I am using an ajax call which is as follows
var ID=$(this).attr('id');
var input=$("#input_"+ID).val();
var dataString = {id: ID, value: input};
$("#span_"+ID).html(input);
if(input.length>0)
{
$.ajax({
type: "POST",
url: "/apps/worker_app.php",
data: dataString,
cache: false,
success: function(html)
{
$("#span_"+ID).html(span);
}
});
}
How can I get the data in my php function edit_ajax() which is inside worker_app
I use post but the array come to be empty
Thanks
Add dataType: 'json',
var ID=$(this).attr('id');
var input=$("#input_"+ID).val();
var dataString = {id: ID, value: input};
$("#span_"+ID).html(input);
if(input.length>0)
{
$.ajax({
type: "POST",
url: "/apps/worker_app.php",
data: dataString,
dataType: 'json',
cache: false,
success: function(msg)
{
$("#span_"+ID).html(span);
}
});
}
and in worker_app.php get id and value by
$id=$_POST['id'];
$value=$_POST['value'];
edit_ajax($id,$value);
function edit_ajax($id,$value){
$sql="update from ..........";
}
Is this what u want?
Have you tried setting the dataType in your ajax call, like this:
$.ajax({
type:"POST",
url: "/apps/worker_app.php",
data: dataString,
cache: false,
success:function(html) {
},
dataType:"json"
});
Also, you could use a tool like firebug to see so the data is passed correctly in your ajax request.
Here is an example:
file1.php:
<html>
<head>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.0/jquery.min.js" type="text/javscript"></script>
<script type="text/javascript">
$(document).ready(function() {
// catch submit event for your form
$('#form1').submit(function() {
// serialize data from form
var data = $(this).serialize();
$.ajax({
type="POST",
url:"file2.php",
data:data,
success:function(resp) {
// output response
$('#output').html(resp);
}
});
return false;
});
]);
</script>
</head>
<body>
<div id="output"></div>
<form method="POST" id="form1">
<input type="text" name="name" />
<input type="submit" value="send" />
</form>
</body>
</html>
file2.php:
<?php
if(isset($_POST['name'])) {
header('Content-type: text/html');
echo '<p>Response From Server - Your name is: '.$_POST['name'].'</p>';
}
?>
in /apps/worker_app.php get those posted values
$id=$_POST['id'];
$value=$_POST['value'];
and now use the values with your php function
function edit_ajax($id,$value)
{
echo $id;
echo $value;
}
echo $return=edit_ajax($id,$value);
now you can get the values in the current page as
success: function(html)
{
$("#span_"+ID).html(span);
}
> $.ajax({
> type: "POST",
> url: "/apps/worker_app.php",
> data: dataString,
> cache: false,
> success: function(msg)
> {
> $("#span_"+ID).html(msg);
> }
> });
on success, the argument function is receiving should be same used with
$("#span_"+ID).html(msg);
Related
I am trying to send the form data to PHP with AJAX but when dumping data on PHP page it is returning null values.
Actually, I want a program where I can upload an image than in pop up crop it and then save the cropped image in the database.
My code is given below:
$('#fileinput').on('change', function() {
var formD = new FormData();
var file = $('#fileinput')[0].files[0];
// var nfile = file.serializeArray();
// console.log(file);
formD.append('file', file);
formD.append("clientID", 2993);
console.log(formD);
$.ajax({
url: 'croped.php',
type: 'POST',
data: {
'ff': formD
},
processData: false,
contentType: true,
// dataType: 'json',
success: function(data) {
console.log(data);
}
});
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<form action="" method="POST" enctype="multipart/form-data">
<input type="file" name="image" id="fileinput" />
<input type="submit" name="upld" id="upldbtn" />
</form>
croped.php
<?php
var_dump($_POST);
echo "ll";
?>
Please let me know what I am missing or doing wrong.
Try this, it works for me :)
contentType: false,
data: formD,
For example
$.ajax({
url: "croped.php",
type: "POST",
data: formD,
processData: false,
contentType: false,
// dataType: 'json',
success: function (data) {
console.log(data);
},
});
For the php code
<?php
echo "<pre>";
print_r($_FILES);
<form action = "" method = "POST" enctype = "multipart/form-data"
id="formdata">
<input type = "file" name = "image" id="fileinput" />
<input type = "submit" name="upld" id="upldbtn" />
</form>
$("#formdata").submit(function(e) {
e.preventDefault();
var formData = new FormData(this);
$.ajax({
url: window.location.pathname,
type: 'POST',
data: formData,
success: function (data) {
alert(data)
},
cache: false,
contentType: false,
processData: false
});
});
for the PHP is just simple check:
All in the same page named **
page.php
if(isset($_POST['XYZ']){
echo "WORKING";
}
then for my HTML:
<h1 id='XYZ'>CLICK ME</h1>
now at the same page i'm trying to do the AJAX POST request like this
$('#XYZ').click(function(){
var XYZ = 'XYZ';
$.post('page.php',{
XYZ: XYZ
})
})
and the request didn't work, How do i just pass the $_POST data? I removed success function since i didn't think it is useful in this case.
What i want is when i click on the <h1> the echo appears.
try this:
var my_string = 'xyz';
$.ajax({
url: 'ajax.php',
type: 'post',
data: { "action":my_string},
dataType: "json",
success: function(response) {
if(response['state'] == 'ok'){
console.log("ok");
}
}
});
ajax.php
<?php echo $_POST['action']; ?>
<head>
<script>
var my_string = 'xyz';
$.ajax({
method: 'POST',
url: './giveposts',
dataType: "text",
contentType: "application/json; charset=utf-8",
data:my_string,
success: function(data) {
{
$("#test").html(data);
}
}
});
</head>
</script>
<body>
<div id="test>
/*echo
</div>
</body>
i want to populate two fields when some one fill up card no.
<input type="text" id="name" name="name" class="form-control" Value="<?php echo $grow['name']; ?>">
<input type="text" id="address" name="address" class="form-control" Value="<?php echo $grow['address']; ?>">
but this code populate field one by one. can any one suggest be better code for populating two fields from database. Thank you
jquery
<script type="text/javascript">
$(document).ready(function()
{
$("#krishi").keyup(function()
{
var k=$(this).val();
var q="name";
$.ajax
({
type: "POST",
url: "getresult.php",
data: 'k='+k+'&q='+q,
cache: false,
success: function(data)
{
if(data){
$("#name").val(data);
$.ajax
({
type: "POST",
url: "getresult.php",
data: 'k='+k+'&q=address',
cache: false,
success: function(data)
{
if(data){
$("#address").val(data);
}
}
});
}else
$("#name").val("");
$("#address").val("");
}
});
});
});
</script>
getresult.php
<?php
define('INCLUDE_CHECK',true);
include("mysql.php");
$k=$_POST['k'];
$q=$_POST['q'];
$sql=mysql_query("select * from inward where krishi='$k'");
$row=mysql_fetch_array($sql);
echo $row[$q];
?>
Try to extract both name and address from database and json them
$k=$_POST['k'];
$q=$_POST['q'];
$sql=mysql_query("select address,name from inward where krishi='$k'");
$row=mysql_fetch_array($sql);
$result = array(
'name'=>$row['name'],
'address'=>$row['address']);
echo json_encode($result);
After that parse them via jquery
$.ajax
({
type: "POST",
url: "getresult.php",
data: 'k='+k+'&q=address',
cache: false,
success: function(data)
{
if(data){
var parsedData = jQuery.parseJSON(data);
$("#name").val(parsedData.name);
$("#address").val(parsedData.address);
}
}
});
Javascript Code:
<script type="text/javascript">
$(document).ready(function()
{
$("#krishi").keyup(function()
{
var k = $(this).val();
var q = "name";
$.ajax({
type: 'POST',
url: "getresult.php",
data: 'k='+k,
cache: false,
success: function(data)
{
var jsonArr = $.parseJSON(data);
if(typeof response =='object')
{
$("#name").val(jsonArr.name);
$("#address").val(jsonArr.address);
}
else
{
$("#name").val("");
$("#address").val("");
}
}
});
});
});
</script>
PHP Code:
<?php
define('INCLUDE_CHECK',true);
include("mysql.php");
$k = $_POST['k'];
$sql = mysql_query("select * from inward where krishi='$k'");
$row = mysql_fetch_assoc($sql);
echo json_encode(array('name' => $row['name'], 'address' => $row['address']);
?>
<script type="text/javascript">
$(document).ready(function() {
$("a").click(function() {
var content = $('#content').html();
var data = {"content":content};
$.ajax({
type: "POST",
dataType: "json",
url: "ajax.php",
data: {content:content},
success function (data) {
alert('Hello!');
}
});
});
});
</script>
<div id="content"><?php echo $content; ?></div>
ajax.php
echo json_encode($_POST['content']); ?>
Nothing happens... WhatI really want to achieve is to get that alert box and get the return data, but I am lost since I don't get any errors or nothing.
You miss " : " after success
<script type="text/javascript">
$(document).ready(function() {
$("a").click(function() {
var content = $('#content').html();
var data = {"content":content};
$.ajax({
type: "POST",
dataType: "json",
url: "ajax.php",
data: {content:content},
success: function (data) {
alert('Hello!');
}
});
});
});
</script>
<div id="content"><?php echo $content; ?></div>
As #sofl said, if you change it to success:function (data) { it will work!
Just remember that the $("a") from $("a").click(function() { called when click in a link tag like <a href"">.
If you are using an input ou button with a class="a" you should change the code to $(".a").click(function() {
(just add a . before a)
PS: If you're using a link, you should set the href="" to href="#" to work.
I'm trying to use the POST method in jQuery to make a data request. So this is the code in the html page:
<form>
Title : <input type="text" size="40" name="title"/>
<input type="button" onclick="headingSearch(this.form)" value="Submit"/><br /><br />
</form>
<script type="text/javascript">
function headingSearch(f)
{
var title=f.title.value;
$.ajax({
type: "POST",
url: "edit.php",
data: {title:title} ,
success: function(data) {
$('.center').html(data);
}
});
}
</script>
And this is the php code on the server :
<?php
$title = $_POST['title'];
if($title != "")
{
echo $title;
}
?>
The POST request is not made at all and I have no idea why. The files are in the same folder in the wamp www folder so at least the url isn't wrong.
You need to use data: {title: title} to POST it correctly.
In the PHP code you need to echo the value instead of returning it.
Check whether title has any value or not. If not, then retrive the value using Id.
<form>
Title : <input type="text" id="title" size="40" name="title" value = ''/>
<input type="button" onclick="headingSearch(this.form)" value="Submit"/><br /><br />
</form>
<script type="text/javascript">
function headingSearch(f)
{
var title=jQuery('#title').val();
$.ajax({
type: "POST",
url: "edit.php",
data: {title:title} ,
success: function(data) {
$('.center').html(data);
}
});
}
</script>
Try this code.
In php code, use echo instead of return. Only then, javascript data will have its value.
try this
$(document).on("submit", "#form-data", function(e){
e.preventDefault()
$.ajax({
url: "edit.php",
method: "POST",
data: new FormData(this),
contentType: false,
processData: false,
success: function(data){
$('.center').html(data);
}
})
})
in the form the button needs to be type="submit"
Id advice you to use a bit simplier method -
$.post('edit.php', {title: $('input[name="title"]').val() }, function(resp){
alert(resp);
});
try this one, I just feels its syntax is simplier than the $.ajax's one...
function signIn()
{
var Username = document.getElementById("Username").value;
var Password = document.getElementById("Password").value;
$.ajax({
type: 'POST',
url: "auth_loginCode.jsp",
data: {Username: Username, Password: Password},
success: function (data) {
alert(data.trim());
window.location.reload();
}
});
}
contentType: 'application/x-www-form-urlencoded'