im fairly new to ajax.. but I have this function which takes and order id and query sql from it in a php page. I can set the returned data to a div but what id like to do is set the return from the function to the returned data in html or text form so i can email it later. also any better ways to do this code would be greatly appreciated
<script>
function getInvoice(id){
var request = "driverInvoice.php"+"?oID="+id;
$.ajax({
url: request,
success: function(data){
$("#theDiv").html(data);
}
});
}
</script>
Use data: {oID: id} to send with POST.
function getInvoice(id){
$.ajax({
async: false,
type: "POST", //Method POST, you can edit and use GET
url: 'driverInvoice.php', //URL
data: {oID:id}, //Variables
success: function(data){
$('#theDiv').html(data);
},
error: function(data) {
}
});
}
If I undestend, you want transform data in text, like a strip_tags php function?
...
success: function(data){
$("#theDiv").html(data);
var text = $("#theDiv").text()
}
Related
I am trying to get my search bar working, however I am having issues with an ajax post.
For whatever reason, none of the data is being appended to the URL. I have tried various things with no success. I am attempting to send the data to the same page (index.php).
Here is my jquery:
$(function(){
$(document).on({
click: function () {
var tables = document.getElementsByTagName("TABLE");
for (var i = tables.length-1; i >= 0; i-= 1) {
if (tables[i]) tables[i].parentNode.removeChild(tables[i]);
}
var text = $('#searchBar').val();
var postData = JSON.stringify({ searchTerm: text });
$.ajax({
type: 'POST',
url: 'index.php',
dataType: 'json',
data: postData,
success: function() {
alert("Test");
}
});
}
}, "#searchButton");
});
And here is the php which I have with index.php:
<?php
require('course.php');
if(isset($_POST['searchTerm'])) {
echo $_POST['searchTerm'];
}
?>
No matter what I try, I am unable to get anything to post. I have checked the network tab in chrome, and I'm not seeing anything that indicates it's working correctly.
Any ideas?
EDIT:
I've changed my code to this, and it seems I'm getting closer:
$(document).on({
click: function () {
$("TABLE").remove()
var text = $('#searchBar').val();
$.ajax({
type: 'GET',
url: 'index.php',
dataType: 'text',
data: { searchTerm: text },
success: function() {
alert("Test");
}
});
}
}, "#searchButton");
And:
<?php
require('course.php');
if(isset($_GET['searchTerm'])) {
echo $_GET['searchTerm'];
}
?>
Now I am getting ?searchTerm=theTextIEnter as expected, however it's still not being echoed in index.php.
Do not use JSON.stringify() to convert object to string. data passed to $.ajax must be an object and not JSON.
For whatever reason, none of the data is being appended to the URL.
You are making a POST request. POST data is sent in the request body, not in the query string component of the URL.
If you change it to a GET request (and inspect it in the correct place, i.e. the Network tab of your browser's developer tools and not the address bar for the browser) then you would see the data in the query string.
This will work for you use data: { postData } on place of data:postData and you will receive your data in $_POST['postData']
$(function(){
$(document).on({
click: function () {
var tables = document.getElementsByTagName("TABLE");
for (var i = tables.length-1; i >= 0; i-= 1) {
if (tables[i]) tables[i].parentNode.removeChild(tables[i]);
}
var text = $('#searchBar').val();
var postData = JSON.stringify({ 'searchTerm' : text });
$.ajax({
type: 'POST',
url: 'index.php',
dataType: 'json',
data: { postData },
success: function(data) {
alert(data.searchTerm);
}
});
}
}, "#searchButton");
});
In index.php
<?php
if(isset($_POST['postData'])) {
echo $_POST['postData'];
die;
}
?>
If you want to send data via the URL you have to use a GET request. To do this, change the type of the request to GET and give the object directly to the data parameter in your jQuery, and use $_GET instead of $_POST in your PHP.
Finally note that you're not returning JSON - you're returning text. If you want to return JSON, use json_encode and get the value in the parameter of the success handler function.
Try this:
$(document).on({
click: function () {
$('table').remove();
$.ajax({
type: 'GET',
url: 'index.php',
dataType: 'json',
data: { searchTerm: $('#searchBar').val() },
success: function(response) {
console.log(response.searchTerm);
}
});
}
}, "#searchButton");
<?php
require('course.php');
if(isset($_GET['searchTerm'])) {
echo json_encode(array('searchTerm' => $_GET['searchTerm']));
}
?>
Remove dataType: 'json', from your AJAX. That is all.
Your response type is not JSON, yet by setting dataType: 'json' you're implying that it is. So when no JSON is detected in the response, nothing gets sent back to the method handler. By removing dataType it allows the API to make an educated decision on what the response type is going to be, based on the data you're returning in the response. Otherwise, you can set dataType to 'text' or 'html' and it will work.
From the manual:
dataType: The type of data that you're expecting back from the server.
This is NOT the type of data you're sending/posting, it's what you're expecting back. And in your index.php file you're not sending back any JSON. This is why the success() method is not satisfying. Always set the dataType to the type of data you're expecting back in the response.
With POST Request:
Please comment below line in your code:
//var postData = JSON.stringify({ searchTerm: text });
And use below ajax code to get the post-data:
$.ajax({
type: 'POST',
url: 'index.php',
dataType: 'json',
data: { searchTerm: text },
success: function() {
alert("Test");
}
});
With GET Request:
You can convert your data to query string parameters and pass them along to the server that way.
$.ajax({
type: 'GET',
url: 'index.php?searchTerm='+text,
dataType: 'json',
success: function(response) {
alert(response);
}
});
In response, You can get the data with alert, so you may get idea about the same.
I want to pass values to a PHP script so i am using AJAX to pass those, and in the same function I am using another AJAX to retrieve those values.
The problem is that the second AJAX is not retrieving any value from the PHP file. Why is this? How can I store the variable passed on to the PHP script so that the second AJAX can retrieve it?
My code is as follows:
AJAX CODE:
$(document).ready(function() {
$("#raaagh").click(function(){
$.ajax({
url: 'ajax.php', //This is the current doc
type: "POST",
data: ({name: 145}),
success: function(data){
console.log(data);
}
});
$.ajax({
url:'ajax.php',
data:"",
dataType:'json',
success:function(data1){
var y1=data1;
console.log(data1);
}
});
});
});
PHP CODE:
<?php
$userAnswer = $_POST['name'];
echo json_encode($userAnswer);
?>
Use dataType:"json" for json data
$.ajax({
url: 'ajax.php', //This is the current doc
type: "POST",
dataType:'json', // add json datatype to get json
data: ({name: 145}),
success: function(data){
console.log(data);
}
});
Read Docs http://api.jquery.com/jQuery.ajax/
Also in PHP
<?php
$userAnswer = $_POST['name'];
$sql="SELECT * FROM <tablename> where color='".$userAnswer."'" ;
$result=mysql_query($sql);
$row=mysql_fetch_array($result);
// for first row only and suppose table having data
echo json_encode($row); // pass array in json_encode
?>
No need to use second ajax function, you can get it back on success inside a function, another issue here is you don't know when the first ajax call finished, then, even if you use SESSION you may not get it within second AJAX call.
SO, I recommend using one AJAX call and get the value with success.
example: in first ajax call
$.ajax({
url: 'ajax.php', //This is the current doc
type: "POST",
data: ({name: 145}),
success: function(data){
console.log(data);
alert(data);
//or if the data is JSON
var jdata = jQuery.parseJSON(data);
}
});
$(document).ready(function() {
$("#raaagh").click(function() {
$.ajax({
url: 'ajax.php', //This is the current doc
type: "POST",
data: ({name: 145}),
success: function(data) {
console.log(data);
$.ajax({
url:'ajax.php',
data: data,
dataType:'json',
success:function(data1) {
var y1=data1;
console.log(data1);
}
});
}
});
});
});
Use like this, first make a ajax call to get data, then your php function will return u the result which u wil get in data and pass that data to the new ajax call
In your PhP file there's going to be a variable called $_REQUEST and it contains an array with all the data send from Javascript to PhP using AJAX.
Try this: var_dump($_REQUEST); and check if you're receiving the values.
you have to pass values with the single quotes
$(document).ready(function() {
$("#raaagh").click(function(){
$.ajax({
url: 'ajax.php', //This is the current doc
type: "POST",
data: ({name: '145'}), //variables should be pass like this
success: function(data){
console.log(data);
}
});
$.ajax({
url:'ajax.php',
data:"",
dataType:'json',
success:function(data1){
var y1=data1;
console.log(data1);
}
});
});
});
try it it may work.......
Ok so I know long hand ajax but trying to use the jQuery short cut. I have two documents
form.php
submit.php
In my "form" page I am calling the "submit" page to process the insert. I am currently using the jquery ajax:
<script type="text/javascript">
jQuery('form').submit(function() {
string = jQuery("form").serializeArray();
jQuery.ajax({
type: "POST",
url: "submit.php",
data: string,
dataType: "json",
})
return false;
});
</script>
When I view firebug it is processing the ajax fine. I am getting 200 and post parameters are set. What I am trying to do is have the ajax return the submit.php file. I know it has something to do with the "success" function but I don't know how to add this. I tried a few things like:
<script type="text/javascript">
jQuery('form').submit(function() {
string = jQuery("form").serializeArray();
jQuery.ajax({
type: "POST",
url: "submit.php",
data: string,
dataType: "json",
success: function(html){
alert(html);
}
})
return false;
});
</script>
and
<script type="text/javascript">
jQuery('form').submit(function() {
string = jQuery("form").serializeArray();
jQuery.ajax({
type: "POST",
url: "submit.php",
data: string,
dataType: "json",
success: function(html){
$('.result').html(data);
}
})
return false;
});
</script>
but neither of these are working. Again I am simply trying to send the ajax request and then return the contents of the submit.php page. Not only does the submit.php page hold the script to process the php/ajax insert but it also display success statements like "insert was successful" so that is why I need to not only run the script in the page but also return the contents of that page. Thank you for any help.
Chagne the dataType:'json' to dataType:'html' for the callback that you wish to display the contents of submit.php.
You were close in your second attempt, but you made a typo. Try:
success: function (data) {
$('.result').html(data);
}
Also, unless your server is returning JSON, you probably want to change the dataType:
dataType: "html"
For some reason this jQuery function is not working properly. Here's my code... the response div is not updating with my response.
WHEN AJAX FUNCTION IS CALLED
if ($action == 'sort') {
echo 'getting a response';
return 0;
}
JQuery FUNCTION
function sort() {
$.ajax({
type: "POST",
url: "contributor_panel.php?action=sort",
data:"sort_by=" + document.getElementById("sort_by").value +
"&view_batch=" + document.getElementById("view_batch").value,
success: function(html){
$("#ajaxPhotographSortResponse").html(html);
}
});
}
DIV TO REPLACE
<div id="ajaxPhotographSortResponse"></div>
Move the action=sort into the data property of the $.ajax function. You're making a POST request, but appending data onto your query string like a GET request. Data only appends to the query string if it's a GET request.
Example:
$.ajax({
type: "POST",
url: "contributor_panel.php",
data: {action:'sort', sort_by: $('#sort_by').val(), view_batch: $('#view_batch').val()},
success: function(html){
$("#ajaxPhotographSortResponse").html(html);
}
});
http://api.jquery.com/jQuery.ajax/
Instead of concatenating the arguments you are passing to your server side script I would recommend you using the following syntax. Also if you already use jQuery you no longer need the document.getElementById function:
$.ajax({
type: "POST",
url: "contributor_panel.php?action=sort",
data: { sort_by: $("#sort_by").val(), view_batch: $("#view_batch").val() },
success: function(html){
$("#ajaxPhotographSortResponse").html(html);
}
});
or even shorter using the .load() function:
$('#ajaxPhotographSortResponse').load('contributor_panel.php?action=sort',
{ sort_by: $("#sort_by").val(), view_batch: $("#view_batch").val() });
I'm writing a simple ajax function and looking to populate two text input fields with the 'success' results. I'm wondering what my php syntax has to be to return an object.
Here is my Javascript function
function editModule(a){
data = {moduleNum:a}
$.ajax({
type: 'POST',
data: data,
url: 'includes/ajaxCalls.php',
success: function(data) {
alert(data['title']); // <-- This is where I'm not sure what to return from php
}
});
}
Here is my php doc (so far, this is where I need to know how to return an object)...
<?php
$data = array('title'=>'this');
echo json_encode($data);
When I run the function I just get the alert "undefined".
Suggestions?
Thanks,
-J
Try this. You can specify that you're expecting a JSON object and then you can interpret data accordingly.
function editModule(a){
data = {moduleNum:a}
$.ajax({
type: 'POST',
data: data,
dataType: 'json',
url: 'includes/ajaxCalls.php',
success: function(data) {
alert(data.title);
}
});
}
I have returned JSON data from the server via a jQuery Ajax call, not in PHP but it should be the same. As long as you set the content-type of your response to application/json jQuery should consider the responseText as a JSON string. Alternatively you can also set dataType: "JSON" in your Ajax call which tells jQuery that you expect JSON.
Your php page returns: {"title":"this"} in this case.
So you can reference the result with:
alert(data.title);
You may need to specify the data type:
function editModule(a){
data = {moduleNum:a}
$.ajax({
type: 'POST',
data: data,
url: 'includes/ajaxCalls.php',
dataType: 'json',
success: function(data) {
alert(data['title']); // <-- This is where I'm not sure what to return from php
}
});
}