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"
Related
here is my simple code in trying to get data from php to ajax.
i just wanted to get a simple data pass back to ajax success. i already searched for this but i cant get it properly. my code is really simple.
<script src="http://code.jquery.com/jquery-1.7.1.min.js"></script>
<SCRIPT TYPE="text/javascript">
$(document).ready(function(){
$("#1").click(function(){
$.ajax({
type: "POST",
url: "ajax.php",
data: { "txt": "try"},
cache: false,
success: function(html)
{
alert(html.mes);
}
});
});
});
</SCRIPT>
<pre><button id="1">try</button></pre>
Then i load a ajax.php that has a simple code too
$var['mes'] = 'message';
echo json_encode($var);
and its alerting me "undefined". i know this is simple but i cant find it out where do am i wrong
You need to tell jQuery that the script is running JSON:
$(document).ready(function(){
$("#1").click(function(){
$.ajax({
type: "POST",
url: "ajax.php",
data: { "txt": "try"},
dataType: 'json',
cache: false,
success: function(html)
{
alert(html.mes);
}
});
});
});
I know this has been asked before, but I can't get it working.
I'm loading a file on a page whose path and name are a variable contained in a PHP script.
My script.php is outputing a variable $filename which contains the path to a file that has to be open in an ajax request.
So, the file can be for example:
'../path/to/file/filea.json' or
'../another/path/fileb.json'
I tried this in my jQuery:
$.ajax({
url:'script.php',
success:$.ajax({
url: ??? // This ($filename) is what I'm trying to get from the 1st Ajax call
sucess: function(data){
//other code here
}
})
);
I know $filename is wrong in the second Ajax call but how do I get the value of that variable?
$.ajax({
url: "script.php",
success: function(data){
$.ajax({
var someParam=data; //response Data from 1st Ajax Call
type: "post",
url: "example.php",
data: 'page='+someParam,
success: function(data){
//Do More Here!
});
});
Try this:
inside script.php
echo json_encode($filename); //somewhere you need to have this echo.
jQuery
$.ajax({
url: 'script.php',
success: function (data) {
alert(data); //or console.log() and doublecheck you have the right info
$.ajax({
url: JSON.parse(data),
success : function (data) {
//other code here
}
})
}
}); // added a extra } to close the ajax
This solved the problem:
$.ajax({
async:false,
url: 'script.php',
success: function(data){
$.ajax({
url: data,
success: //code here
})
})
#Sergio: script.php wasn't echoeing the output of $filename... Thanks for reminding!
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.......
I was wondering if this was making an asynchronous request...write now Im using:
<script type='text/javascript'>
$(document).ready(function() {
var ktitle = $('.hiddentwo').text();
$('div#tab2').load('morefour.php?title=' + encodeURIComponent(ktitle));
});
</script>
what Im doing though is adding text in the first, into the database, on the current php file (addtext.php). Im passing the Id of the current document to the morefour.php and that is loading the added text on the second tab...the thing is, Im having to refresh to see the content again. Im running on localhost btw.
For more clarity, Im running another jquery script that on clicks, retrieves this data to send it to a php file to enter into a database
$(".button").click(function() {
var content = $(this).siblings().outerHTML();
$.ajax({
async: false,
type: "POST",
url: "tosqltwo.php",
data: {
content: content
}
});
});
$(function(){ //shorthand of $(document).ready
$('div#tab2').html($.ajax({
type: "GET", //if you are doin $_GET['title'] in morefour.php
url: "morefour.php",
data : {title:ktitle},
dataType: 'html', //i am not sure about this part
async: false
}).responseText)
});
or you can try
$(function(){
$.ajax({
url : 'morefour.php',
data : {title:ktitle},
type:'GET',
dataType:'html',
success: function(data) {
$('div#tab2').html(data);
}
});
});
you can use $.ajax function with async to false.
$.ajax({
async: false,
url : 'morefour.php',
data : 'title=' + encodeURIComponent(ktitle),
success: function(data) {
$('div#tab2').html(data);
}
});
For some reason this jQuery function is not working properly. Here's my code... the response div is not updating with my response.
WHEN AJAX FUNCTION IS CALLED
if ($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() });