Here is my ajax function in response I am getting result but I don't know how to set that response in my page. Is it possible with json_decode or I have to try something else
JSON file is
<?php
$group_id = $_POST['group_id'];
$query = "SELECT *,group_id FROM contact JOIN addressgroup ON addressgroup.contact_id = contact.contact_id WHERE group_id IN (".$group_id.") GROUP BY contact.contact_id";
$res = mysql_query($query);
$data = array();
$k=0;
while($row = mysql_fetch_array($res))
{
$data[$k][0] = $row['user_id'];
$data[$k][1] = $row['first_name'];
$data[$k][2] = $row['middle_name'];
$data[$k][3] = $row['last_name'];
$k++;
}
echo json_encode(array($data));
?>
AJAX function
var myarray;
function getcon() {
myarray = [];
myarray.push($(".group_id").val());
$.ajax({
dataType: "json",
type: "POST",
url: "getcon.php",
data: 'group_id=' + myarray.join(),
success: function(data) {
totalRecords=data.length;
zone.fnClearTable();
for(var i=0; i < (data.length); i++) {
zone.fnAddData([
data[k][0],
data[k][1],
data[k][2],
data[k][3],
]);
}
return false;
}
});
return false;
}
As your Response is an JSON object..
you have to use it like this:-
replace
echo json_encode(array($data));
by
echo json_encode($data); /* because $data is already an array*/
and keep it same as it was earlier
while($row = mysql_fetch_array($res))
{
$data[] = $row['user_id'];
$data[] = $row['first_name'];
$data[] = $row['middle_name'];
$data[] = $row['last_name'];
$k++;
}
<script>
var myarray;
function getcon() {
myarray = [];
myarray.push($(".group_id").val());
$.ajax({
dataType: "json",
type: "POST",
url: "getcon.php",
data: 'group_id=' + myarray.join(),
success: function(data) {
console.log(data)/* check the structure of data here*/
zone.fnClearTable();
zone.fnAddData([
data.user_id.,
data.first_name,
data.middle_name,
data.last_name,
]);
return false;
}
});
return false;
}
</script>
JSON is an object! Your are getting object, so in your ajax function You must use some loop function like for or $.each JQuery functions.
`success : function (data) {
$.each (data,function(key,val){
$("#someId).html(key + " - " + val)
}
}`
Related
$.ajax({
type: "GET",
url: 'http://localhost/abc/all-data.php',
data: {
data1: "1"},
success: function(response)
{
alert(response);
}
});
return false;
I want to display each element of array one by one in success function of the ajax currently i get all elements to gether
this is my php code
$i=0;
while($row = mysqli_fetch_assoc( $qry )){
$temp[$i]['c_n'] = $row['c_name'];
$temp[$i]['j_t'] = $row['Job_Title'];
$temp[$i]['des'] = $row['description'];
$temp[$i]['req'] = $row['requirments'];
$temp[$i]['dat'] = $row['posted'];
$i++;
}
$data = array('temp'=> $temp);
echo JSON_encode($temp);
I do appreciate your helps
you probably use something like this in your success function :
response.temp.forEach(function(element){
console.log(element.c_n) ;
console.log(element.j_t) ;
console.log(element.des) ;
console.log(element.req) ;
console.log(element.dat) ;
});
In your success function, you need to json parse your response
var data = JSON.parse(response);
You can access to your data:
data['temp']
If you want your response parsed to json automaticallym you can setup your ajax settings like this:
$.ajaxSetup ({
contentType: "application/json",
dataType: 'json'
});
Then you don't need to call JSON.parse anymore.
Your code:
$i=0; while($row = mysqli_fetch_assoc($qry)){
$temp[$i]['c_n'] = $row['c_name'];
$temp[$i]['j_t'] = $row['Job_Title'];
$temp[$i]['des'] = $row['description'];
$temp[$i]['req'] = $row['requirments'];
$temp[$i]['dat'] = $row['posted'];
$i++;
} $data = array('temp'=> $temp); echo JSON_encode($temp);
Please change the last line as
return JSON_encode($data);
Hope this helps you :)
I'm creating an ajax script to update a few fields in the database. I got it to a point where it worked but it sent the user to the php script instead of staying on the page so I did some googling, and people suggested using either return false; or e.preventDefault() however, if I do this, it breaks the php script on the other page and returns a fatal error. I might be missing something being newish to AJAX but it all looks right to me
JS:
$(document).ready(function() {
var form = $('form#edit_child_form'),
data = form.serializeArray();
data.push({'parent_id': $('input[name="parent_id"]').val()});
$('#submit_btn').on('click', function(e) {
e.preventDefault();
$.ajax({
url: form.prop('action'),
dataType: 'json',
type: 'post',
data: data,
success: function(data) {
if (data.success) {
window.opener.$.growlUI(data.msg);
}
},
error: function(data) {
if (!data.success) {
window.opener.$.growlUI(data.msg);
}
}
});
});
})
AJAX:
<?php
//mysql db vars here (removed on SO)
$descriptions = $_GET['descriptions'];
$child_id = $_GET['child_id'];
$parent_id = $_GET['parent_id'];
$get_child_ids = $dbi->query("SELECT child_ids FROM ids WHERE parent = ". $parent_id ." ORDER BY id"); //returns as object
$count = 0;
$res = array();
while ($child_row = $get_child_ids->fetch_row())
{
try
{
$dbi->query("UPDATE ids SET description = '$descriptions[$count]', child_id = '$child_id[$count]' WHERE parent_id = $child_row[0]");
$res['success'] = true;
$res['msg'] = 'Success! DDI(s) updated';
} catch (Exception $e) {
$res['success'] = true;
$res['msg'] = 'Error! '. $e->getMessage();
}
$count++;
}
echo json_encode($res);
it's probably something really small that I've just missed but not sure what - any ideas?
my solution:
I var_dumped $_GET and it returned null - changed to $_REQUEST and it got my data so all good :) thanks for suggestions
Try the following instead.
I moved the form data inside click and enclosed the mysql queries values in single quotes.
JS:
$(document).ready(function() {
var form = $('form#edit_child_form');
$('#submit_btn').on('click', function(e) {
e.preventDefault();
var data = form.serializeArray();
data.push({'parent_id': $('input[name="parent_id"]').val()});
$.ajax({
url: form.prop('action'),
dataType: 'json',
type: 'get',
data: data,
success: function(data) {
if (data.success) {
window.opener.$.growlUI(data.msg);
}
},
error: function(data) {
if (!data.success) {
window.opener.$.growlUI(data.msg);
}
}
});
});
})
AJAX:
<?php
//mysql db vars here (removed on SO)
$descriptions = $_GET['descriptions'];
$child_id = $_GET['child_id'];
$parent_id = $_GET['parent_id'];
$get_child_ids = $dbi->query("SELECT child_ids FROM ids WHERE parent = '". $parent_id ."' ORDER BY id"); //returns as object
$count = 0;
$res = array();
while ($child_row = $get_child_ids->fetch_row())
{
try
{
$dbi->query("UPDATE ids SET description = '$descriptions[$count]', child_id = '$child_id[$count]' WHERE parent_id = '$child_row[0]'");
$res['success'] = true;
$res['msg'] = 'Success! DDI(s) updated';
} catch (Exception $e) {
$res['success'] = true;
$res['msg'] = 'Error! '. $e->getMessage();
}
$count++;
}
echo json_encode($res);
You are using an AJAX POST request so in your PHP you should be using $_POST and not $_GET.
You can just change this:
$descriptions = $_GET['descriptions'];
$child_id = $_GET['child_id'];
$parent_id = $_GET['parent_id'];
to this:
$descriptions = $_POST['descriptions'];
$child_id = $_POST['child_id'];
$parent_id = $_POST['parent_id'];
I'm not sure how to pass the result of mysql query into html page via ajax JSON.
ajax2.php
$statement = $pdo - > prepare("SELECT * FROM posts WHERE subid IN (:key2) AND Poscode=:postcode2");
$statement - > execute(array(':key2' => $key2, ':postcode2' => $postcode));
// $row = $statement->fetchAll(PDO::FETCH_ASSOC);
while ($row = $statement - > fetch()) {
echo $row['Name']; //How to show this in the html page?
echo $row['PostUUID']; //How to show this in the html page?
$row2[] = $row;
}
echo json_encode($row2);
How to pass the above query result to display in the html page via ajax below?
my ajax
$("form").on("submit", function () {
var data = {
"action": "test"
};
data = $(this).serialize() + "&" + $.param(data);
$.ajax({
type: "POST",
dataType: "json",
url: "ajax2.php", //Relative or absolute path to response.php file
data: data,
success: function (data) {
//how to retrieve the php mysql result here?
console.log(data); // this shows nothing in console,I wonder why?
}
});
return false;
});
Your json encoding should be like that :
$json = array();
while( $row = $statement->fetch()) {
array_push($json, array($row['Name'], $row['PostUUID']));
}
header('Content-Type: application/json');
echo json_encode($json);
And in your javascript part, you don't have to do anything to get back your data, it is stored in data var from success function.
You can just display it and do whatever you want on your webpage with it
header('Content-Type: application/json');
$row2 = array();
$result = array();
$statement = $pdo->prepare("SELECT * FROM posts WHERE subid IN (:key2) AND Poscode=:postcode2");
$statement->execute(array(':key2' => $key2,':postcode2'=>$postcode));
// $row = $statement->fetchAll(PDO::FETCH_ASSOC);
while( $row = $statement->fetch())
{
echo $row['Name'];//How to show this in the html page?
echo $row['PostUUID'];//How to show this in the html page?
$row2[]=$row;
}
if(!empty($row2)){
$result['type'] = "success";
$result['data'] = $row2;
}else{
$result['type'] = "error";
$result['data'] = "No result found";
}
echo json_encode($row2);
and in your script:
$("form").on("submit",function() {
var data = {
"action": "test"
};
data = $(this).serialize() + "&" + $.param(data);
$.ajax({
type: "POST",
dataType: "json",
url: "ajax2.php", //Relative or absolute path to response.php file
data: data,
success: function(data) {
console.log(data);
if(data.type == "success"){
for(var i=0;i<data.data.length;i++){
//// and here you can get your values //
var db_data = data.data[i];
console.log("name -- >" +db_data.Name );
console.log("name -- >" +db_data.PostUUID);
}
}
if(data.type == "error"){
alert(data.data);
}
}
});
return false;
});
In ajax success function you can use JSON.parse (data) to display JSON data.
Here is an example :
Parse JSON in JavaScript?
you can save json encoded string into array and then pass it's value to javascript.
Refer below code.
<?php
// your PHP code
$jsonData = json_encode($row2); ?>
Your JavaScript code
var data = '<?php echo $jsonData; ?>';
Now data variable has all JSON data, now you can move ahead with your code, just remove below line
data = $(this).serialize() + "&" + $.param(data);
it's not needed as data variable is string.
And in your ajax2.php file you can get this through
json_decode($_REQUEST['data'])
I would just..
$rows = $statement->fetchAll(FETCH_ASSOC);
header("content-type: application/json");
echo json_encode($rows);
then at javascript side:
xhr.addEventListener("readystatechange",function(ev){
//...
var data=JSON.parse(xhr.responseText);
var span=null;
var i=0;
for(;i<data.length;++i){span=document.createElement("span");span.textContent=data[i]["name"];div.appendChild(span);/*...*/}
}
(Don't rely on web browsers parsing it for you in .response because of the application/json header, it differs between browsers... do it manually with responseText);
I'm new Jquery and AJAX and I've really been struggling with the syntax I've been trying to use other tutorials as reference but nothing seems to work. I feel I have the right idea but syntax is wrong somewhere please help.
Here is the Ajax side
var var_numdatacheck = <?php echo $datacheck; ?>;
var var_numcheck = parseInt(var_numdatacheck);
function activitycheck(){
$.ajax({
type: 'POST',
url: 'feedupdate.php',
data: {function: '3test', datacheck: var_numcheck},
dataType: "json",
success: function(data) {
var json = eval('(' + data + ')');
$('#datacheck').html(json['0']);
var var_numcheck = parseInt(msg);
//setTimeout('activitycheck()',1000)},
error:function(msg) {
console.log(msg);
}
});
}
$(document).ready(function() {
activitycheck();
});
Here is the php the AJAX calls
<?php
require "dbc.php";
$function = $_POST['function'];
$datacheck = $_POST['datacheck'];
$search="SELECT * FROM Feedtest ORDER BY id DESC";
$request = mysql_query($search);
$update= mysql_fetch_array($request);
$updateid = $update['id'];
$updatecheck = mysql_num_rows($request);
$data = array();
if ($function == $datacheck){
echo $updatecheck;
echo $datacheck;
}
if ($function == "3test" && $updatecheck > $datacheck ) {
$updatesearch="SELECT * FROM Feedtest WHERE id = '$updateid' ORDER BY id DESC";
$updatequery = mysql_query($updatesearch);
$data['id'] = $updateid;
while ($row = mysql_fetch_array($updatequery))
{
?>
<?php $data[]= $row['First Name']; ?>
<?php
}
echo json_encode($data);
}
?>
</div>
</ul>
first of all ,always use JSON.parse(data) instead of eval.It is considereda a good practice.
second thing is always try to debug your code by checking it in console or alerting.In your context,this is what is happening-:
$.ajax({
type: 'POST',
url: 'feedupdate.php',
data: {function: '3test', datacheck: var_numcheck},
dataType: "json",
success: function(data) {
var data = eval('(' + data + ')');
console.log("myData"+data)//debugging.check the pattern so that you can acces it the way you want!!!
for(var i=0;i< data.length;i++)
{
alldata += "<li>"+data[i][0]+"<li><hr>";
}
$('#datacheck').html(alldata);
});
}
For JSON.parse:
success: function(data) {
var data = JSON.parse(data);
console.log("myData"+data)//debugging.check the pattern so that you can acces it the way you want!!!
for(var i in data)
{
alldata += "<li>"+data[i].First Name+"<li><hr>";
}
$('#datacheck').html(alldata);
});
How can I pass data from a php of then rows back to ajax ?
PHP
$query = 'SELECT * FROM picture order by rand() LIMIT 10';
$result = mysql_query($query);
while ($rec = mysql_fetch_array($result, MYSQL_ASSOC)) {
$url[]=$rec['pic_location'];
$name[]=$rec['name'];
$age[]=$rec['age'];
$gender[]=$rec['gender'];
}
echo json_encode($url);
echo json_encode($name);
echo json_encode($age);
echo json_encode($gender);
Ajax
$(".goButton").click(function() {
var dir = $(this).attr("id");
var imId = $(".theImage").attr("id");
$.ajax({
url: "viewnew.php",
dataType: "json",
data: {
current_image: imId,
direction : dir
},
success: function(ret){
console.log(ret);
var arr = ret;
alert("first image url: " + arr[0][0] + ", second image url: " + arr[0][1]); // This code isnt working
alert("first image Name: " + arr[1][0] + ", second image name: " + arr[1][1]);
$(".theImage").attr("src", arr[0]);
if ('prev' == dir) {
imId ++;
} else {
imId --;
}
$("#theImage").attr("id", imId);
}
});
});
});
</script>
My question is how can I display the values here ? The Alert message is giving me "Undefined" ?
You can do something along these lines.
PHP
$query = 'SELECT * FROM picture order by rand() LIMIT 10';
$res = mysql_query($query);
$pictures = array();
while ($row = mysql_fetch_array($res)) {
$picture = array(
"pic_location" => $row['pic_location'],
"name" => $row['name'],
"age" => $row['age'],
"gender" => $row['gender']
);
$pictures[] = $picture;
}
echo json_encode($pictures);
JS
...
$.ajax({
...
dataType: "json",
...
success: function(pictures){
$.each(pictures, function(idx, picture){
// picture.pic_location
// picture.name
// picture.age
// picture.gender
});
}
});
...
You can't put multiple echo statements for the AJAX response:
echo json_encode($url);
echo json_encode($name);
echo json_encode($age);
echo json_encode($gender);
Join your arrays and send a single response:
$arr = $url + $name + $age + $gender;
echo json_encode($arr);
You can easily do this using a single Array:
$pics = array();
while ($rec = mysql_fetch_array($result, MYSQL_ASSOC)) {
$pics[$rec['id']]['url'] = $rec['pic_location'];
$pics[$rec['id']]['name']=$rec['name'];
$pics[$rec['id']]['age']=$rec['age'];
$pics[$rec['id']]['gender']=$rec['gender'];
}
echo json_encode($pics);