {
$('#main').html(data.responseText);
}
});
How do I make this into working JSON?
You need to call the
$.parseJSON();
For example:
...
success: function(data){
var json = $.parseJSON(data); // create an object with the key of the array
alert(json.html); // where html is the key of array that you want, $response['html'] = "<a>something..</a>";
},
error: function(data){
var json = $.parseJSON(data);
alert(json.error);
} ...
see this in
http://api.jquery.com/jQuery.parseJSON/
if you still have the problem of slashes:
search for security.magicquotes.disabling.php
or:
function.stripslashes.php
Note:
This answer here is for those who try to use $.ajax with the dataType property set to json and even that got the wrong response type. Defining the header('Content-type: application/json'); in the server may correct the problem, but if you are returning text/html or any other type, the $.ajax method should convert it to json. I make a test with older versions of jQuery and only after version 1.4.4 the $.ajax force to convert any content-type to the dataType passed. So if you have this problem, try to update your jQuery version.
Firstly, it will help if you set the headers of your PHP to serve JSON:
header('Content-type: application/json');
Secondly, it will help to adjust your ajax call:
$.ajax({
url: "main.php",
type: "POST",
dataType: "json",
data: {"action": "loadall", "id": id},
success: function(data){
console.log(data);
},
error: function(error){
console.log("Error:");
console.log(error);
}
});
If successful, the response you receieve should be picked up as true JSON and an object should be logged to console.
NOTE: If you want to pick up pure html, you might want to consider using another method to JSON, but I personally recommend using JSON and rendering it into html using templates (such as Handlebars js).
Since you are creating a markup as a string you don't have convert it into json. Just send it as it is combining all the array elements using implode method. Try this.
PHP change
$response = array();
$response[] = "<a href=''>link</a>";
$response[] = 1;
echo implode("", $response);//<-----Combine array items into single string
JS (Change the dataType from json to html or just don't set it jQuery will figure it out)
$.ajax({
type: "POST",
dataType: "html",
url: "main.php",
data: "action=loadall&id=" + id,
success: function(response){
$('#main').html(response);
}
});
Connect your javascript clientside controller and php serverside controller using sending and receiving opcodes with binded data. So your php code can send as response functional delta for js recepient/listener
see https://github.com/ArtNazarov/LazyJs
Sorry for my bad English
Try this code. You don't require the parse function because your data type is JSON so it is return JSON object.
$.ajax({
url : base_url+"Login/submit",
type: "POST",
dataType: "json",
data : {
'username': username,
'password': password
},
success: function(data)
{
alert(data.status);
}
});
Related
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
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!
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));
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);
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.
I'm having trouble getting a response from my php jquery / json / ajax. I keep combining all these different tutorials together but I still can't seem to pull it all together since no one tutorial follow what I'm trying to do.
Right now I'm trying to pass two arrays (since there's no easy way to pass associative arrays) to my jquery ajax function and just alert it out. Here's my code:
PHP
$names = array('john doe', 'jane doe');
$ids = array('123', '223');
$data['names'] = $names;
$data['ids'] = $ids;
echo json_encode($data);
Jquery
function getList(){
$.ajax({
type: "GET",
url: 'test.php',
data: "",
complete: function(data){
var test = jQuery.parseJSON(data);
alert(test.names[0]);
alert("here");
}
},
"json");
}
getList();
In my html file all I'm really calling is my javascript file for debugging purposes. I know i'm returning an object but I'm getting an error with null values in my names section, and i'm not sure why. What am I missing?
My PHP file returns
{"names":["john doe","jane doe"],"ids":["123","223"]}
It seems to be just ending here
Uncaught TypeError: Cannot read property '0' of undefined
so my sub0 is killing me.
You could prob use the $.getJSON facade that jQuery provides, this will setup all the required ajax params for a standard JSON request:
$.getJSON('test.php', function(response) {
alert(response.names[0]); // john doe
});
However i think the route of the issue is that 1) your server may not be returning the correct response codes and/or the correct headers (ie: JSON data) - however the above method at least for the latter should force this conclusion.
See: http://api.jquery.com/jQuery.getJSON
It looks like the problem is that you're using the complete callback instead of the success callback:
function getList(){
$.ajax({
type: "GET",
url: 'test.php',
data: "",
success: function(data) { /* success callback */
var test = jQuery.parseJSON(data);
alert(test.names[0]);
alert("here");
}
},
"json");
}
getList();
From jQuery AJAX Docs:
success(data, textStatus, jqXHR)
A function to be called if the request succeeds. The function gets passed three arguments: The data returned from the server, formatted according to the dataType parameter; a string describing the status; and the jqXHR (in jQuery 1.4.x, XMLHttpRequest) object. As of jQuery 1.5, the success setting can accept an array of functions. Each function will be called in turn. This is an Ajax Event.
complete(jqXHR, textStatus)
A function to be called when the request finishes (after success and error callbacks are executed). The function gets passed two arguments: The jqXHR (in jQuery 1.4.x, XMLHTTPRequest) object and a string categorizing the status of the request ("success", "notmodified", "error", "timeout", "abort", or "parsererror"). As of jQuery 1.5, the complete setting can accept an array of functions. Each function will be called in turn. This is an Ajax Event.
jQuery wants to know what kind of data to expect as a response, otherwise it wont know how to parse it.
So, as has been said before here, you tell jQuery using the dataType = 'json' attribute.
function getList() {
$.ajax({
type: "GET",
url: 'test.php',
data: "",
dataType: "json",
success: function(data) {
console.log(data);
}
});
}
On top of this it is a good idea to have PHP present its content as json rather than html. You use the header for this by setting header('Content-type: application/json'); before any output in your PHP script. So:
$names = array('john doe', 'jane doe');
$ids = array('123', '223');
$data['names'] = $names;
$data['ids'] = $ids;
header('Content-type: application/json');
echo json_encode($data);
You should pass all parameters for ajax() function in single object. So, there should be "dataType" option. Also, if you set data type explicitly, jQuery will parse JSON data for you. Complete callback will receive parsed JavaScript object as parameter.
function getList() {
$.ajax({
type: "GET",
url: 'test.php',
data: "",
dataType: "json",
success: function(test) {
alert(test.names[0]);
alert("here");
}
});
}
By using Alert on the response paramater after a jquery success will display the values I need, and the problem is picking them out with k/v. I don't know if this is some incompatibily issue with json format or what, from php. Either nothing happens (no Alert) or the alert will say 'undefined' if I attempt to get values by using the keys to them.
Relevant code:
JQuery:
var curr = $(this).val();
// alert(curr);
$.ajax({
type: 'GET',
url: 'CurrencyResponse.php?q='+curr,
contentType: "application/json; charset=utf-8",
success: function(response) {
var items = response.d;
// alert(response); this will display some json key value from server
$.each(items, function (index, item) {
// alert(item.msg); or updating some div tag here, eventless
});
},
PHP:
<?php
// $query = parse_url($_SERVER['REQUEST_URI'], PHP_URL_QUERY);
// $query = $_GET["q"];
$response = array();
$response["msg"] = "Hjksa!";
$response["nok"] = "9.32";
echo json_encode($response);
?>
Help here will be much appreciated! =)
Use the dataType parameter for jQuery's AJAX, like so:
$.ajax({
type: 'GET',
url: 'CurrencyResponse.php?q='+curr,
dataType: 'json',
success: function( json) {
alert( json.msg); // Will output Hjksa!
}
});
This tells jQuery that the response from the server is JSON, and that it should return a JSON object to the callback functions that take the server response as their parameters.
Read more about dataType on jQuery's site.
Judging from the documentation, it'll read the MIME type from the response header, and use that, unless you explicitly specify it. So either set the headers on the PHP script to application/json or set the "dataType" param to "json".
I recommend sending an object to the client script. It makes it easier to read the data.
echo json_encode((object) $array);
I was working through a similar problem for a couple of days, I had set
dataType: "json",
but it was still coming back as a string. Specifying json in the header of my php file in addition to in the jquery request resolved the issue for me.
header("Content-type: application/json");
Hopefully that helps someone!
I am trying to send simple data to theservre, and I need a "rough and ready" way to do this.
This is what I have so far:
var emails = ['a#123.com', 'b#123.com', 'c#123.com'];
var ruff_json = "{ 'emails': [";
for (i in emails)
ruff_json += ((i == 0) ? '' : ', ') + '\''+emails[i]+'\'';
ruff_json += '] }';
jQuery.ajax({
type: 'POST',
url: '1.php',
data: ruff_json,
dataType: "json",
timeout: 2000,
success: function(result){
//do something
},
error: function (xhr, ajaxOptions, thrownError){
//do something
}
});
Using Firebug, I can see that the data is POSTed to the server - however, at the server, there is no data ($_POST is empty) - what am I doing wrong?
We post all of our data with json.
var myobj = { this: 'that' };
$.ajax({
url: "my.php",
data: JSON.stringify(myobj),
processData: false,
dataType: "json",
success:function(a) { },
error:function() {}
});
then in php we do
<?php
$json = json_decode(file_get_contents("php://input"), true);
// Access your $json['this']
// then when you are done
header("Content-type: application/json");
print json_encode(array(
"passed" => "back"
));
?>
This way we don't even mess with the post variables, and in general, its faster than having jQuery process them.
Your data field should contain an object with key-value pairs, because it gets encoded as POST key-values pairs.
data = {my_json: encoded_string};
Then on the PHP side you can access the data as:
$data = json_decode($_POST['my_json']);
PHP populates $_POST by parsing the data received. However, it only knows form-encoded data, JSON data cannot be parsed automatically. So $_POST will be useless in this case. You need to get the raw post data and parse it with json_decode.
Sorry if this is basic, but I have been dealing with figuring this out all day and have gotten to where I can do everything I need with Jquery and cakephp (not sure if cakephp matters in this or if its same as any PHP), I want to return a variable from a cakephp function to jquery, I had read about how to do it, like here:
the cakephp:
$test[ 'mytest'] = $test;
echo json_encode($test);
and the jquery:
$.ajax({
type: 'POST',
url: 'http://localhost/site1/utilities/ajax_component_call_handler',
data: {
component_function: component_function,
param_array: param_array
},
dataType: "json",
success: function(data) {
// how do i get back the JSON variables?
}
});
I just can't figure out how to get one or more variables back into usable form within jquery, I just want the variable so I can do whatever else with it, I've been looking at what I can find through searching but its not making it fully clear to me.. thanks for any advice.
The JSON variables are in the data variable. In your case, it'll look like this:
var data = {
myTest: "Whatever you wrote here"
};
... so you can read it from data.myTest.
(Not sure whether it's relevant but you can remove the http://localhost/ part from the URL;
AJAX does not allow cross-domain requests anyway.)
Your variables are in data.
$.ajax({
type: 'POST',
url: 'http://localhost/site1/utilities/ajax_component_call_handler',
data: {
component_function: component_function,
param_array: param_array
},
dataType: "json",
success: function(data) {
// how do i get back the JSON variables?
var values = eval( data ); //if you 100 % trust to your sources.
}
});
Basically data variable contain the json string. To parse it and convert it again to JSON, you have to do following:
$.ajax({
type: 'POST',
url: 'http://localhost/site1/utilities/ajax_component_call_handler',
data: {
component_function: component_function,
param_array: param_array
},
dataType: "json",
success: function(data) {
json = $.parseJSON(data);
alert(json.mytest);
}
I haven't test it but it should work this way.
Note that when you specify dataType "json" or use $.getJSON (instead of $.ajax) jQuery will apply $.parseJSON automatically.
So in the "success" callback you do not need to parse the response using parseJSON again:
success: function(data) {
alert(data.mytest);
}
In case of returning a JSON variable back to view files you can use javascript helper:
in your utilities controller:
function ajax_component_call_handler() {
$this->layout = 'ajax';
if( $this->RequestHandler->isAjax()) {
$foobar = array('Foo' => array('Bar'));
$this->set('data', $foobar);
}
}
and in your view/utilities/ajax_component_call_handler.ctp you can use:
if( isset($data) ) {
$javascript->object($data); //converts PHP var to JSON
}
So, when you reach the stage in your function:
success: function(data) {
console.log(data); //it will be a JSON object.
}
In this case you will variable type handling separated from controllers and view logic (what if you'll need something else then JSON)...