I have working code that marks waypoints on google map. I wanted to change the waypoint marking according to the Ajax response.
Here is my code - http://fiddle.jshell.net/TechmazeSolution/8fgobauz/
Your code doesn't show an attempt to use the response to update the waypoints. But anyway:
in the AJAX request, $(this) is the AJAX object and doesn't refer to $("#calcRoute")
you must call findroute() after you update the waypts array
Updated code:
$("#calcRoute").change(function () {
var self = $(this);
$.ajax({
type: "POST",
url: "ajax-response.php?" + Math.random(),
dataType: "json",
success: function (response) { console.log($(this));
if (self.val() == 1) {
start = new google.maps.LatLng(1.30365, 103.85256);
end = new google.maps.LatLng(1.29411, 103.84631);
waypts = [{
location: new google.maps.LatLng(1.28644, 103.84663),
stopover: true
}, {
location: new google.maps.LatLng(1.28627, 103.85927),
stopover: true
}];
findroute();
}
}
});
});
Updated fiddle:
http://fiddle.jshell.net/8fgobauz/1/
Use your javascript console to debug your code. You could find these errors easily.
Related
Sorry for my terms incase they are incorrect, but I have a php function that I am going to interop with ajax so I can use a php function to get a value for a jquery variable.
So now I am just at the point where I have received the ajax callback and would like to set the jquery variable to the response value I got back. Here is my code as an example:
$(document).ready(function() {
$('body').videoBG({
position:"fixed",
zIndex:0,
mp4:'', //This is where I want to use the ajax response value
opacity:1,
});
})
jQuery.ajax({
type : "POST",
url : "index.php",
data : {
request : "getvideo_Action"
},
success : function(response) {
alert(response);
//Do my response stuff
}
});
So basically what I want to do is set the 'Mp4' var(or is it property?) with the value that I get from the response. Can anyone help me out with this? Thanks.
You can put the entire thing inside the success function, like this:
jQuery.ajax({
type: "POST",
url: "index.php",
data: {
request: "getvideo_Action"
},
success: function (response) {
$('body').videoBG({
position: "fixed",
zIndex: 0,
mp4: response,
opacity: 1
});
}
});
I have a PHP populated table from Mysql and I am using JQuery to listen if a button is clicked and if clicked it will grab notes on the associated name that they clicked. It all works wonderful, there is just one problem. Sometimes when you click it and the dialog(JQuery UI) window opens, there in the text area there is nothing. If you are to click it again it will pop back up. So it seems sometimes, maybe the value is getting thrown out? I am not to sure and could use a hand.
Code:
$(document).ready(function () {
$(".NotesAccessor").click(function () {
notes_name = $(this).parent().parent().find(".user_table");
run();
});
});
function run(){
var url = '/pcg/popups/grabnotes.php';
showUrlInDialog(url);
sendUserfNotes();
}
function showUrlInDialog(url)
{
var tag = $("#dialog-container");
$.ajax({
url: url,
success: function(data) {
tag.html(data).dialog
({
width: '100%',
modal: true
}).dialog('open');
}
});
}
function sendUserfNotes()
{
$.ajax({
type: "POST",
dataType: "json",
url: '/pcg/popups/getNotes.php',
data:
{
'nameNotes': notes_name.text()
},
success: function(response) {
$('#notes_msg').text(response.the_notes)
}
});
}
function getNewnotes(){
new_notes = $('#notes_msg').val();
update(new_notes);
}
// if user updates notes
function update(new_notes)
{
$.ajax({
type: "POST",
//dataType: "json",
url: '/pcg/popups/updateNotes.php',
data:
{
'nameNotes': notes_name.text(),
'newNotes': new_notes
},
success: function(response) {
alert("Notes Updated.");
var i;
$("#dialog-container").effect( 'fade', 500 );
i = setInterval(function(){
$("#dialog-container").dialog( 'close' );
clearInterval(i);
}, 500);
}
});
}
/******is user closes notes ******/
function closeNotes()
{
var i;
$("#dialog-container").effect( 'fade', 500 );
i = setInterval(function(){
$("#dialog-container").dialog( 'close' );
clearInterval(i);
}, 500);
}
Let me know if you need anything else!
UPDATE:
The basic layout is
<div>
<div>
other stuff...
the table
</div>
</div>
Assuming that #notes_msg is located in #dialog-container, you would have to make sure that the actions happen in the correct order.
The best way to do that, is to wait for both ajax calls to finish and continue then. You can do that using the promises / jqXHR objects that the ajax calls return, see this section of the manual.
You code would look something like (you'd have to test it...):
function run(){
var url = '/pcg/popups/grabnotes.php';
var tag = $("#dialog-container");
var promise1 = showUrlInDialog(url);
var promise2 = sendUserfNotes();
$.when(promise1, promise2).done(function(data1, data2) {
// do something with the data returned from both functions:
// check to see what data1 and data2 contain, possibly the content is found
// in data1[2].responseText and data2[2].responseText
// stuff from first ajax call
tag.html(data1).dialog({
width: '100%',
modal: true
}).dialog('open');
// stuff from second ajax call, will not fail because we just added the correct html
$('#notes_msg').text(data2.the_notes)
});
}
The functions you are calling, should just return the result of the ajax call and do not do anything else:
function showUrlInDialog(url)
{
return $.ajax({
url: url
});
}
function sendUserfNotes()
{
return $.ajax({
type: "POST",
dataType: "json",
url: '/pcg/popups/getNotes.php',
data: {
'nameNotes': notes_name.text()
}
});
}
It's hard to tell from this, especially without the mark up, but both showUrlInDialog and sendUserfNotes are asynchronous actions. If showUrlInDialog finished after sendUserfNotes, then showUrlInDialog overwrites the contents of the dialog container with the data returned. This may or may not overwrite what sendUserfNotes put inside #notes_msg - depending on how the markup is laid out. If that is the case, then it would explains why the notes sometimes do not appear, seemingly randomly. It's a race condition.
There are several ways you can chain your ajax calls to keep sendUserOfNotes() from completing before ShowUrlInDialog(). Try using .ajaxComplete()
jQuery.ajaxComplete
Another ajax chaining technique you can use is to put the next call in the return of the first. The following snippet should get you on track:
function ShowUrlInDialog(url){
$.get(url,function(data){
tag.html(data).dialog({width: '100%',modal: true}).dialog('open');
sendUserOfNotes();
});
}
function sendUserOfNotes(){
$.post('/pcg/popups/getNotes.php',{'nameNotes': notes_name.text()},function(response){
$('#notes_msg').text(response.the_notes)
},"json");
}
James has it right. ShowUrlInDialog() sets the dialog's html and sendUserOfNotes() changes an element's content within the dialog. Everytime sendUserOfNotes() comes back first ShowUrlInDialog() wipes out the notes. The promise example by jeroen should work too.
The code I want to work:
$.ajax({
type: "POST",
url: "_Source/ajap/ajap.nlSrch.php",
data: { sndJson : jsonData },
dataType: "json",
processData: false,
success: function(html) {
$("#srchFrm").append(html);}
});
The code that works:
$.ajax({
type: "POST",
url: "_Source/ajap/ajap.nlSrch.php",
data: { sndJson : jsonData },
success: function(html) {
$("#srchFrm").append(html);}
});
Unfortunately when I send the first one my post data looks like this "Array ()" and when I use the later I get this "Array ( [sndJson] => [\"8\",\"3\",\"6\",\"7\"] )".
I know that there has to be a simple explanation but I haven't been able to figure it out.
Help please!
Try sending your data in a query string...
$.ajax({
type:"POST",
url:"_Source/ajap/ajap.nlSrch.php?json="+jsonData,
dataType:"json",
success: function(data) {
$("#srchFrm").append(data);}
error: function(xhr, ajaxOptions, thrownError)
{alert("Error!");}
});
You can use shorthand $.post instead of using low level ajax class --- because you don't need to advanced handling. So, this one will be great enough.
$(document.ready(function(){
$("#submit_button").click(function(){
$.post('php_script.php', {
// here's what you want to send
// important -- double quotes, 'cause It's evals as valid JSON
"var1" : "val1"
"var2" : "val2"
}, function (respond){
try {
var respond = JSON.parse(respond);
} catch(e){
//error - respond wasn't JSON
}
});
});
});
PHP code:
<?php
/**
* Here you can handle variable or array you got from JavaScript
* and send back if need.
*/
print_r($_POST); // var1 = val1, var2 = val2
?>
Back to your question,
Why my .ajax request doesn't work?
This is because JavaScript throws fatal error and stops further code execution.
You can catch and determine the error occasion, simply by adding
try {} catch(){} block to the statement you think may occur any error
When you specify dataType: json, jQuery will automatically evaluate the response and return a Javascript object, in this case an array. You're taking the result and adding it as html to #srchForm, so it does not make sense to convert it to a javascript object. Use dataType: html, or none at all.
http://api.jquery.com/jQuery.ajax/
The following examples above are not reusable. I am a huge fan of reuseable code. here is my solution.
Software design 101:
DRY Don't repeat your self. You should wrap your code into an object. This way you can call it from anywhere.
var Request = {
version: 1.0, //not needed but i like versioning things
xproxy: function(type, url, data, callback, timeout, headers, contentType)
{
if (!timeout || timeout <= 0) { timeout = 15000; }
$.ajax(
{
url: url,
type: type,
data: data,
timeout: timeout,
contentType: contentType,
success:function(data)
{
if (callback != undefined) { callback(data); }
},
error:function(data)
{
if (callback != undefined) { callback(data); }
},
beforeSend: function(xhr)
{
//headers is a list with two items
if(headers)
{
xhr.setRequestHeader('secret-key', headers[0]);
xhr.setRequestHeader('api-key', headers[1]);
}
}
});
}
};
Usage:
<script type="text/javascript">
var contentType = "applicaiton/json";
var url = "http://api.lastfm.com/get/data/";
var timeout = 1000*5; //five seconds
var requestType = "POST"; //GET, POST, DELETE, PUT
var header = [];
header.push("unique-guid");
header.push("23903820983");
var data = "{\"username\":\"james\"}"; //you should really deserialize this w/ a function
function callback(data)
{
//do logic here
}
Request.xproxy(requestType, url, data, callback, timeout, header, contentType);
</script>
Everything was going great in my previous help request thread. I was on the correct track to get around a CSRF, but needed to be pointed in the right direction. I received great help and even an alternate script used to log into Google's Android Market. Both my script and the one I altered to match my form is get hung up at the same point. Apparently cURL cannot process JS, is there any way to work around the form being submitted with submitForm() without changing the form?
Here is the code for the SubmitForm function
function submitForm(formObj, formMode) {
if (!formObj)
return false;
if (formObj.tagName != "FORM") {
if (!formObj.form)
return false;
formObj = formObj.form;
}
if (formObj.mode)
formObj.mode.value = formMode;
formObj.submit();
}
Here is the code for the submit button -
<a class="VertMenuItems" href="javascript: document.authform.submit();">Submit</a>
Here is a link to my last question in case more background information is needed.
PHP service...
<?php
// PHP service file
// Get all data coming in via GET or POST
$vars = $_GET + $_POST;
// Do something with the data coming in
?>
Javascript elsewhere...
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
function sendData(data)
{
var response;
$.ajax({
url: 'phpservice.php',
data: data,
type: 'POST',
dataType: 'json',
async: false,
success: function(response_from_service)
{
response = response_from_service;
},
error: function()
{
}
});
return response;
};
function getData(data)
{
var response;
$.ajax({
url: 'phpservice.php',
data: data,
type: 'GET',
dataType: 'json',
async: false,
success: function(response_from_service)
{
response = response_from_service;
},
error: function()
{
}
});
return response;
};
});
</script>
Is it possibe to simply load a php script with a url with js?
$(function() {
$('form').submit(function(e) {
e.preventDefault();
var title = $('#title:input').val();
var urlsStr = $("#links").val();
var urls = urlsStr.match(/\bhttps?:\/\/[^\s]+/gi);
var formData = {
"title": title,
"urls": urls
}
var jsonForm = JSON.stringify(formData);
$.ajax({
type: 'GET',
cache: false,
data: { jsonForm : jsonForm },
url: 'publishlinks/publish'
})
//load php script
});
});
Edit:
function index() {
$this->load->model('NewsFeed_model');
$data['queryMovies'] = $this->NewsFeed_model->getPublications();
$this->load->view('news_feed_view', $data);
}
simple
jQuery and:
<script>
$.get('myPHP.php', function(data) {});
</script>
Later edit:
for form use serialize:
<script>
$.post("myPHP.php", $("#myFormID").serialize());
</script>
like this ?
$.get('myPHP.php', function(data) {
$('.result').html(data);
alert('Load was performed.');
});
There are various ways to execute a server side page using jQuery. Every method has its own configuration and at the minimum you have to specify the url which you want to request.
$.ajax
$.ajax({
type: "Get",//Since you just have to request the page
url:"test.php",
data: {},//In case you want to provide the data along with the request
success: function(data){},//If you want to do something after the request is successfull
failure: function(){}, //If you want to do something if the request fails
});
$.get
$.get("test.php");//Simplest one if you just dont care whether the call went through or not
$.post
var data = {};
$.post("test.php", data, function(data){});
You can get the form data as a json object as below
var data = $("formSelector").searialize();//This you can pass along with your request