I am using a jquery ajax get method to fetch information from the server however I am having trouble parsing the information so that I may use it. My website has a gallery of products that will filter its items based on category.
Here is the jQuery ajax function:
$('.category').click(function() {
var category;
if ($(this).hasClass('Shirts')) {
category = 'shirts';
}
if ($(this).hasClass('Hats')) {
category = 'hats';
}
if ($(this).hasClass('Acc')) {
category = 'acc';
}
$.ajax({
type: 'GET',
url: 'galleryfetch.php',
data: { 'category' : category },
dataType: 'json',
success: function(data) {
arr = $.parseJSON(data);
alert(arr);
}
});
});
This is the php script that the information is posted to:
<?php
if ($_SERVER['REQUEST_METHOD'] == 'GET') {
$category = $_GET['category'];
$conn = mysqli_connect('localhost', '*****', '*****', 'clothing');
$rows = mysqli_query($conn, "SELECT * FROM products WHERE category = '".$category."'");
while ($row = mysqli_fetch_array($rows)) {
$arr[] = $row;
}
echo json_encode(array('data' => $arr));
}
I using the alert in the success function to see if the information is passed succesfully but at the moment there is no alert and i get an:
Unexpected token o error.
I'm not sure if I'm parsing the information correctly or if Im not correctly using JSON
tl;dr: $.parseJSON(data); should be removed.
Your server is returning JSON (but claiming it is sending HTML, you should have header("Content-Type: application/json")).
You have told jQuery to ignore the claim that it is HTML and parse it as JSON. (This would be redundant if you fixed the above problem)
dataType: 'json',
The parsed data is passed to your success function.
You then pass that data to JSON.parse so it gets converted to a string (which will look something like [ [Object object], ... and is not valid JSON) and then errors.
Remove:
arr = $.parseJSON(data);
And just work with data.
Related
I want to send an array to PHP (POST method) using jQuery.
This is my code to send a POST request:
$.post("insert.php", {
// Arrays
customerID: customer,
actionID: action
})
This is my PHP code to read the POST data:
$variable = $_POST['customerID']; // I know this is vulnerable to SQLi
If I try to read the array passed with $.post, I only get the first element.
If I inspect the POST data with Fiddler, I see that the web server answers with "500 status code".
How can I get the complete array in PHP?
Thanks for your help.
To send data from JS to PHP you can use $.ajax :
1/ Use "POST" as type, dataType is what kind of data you want to receive as php response, the url is your php file and in data just send what you want.
JS:
var array = {
'customerID': customer,
'actionID' : action
};
$.ajax({
type: "POST",
dataType: "json",
url: "insert.php",
data:
{
"data" : array
},
success: function (response) {
// Do something if it works
},
error: function(x,e,t){
// Do something if it doesn't works
}
});
PHP:
<?php
$result['message'] = "";
$result['type'] = "";
$array = $_POST['data']; // your array
// Now you can use your array in php, for example : $array['customerID'] is equal to 'customer';
// now do what you want with your array and send back some JSON data in your JS if all is ok, for example :
$result['message'] = "All is ok !";
$result['type'] = "success";
echo json_encode($result);
Is it what you are looking for?
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>
When call php from jquery via ajax ,have any response .I changed the dataTypeand put jsoninstead html.I´m thinking the issue is that,for the ajax call never trigger the php code,it seems $_POST['retriveForm'] never carries a value.
PHP:
if(isset($_POST["retriveForm"])) {
$data_json =array();
$id = $_POST['retriveForm'];
$sql = "SELECT * FROM mytable WHERE Id = $id";
while ($row = mysqli_fetch_array($db->consulta($sql)) {
$data_json = array('item1' => $row['item1'],'item2' => $row['item2']) ;
}
$data_json['item_array'] = call_a_function_return_array();//this works
echo json_encode($data_json);
}
and jQuery :
$(document.body).on('click', '.edit', function() {
var id = $(this).data('id');
$.ajax({
type: "POST",
url: "is_the_same_page.php",
data: {
retriveForm: id
},
dataType: "json",
success: function(response) {
$('#myForm').find('input').eq(1).val(response.item1);
}
});
});
Code is all in the same page if that may be important.
Since the AJAX code is in the same script, make sure you don't output any of the normal HTML in this case. Your PHP code should be at the very beginning of the script, before any HTML is output. And you should exit the script after echoing the JSON. Otherwise your JSON will be mixed together with HTML, and jQuery won't be able to parse the response.
Also, you're not correctly adding to $data_json in your loop. You're overwriting the variable each time instead of pushing onto it.
<?php
// code here to set up database connection
if(isset($_POST["retriveForm"])) {
$data_json =array();
$id = $_POST['retriveForm'];
$sql = "SELECT * FROM mytable WHERE Id = $id";
while ($row = mysqli_fetch_array($db->consulta($sql)) {
$data_json[] = array('item1' => $row['item1'],'item2' => $row['item2']) ;
}
$data_json['item_array'] = call_a_function_return_array();//this works
echo json_encode($data_json);
exit();
}
?>
<html>
<head>
...
Then in the success function, you'll need to index the elements of response, since it's an array, not a single object.
I am trying to figure out how to retrieve data from a MySQL database using an AJAX call to a PHP page. I have been following this tutorial
http://www.ryancoughlin.com/2008/11/04/use-jquery-to-submit-form/
But i cant figure out how to get it to send back json data so that i can read it.
Right now I have something like this:
$('h1').click(function() {
$.ajax({
type:"POST",
url: "ajax.php",
data: "code="+ code,
datatype: "xml",
success: function() {
$(xml).find('site').each(function(){
//do something
});
});
});
My PHP i guess will be something like this
<?php
include ("../../inc/config.inc.php");
// CLIENT INFORMATION
$code = htmlspecialchars(trim($_POST['lname']));
$addClient = "select * from news where code=$code";
mysql_query($addClient) or die(mysql_error());
?>
This tutorial only shows how to insert data into a table but i need to read data. Can anyone point me in a good direction?
Thanks,
Craig
First of all I would highly recommend to use a JS object for the data variable in ajax requests. This will make your life a lot simpler when you will have a lot of data. For example:
$('h1').click(function() {
$.ajax({
type:"POST",
url: "ajax.php",
data: { "code": code },
datatype: "xml",
success: function() {
$(xml).find('site').each(function(){
//do something
});
});
});
As for getting information from the server, first you will have to make a PHP script to pull out the data from the db. If you are suppose to get a lot of information from the server, then in addition you might want to serialize your data in either XML or JSON (I would recomment JSON).
In your example, I will assume your db table is very small and simple. The available columns are id, code, and description. If you want to pull all the news descriptions for a specific code your PHP might look like this. (I haven't done any PHP in a while so syntax might be wrong)
// create data-structure to handle the db info
// this will also make your code more maintainable
// since OOP is, well just a good practice
class NewsDB {
private $id = null;
var $code = null;
var $description = null;
function setID($id) {
$this->id = $id;
}
function setCode($code) {
$this->code = $code;
}
function setDescription($desc) {
$this->description = $desc;
}
}
// now you want to get all the info from the db
$data_array = array(); // will store the array of the results
$data = null; // temporary var to store info to
// make sure to make this line MUCH more secure since this can allow SQL attacks
$code = htmlspecialchars(trim($_POST['lname']));
// query
$sql = "select * from news where code=$code";
$query = mysql_query(mysql_real_escape_string($sql)) or reportSQLerror($sql);
// get the data
while ($result = mysql_fetch_assoc($query)) {
$data = new NewsDB();
$data.setID($result['id']);
$data.setCode($result['code']);
$data.setDescription($result['description']);
// append data to the array
array_push($data_array, $data);
}
// at this point you got all the data into an array
// so you can return this to the client (in ajax request)
header('Content-type: application/json');
echo json_encode($data_array);
The sample output:
[
{ "code": 5, "description": "desc of 5" },
{ "code": 6, "description": "desc of 6" },
...
]
So at this stage you will have a PHP script which returns data in JSON. Also lets assume the url to this PHP script is foo.php.
Then you can simply get a response from the server by:
$('h1').click(function() {
$.ajax({
type:"POST",
url: "foo.php",
datatype: "json",
success: function(data, textStatus, xhr) {
data = JSON.parse(xhr.responseText);
// do something with data
for (var i = 0, len = data.length; i < len; i++) {
var code = data[i].code;
var desc = data[i].description;
// do something
}
});
});
That's all.
It's nothing different. Just do your stuff for fetching data in ajax.php as usually we do. and send response in your container on page.
like explained here :
http://openenergymonitor.org/emon/node/107
http://www.electrictoolbox.com/json-data-jquery-php-mysql/
I'm trying to populate a form with jquery's populate plugin, but using $.ajax
The idea is to retrieve data from my database according to the id in the links (ex of link: get_result_edit.php?id=34), reformulate it to json, return it to my page and fill up the form up with the populate plugin. But somehow i cannot get it to work. Any ideas:
here's the code:
$('a').click(function(){
$('#updatediv').hide('slow');
$.ajax({
type: "GET",
url: "get_result_edit.php",
success: function(data)
{
var $response=$(data);
$('#form1').populate($response);
}
});
$('#updatediv').fadeIn('slow');
return false;
whilst the php file states as follow:
<?php
$conn = new mysqli('localhost', 'XXXX', 'XXXXX', 'XXXXX');
#$query = 'Select * FROM news WHERE id ="'.$_GET['id'].'"';
$stmt = $conn->query($query) or die ($mysql->error());
if ($stmt)
{
$results = $stmt->fetch_object(); // get database data
$json = json_encode($results); // convert to JSON format
echo $json;
}
?>
Now first thing is that the mysql returns a null in this way: is there something wrong with he declaration of the sql statement in the $_GET part? Second is that even if i put a specific record to bring up, populate doesn't populate.
Update:
I changed the populate library with the one called "PHP jQuery helper functions" and the difference is that finally it says something. finally i get an error saying NO SUCH ELEMENT AS
i wen into the library to have a look and up comes the following function
function populateFormElement(form, name, value)
{
// check that the named element exists in the form
var name = name; // handle non-php naming
var element = form[name];
if(element == undefined)
{
debug('No such element as ' + name);
return false;
}
// debug options
if(options.debug)
{
_populate.elements.push(element);
}
}
Now looking at it one can see that it should print out also the name, but its not printing it out. so i'm guessing that retrieving the name form the json is not working correctly.
Link is at http://www.ocdmonline.org/michael/edit_news.php with username: Testing and pass:test123
Any ideas?
First you must set the dataType option for the .ajax call to json:
$.ajax({dataType: 'json', ...
and then in your success function the "data" parameter will already be a object so you just use it, no need to do anything with it (I don't know why you are converting it into a jQuery object in your code).
edit:
$( 'a' ).click ( function () {
$( '#updatediv' ).hide ( 'slow' );
$.ajax ( {
type: "GET",
url: "get_result_edit.php",
success: function ( data ) {
$( '#form1' ).populate ( data );
},
dataType: 'json'
} );
$( '#updatediv' ).fadeIn ( 'slow' );
return false;
}
also consider using $.getJSON instead of $.ajax so you don't have to bother with the dataType
Try imPagePopulate (another jquery plugin). It may be easier to use:
http://grasshopperpebbles.com/ajax/jquery-plugin-impagepopulate/