Iterate over json array using jquery - php

There have been some post with my similar problem: How do I iterate over a JSON array using jQuery/AJAX call from PHP? but not quite the same.
I'm getting and error from jquery:
a is null
It is because of the code I've added to loop through the json data:
$(function ()
{
$.ajax({
url: 'ajax_dashboard/api.php', //the script to call to get data
data: "",
dataType: 'json',
success: function(data)
{
$.each(data, function() {
$.each(this, function(k, v) {
$('#output').append("<b>key: </b>"+k+"<b> value: </b>"+v)
.append("<hr />");
});
});
}
});
});
And here is the php file (which I did verify gives valid JSON format):
$query_camera_name = "SELECT camera_name, camera_status, camera_quality, email_notice, camera_hash, camera_type FROM #__cameras WHERE user_id=".$user->id." AND camera_status!='DELETED'";
$db->setQuery($query_camera_name);
//get number of cameras so we can build the table accordingly
$db->query();
$num_rows = $db->getNumRows();
// We can use array names with loadAssocList.
$result_cameras = $db->loadAssocList();
echo json_encode($result_cameras);
?>
This returns this json formatted data:
[
{
"camera_name": "ffgg",
"camera_status": "DISABLED",
"camera_quality": "MEDIUM",
"email_notice": "DISABLED",
"camera_hash": "0d5a57cb75608202e64b834efd6a4667a71f6dee",
"camera_type": "WEBCAM"
},
{
"camera_name": "test",
"camera_status": "ENABLED",
"camera_quality": "HIGH",
"email_notice": "ENABLED",
"camera_hash": "6ab000ef7926b4a182f0f864a0d443fc19a29fdd",
"camera_type": "WEBCAM"
}
]
If I remove the loops the "a is null" error is gone. What am I doing wrong?

Your iteration code works just fine: http://jsfiddle.net/SuyMj/
The error is elsewhere.
Edit:
Try this to help debug.
success: function(data, textStatus, xhr) {
console.log(xhr);
...
}
xhr will contain a lot of information about the request being made. What does the responseText contain? What is the statusText?

Your code works fine:
http://jsfiddle.net/QSvNy/
So the error is not there.

I don't see that you set the Content-Type of your response from php. Possibly the mime type of your response is incorrect and so jQuery does not parse the response as json.
Try this in your php before you echo your json:
header('Content-Type: application/json');

Related

my ajax call is getting correct response but isn't doing anything

I'm trying to make a like/dislike button in ajax. The ajax is sending my data to a separate file where it is saved in a database and that file sends back the successful response {"status":"success","message":"Like has been saved.","data":{"like":"1"}} that I got from the chrome network response window. However the code in $ajax(...).done isn't working
I have console.logged and var.dumped every bit of code i possibly could. my data IS being sent to my database which should mean that the SQL and the like class is correct. I've also tried simply console.log 'ging the response "res" and putting the rest in comments, but that again gives me nothing
<div>
Like
Dislike
<span class='likes' data-id="<?php echo $post->id ?>"><?php echo $post->getLikes(); ?></span> people like this
</div>
$("a.like, a.dislike").on("click",function(e){
var postId = $(this).data("id");
if($("a.like")){
var type = 1;
}else if($("a.dislike")){
var type = 0;
}
var elLikes = $(this).siblings(".likes");
var likes=elLikes.html();
$.ajax({
method: "POST",
url: "ajax/postlike.php",
data: {postId: postId, type:type},
dataType: "json",
})
.done(function( res ) {
console.log(res);
if(res.status=="succes"){
console.log(res);
if(res.data.like=="1"){
likes++;
elLikes=html(likes);
$("a.like").css("display","none");
$("a.dislike").css("display","inline-block");
} else if(res.data.like=="0"){
likes--;
elLikes=html(likes);
$("a.dislike").css("display","none");
$("a.like").css("display","inline-block");
}
}
});
e.preventDefault();
});
if(!empty($_POST)){
try {
$postId=$_POST['postId'];
$type=htmlspecialchars($_POST['type']);
$userId=$_SESSION['user_id'];
$l = new Like();
$l->setPostId($postId);
$l->setUserId($userId);
$l->setType($type);
$l->save();
$res = [
"status" => "success",
"message" => "Like has been saved.",
"data" =>[
"like" => $type
]
];
}catch (trowable $t) {
$res = [
'status' => 'failed',
'message' => $t->getMessage()
];
}
echo json_encode($res);
var_dump($res);
}
what I expected to happen was that Ajax sent the JSON data to the php code, that put it in a database, which works. Then gives a successful response to the Ajax, also works. The Ajax would then switch out the like/dislike buttons whilst adding or taking 1 like from the span "likes". It however does absolutely nothing
I'm almost 100% certain that the problem is something stupid that I'm overlooking, but i really can't find it.
Typo in 'success' in on line: if(res.status=="succes"){
you can try with this:
error: function(xhr, status, error) {
console.log(error)
},
success: function(response) {
console.log(response)
}
in your Ajax function, to know what happen in the server side with the response.
If you specify a return data type for the ajax request to expect, and the actual returned value isn't what you specified, then your error/fail function will be triggered if you have one. This is because adding dataType: "json" causes you're ajax try and parse your return value as json and when it fails, it triggers your error handler. It's best to omit the dataTaype and then add a try catch with JSON.parse in your done function, to get around this.
E.G
.done(function (string_res) {
console.log(string_res);
try {
var json_obj = JSON.parse(string_res);
console.log(json_obj);
} catch (e) {
console.log('failed to parse');
}
// do work/operations with json_obj not string_res
})
.fail(function (jqXHR, textStatus) {
console.log('failed')
});

how can I access json data from another url inside a html page

here is my php code which would return json datatype
$sql="SELECT * FROM POST";
$result = mysqli_query($conn, $sql);
$sJSON = rJSONData($sql,$result);
echo $sJSON;
function rJSONData($sql,$result){
$sJSON=array();
while ($row = mysqli_fetch_assoc($result))
{
$sRow["id"]=$row["ID"];
$sRow["fn"]=$row["posts"];
$sRow["ln"]=$row["UsrNM"];
$strJSON[] = $sRow;
}
echo json_encode($strJSON);
}
this code would return
[{"id":"1","fn":"hi there","ln":"karan7303"},
{"id":"2","fn":"Shshhsev","ln":"karan7303"},
{"id":"3","fn":"karan is awesome","ln":"karan7303"},
{"id":"4","fn":"1","ln":"karan7303"},
{"id":"5","fn":"asdasdas","ln":"karan7303"}]
But how can I access this data in html, that is, I want particular data at particular position for example i want to show 'fn' in my div and 'ln' in another div with another id
Before trying anything else I tried this
$.ajaxSetup({
url: 'exm1.php',
type: 'get',
dataType: 'json',
success: function(data){
console.log(data);
}
});
but it shows that data is undefined I don't know what I am doing wrong
What you've got should kind-of work if you swapped $.ajaxSetup (which is a global configuration method) with $.ajax. There are some significant improvements you could make though.
For example, your PHP does some odd things around the value returned by rJSONData. Here's some fixes
function rJSONData($result) {
$sJSON = array();
while ($row = mysqli_fetch_assoc($result)) {
$sJSON[] = array(
'id' => $row['ID'],
'fn' => $row['posts'],
'ln' => $row['UsrNM']
);
}
return json_encode($sJSON);
}
and when you call it
header('Content-type: application/json');
echo rJSONData($result);
exit;
Also make sure you have not output any other data via echo / print or HTML, eg <html>, etc
In your JavaScript, you can simplify your code greatly by using
$.getJSON('exm1.php', function(data) {
console.info(data);
}).fail(function(jqXHR, textStatus, errorThrown) {
console.error(jqXHR, textStatus, errorThrown);
});
Use $.ajax instead of $.ajaxSetup function.
Here is a detailed answer from another SO post how to keep running php part of index.php automatically?
<script>
$.ajax({
// name of file to call
url: 'fetch_latlon.php',
// method used to call server-side code, this could be GET or POST
type: 'GET'
// Optional - parameters to pass to server-side code
data: {
key1: 'value1',
key2: 'value2',
key3: 'value3'
},
// return type of response from server-side code
dataType: "json"
// executes when AJAX call succeeds
success: function(data) {
// fetch lat/lon
var lat = data.lat;
var lon = data.lon;
// show lat/lon in HTML
$('#lat').text(lat);
$('#lon').text(lon);
},
// executes when AJAX call fails
error: function() {
// TODO: do error handling here
console.log('An error has occurred while fetching lat/lon.');
}
});
</script>

parsing and looping through json data

I have been working on this code for some time now and have researched on various solutions on how to solve it, but i have found none.
I am getting the following json data:
{"id":"4","name":"bugatti","details":"the bugatti v8 simplly the best","seats":"2","status":"0","image"
:"","category":"sedean"}{"id":"5","name":"bently","details":"the bently the best of the best","seats"
:"5","status":"0","image":"","category":"SUV"}
I'm having trouble parsing that json data which I get from the jquery ajax $.post request as shown below:
$.post('response.php',
{
task : 'fleet'
},
function(data)
{
data = $.parseJSON(data);
$.each(data , function (key,value)
{
console.log(value);
});
console.log(data.id);
}
).error(
function()
{
}
).success(
);
Here is the response.php data:
if(isset($_POST['task']) && $_POST['task'] == 'fleet')
{
$car = new Cars();
$fleet = $car->getAll();
foreach( $car->data() as $cars )
{
echo json_encode($cars);
}
}
What could be the problem?
I would do something like this
You can also add dataType: 'json' to your $.post function.
// making sure we pass the header for the json output
header('Content-Type: application/json');
// if $car->data() (is the array, just encode it to json)
echo json_encode($car->data());
I ran some examples with your code (without the ajax)
var data = '{"id":"4","name":"bugatti","details":"the bugatti v8 simplly the best"}';
data = $.parseJSON(data);
$.each(data , function (key,value)
{
console.log(value);
});
// key will output id, name, details..
// value will output (what is below)
output :
4
bugatti
the bugatti v8 simplly the best
(remove value.id and just say value)
If you wanted to confirm the param, you can easily do
if (key == 'id')

JQuery Post And Get

I have a method I'm trying to write that can post data to a php file and get the results and return the output in a variable. For some reason, my block of code is not working.
function post_get(){
var result = null;
$.post("modules/data.php", { "func": "getNameAndTime" },
function(data){
result = JSON.parse(data);
}, "json");
return result;
}
I get this error when using this method
SyntaxError: JSON Parse error: Unexpected identifier "undefined"
Ajax is Asynchronous.
Is your PHP returning valid JSON?
This is how your code should be written to take advantage of the asynchronous nature of ajax.
function post_get(){
return $.post("modules/data.php", { "func": "getNameAndTime" }, "json");
}
post_get().done(function(data){
// do stuff with data
console.log(data);
}).fail(function(){
console.log(arguments);
alert("FAIL.\nCheck the console.");
});
// Do not attempt to bring data from inside the above function to out here.
If your server returns proper JSON encoded output and sets the correct headers (Content-Type: application/json), you can use data immediately:
$.post("modules/data.php", {
"func": "getNameAndTime"
},
function(data){
console.log(data);
}, "json");
// btw, at this point in the code you won't have access to the return value
In fact, even if it didn't return proper data, the console.log(data) should provide you with enough information to figure out why it wasn't working in the first place.

JSON.parse: unexpected character

I'm trying to pass a json from php jquery, after getting into an array of sql query and get the following javascript error.
JSON.parse: unexpected character
The function to return result of sql:
public function selectassocSql($sql){
$i = 0;
$resSelect = array();
mysql_query("SET NAMES 'utf8'");
$result = mysql_query($sql);
while ( $row = mysql_fetch_assoc($result) )
{
$resSelect[$i] = $row;
$i++;
}
mysql_free_result($result);
return $resSelect;
}
After use this function in this way,
$sql = "SELECT id, code, name FROM table WHERE code LIKE '%$codcli%' ";
$v = $data->selectassocSql($sql);
echo json_encode($v, JSON_FORCE_OBJECT);
And the javascript code is this:
$('#formclientes').submit(function(e){
e.preventDefault();
$.ajax({
type: 'POST',
url:$(this).attr('action'),
data:$(this).serialize(),
success:function(data)
{
//console.log("SUCCESS " + data);
var json_cli = $.parseJSON(data);
}
})
})
How I can correct this error and how I can read a json from jquery?
You don't need the $.parseJSON call as jQuery automatically does it because if you don't specify a dataType property jQuery tries to guess it and calls the correct function to parse the response before the data is handled to the success function
$.ajax({
type: 'POST',
url:$(this).attr('action'),
data:$(this).serialize(),
success:function(data)
{
//console.log("SUCCESS " + data);
var json_cli = data;
}
})
check out also this question Why is 'jQuery.parseJSON' not necessary?
I just ran into this in FF10.0.2 with data that looked like:
[ { "firstName": 'Joe', "lastName": 'Smith' } ]
(with multiple objects in the array - shortened for clarity)
It actually parsed OK using eval, though, instead of JSON.parse. (I'm not using jQuery here.)
The problem went away when I changed ' to " for the values:
[ { "firstName": "Joe", "lastName": "Smith" } ]
I thought the " requirement was only for property names, not data values.

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