Here in the following code in jquery i get the alert test1 but when i try to get alert test2 i dint got that... there is some issue in code please help me to resolve.
$(document).ready(function() {
$(".delete-item-details").click(function() {
alert('test1');
var id = $(this).attr('ids');
alert('test2');
var order_id = $("#order_id").val();
goto_url = "/order/DeleteOrderDetailItems/" + id;
dataString = 'id=' + id + '&order_id=' + order_id;
$.ajax({
type: "POST",
url: goto_url,
data: dataString,
cache: false,
success: function(html) {
$("#row-id-" + id).fadeOut();
$("#total_span").html(html);
}
});
});
});
Please change this three lines in code and you will definitely get alert !!
alert('test1');
var id=$(this).attr('id'); // here you have written ids
alert('test2');
The default attribute is 'id' not 'ids'
If still now clear let me know i will help you out.
TRY
var id = $(this).prop('ids');
Related
I have a fontawesome icon in my HTML as a button and i'd like to use javascript and trigger it AJAX style
<i id="heart" class="jam jam-heart-f"></i> Like
Here is the attempt at javascript to trigger it - but I dont get any errors to follow up on. I try to post the like attempt to a PHP page like.php to add the link to a database.
$(document).ready(function()
{
$('body').on("click",'#heart',function()
{
var videoId = "<?php echo $video_id; ?>";
var A=$(this).attr("id");
var B=A.split("like");
var messageID=B[1];
var C=parseInt($("#likeCount"+messageID).html());
$.ajax({
method: 'POST',
url: 'like.php',
data: {videoId: videoId},
cache: false,
success: function(result){
likeInfo = JSON.parse(result);
$("#likeCount1").html("Likes:" + likeInfo.likeCount);
//document.getElementById("likeCount1").value = likeInfo.likeCount;
//$("#likeCount1").html(likeCount);
}
});
}
});
I dont think #heart seems to be triggered in JS by the id="heart" with the font awesome icon. Any ideas how I can rig this together
You forgot to add closing parenthesis and semicolon for your $('body').on... statement
Try this:
$(document).ready(function()
{
$('body').on("click",'#heart',function()
{
var videoId = "<?php echo $video_id; ?>";
var A=$(this).attr("id");
var B=A.split("like");
var messageID=B[1];
var C=parseInt($("#likeCount"+messageID).html());
$.ajax({
method: 'POST',
url: 'like.php',
data: {videoId: videoId},
cache: false,
success: function(result){
likeInfo = JSON.parse(result);
$("#likeCount1").html("Likes:" + likeInfo.likeCount);
//document.getElementById("likeCount1").value = likeInfo.likeCount;
//$("#likeCount1").html(likeCount);
}
});
});
});
Your code triggers the post-request correctly, but you are not closing your functions and scopes correctly.
I tried it out here:
http://jsfiddle.net/4cohrz5p/
And code to keep stackoverflow happy:
$(document).ready(function() {
$('body').on("click", '#heart', function() {
var videoId = "<?php echo $video_id; ?>";
var A = $(this).attr("id");
var B = A.split("like");
var messageID = B[1];
var C = parseInt($("#likeCount" + messageID).html());
$.ajax({
method: 'POST',
url: 'like.php',
data: {
videoId: videoId
},
cache: false,
success: function(result) {
likeInfo = JSON.parse(result);
$("#likeCount1").html("Likes:" + likeInfo.likeCount);
//document.getElementById("likeCount1").value = likeInfo.likeCount;
//$("#likeCount1").html(likeCount);
}
});
});
});
Besides, the javascript console shows Uncaught SyntaxError: missing ) after argument list for your code. And you open the network-tab when you click the heart to see outgoing requests and can inspect them to see that they send the correct data (and the response too!).
Any decent js editor would have shown this error before even running the code. Try VS Code. Free and lightweight and pretty great overall.
I have a feeling there is something wrong with my for loop. When my websites event is activated the first time, I get no response. It works as intended every time after that. I have tried tuning the numbers in the for loop looking for mistakes but as far as what I've tried. It works best as is.
For the full app: https://codepen.io/xcidis/full/KvKVZb/
var reference = [];
function random() {
$.ajax({
url: "https://api.forismatic.com/api/1.0/?",
dataType: "jsonp",
data: "method=getQuote&format=jsonp&lang=en&jsonp=?",
success: function(quote) {
reference.push([quote.quoteText + "<br/><br/><br/><div align='right'>~" + quote.quoteAuthor + "</div>"]);
}
});
}
$("button").click(function(){
random();
for(i=0;i<4; i++){
if(reference[reference.length-1] == undefined){continue}else{
var boxes = $("<div id='boxes'></div>").html("<p>" + reference[reference.length-1] + "</p>");
$('body').append(boxes);
break;
};
};
});
Your rest of the code ran before your ajax push the value to reference variable.
https://www.w3schools.com/xml/ajax_intro.asp
You can either put your page rendering code within the ajax or use some tips to run the rederer synchronously
$("button").click(function(){
$.when( $.ajax({
url: "https://api.forismatic.com/api/1.0/?",
dataType: "jsonp",
data: "method=getQuote&format=jsonp&lang=en&jsonp=?",
success: function(quote) {
reference.push([quote.quoteText + "<br/><br/><br/><div class='tweet' align='left'></div><div align='right'>~" + quote.quoteAuthor + "</div>"]);
}
})).then(function() {
console.log(reference)
for(i=0;i<4; i++){
if(reference[reference.length-1] == undefined){continue}else{
var boxes = $("<div id='boxes'></div>").html("<p>" + reference[reference.length-1] + "</p>");
$('body').append(boxes);
break;
};
};
});
});
I've written one jQuery function to get the city and state code based upon the zip code value but facing some issue with some errors. Can someone please help me in correcting the mistakes I'm making here.
Following is my code :
$(document).ready(function() {
$("#zip_code").keyup(function() {
var el = $(this);
var module_url = $('#module_url').val();
if (el.val().length === 5) {
$.ajax({
url : module_url,
cache: false,
dataType: "json",
type: "GET",
data: {'request_type':'ajax', 'op':'get_test_category_list','zip_code =' + el.val()},
success: function(result, success) {
$("#city").val(result.city);
$("#state_code").val(result.state);
}
});
}
});
});
Thanks in advance.
The issue is in your data object, you have invalid syntax. Change this:
'zip_code =' + el.val()
To this:
'zip_code': el.val()
The full object should look something like this:
data: {
'request_type': 'ajax',
'op': 'get_test_category_list',
'zip_code': el.val()
},
I think the problem is with data part of the ajax
Change it like this
data: {request_type:"ajax", op:"get_test_category_list",zip_code : el.val()},
After searching here on SO and google, didn't find an answer to my problem.
The animation doesn't seem to trigger, tried a simple alert, didn't work either.
The function works as it is supposed (almost) as it does what i need to, excluding the success part.
Why isn't the success event being called?
$(function() {
$(".seguinte").click(function() {
var fnome = $('.fnome').val();
var fmorada = $('.fmorada').val();
var flocalidade = $('.flocalidade').val();
var fcodigopostal = $('.fcodigopostal').val();
var ftelemovel = $('.ftelemovel').val();
var femail = $('.femail').val();
var fnif = $('.fnif').val();
var fempresa = $('.fempresa').val();
var dataString = 'fnome='+ fnome + '&fmorada=' + fmorada + '&flocalidade=' + flocalidade + '&fcodigopostal=' + fcodigopostal + '&ftelemovel=' + ftelemovel + '&femail=' + femail + '&fnif=' + fnif + '&fempresa=' + fempresa;
$.ajax({
type: "GET",
url: "/ajaxload/editclient.php",
data: dataString,
success: function() {
$('.primeirosector').animate({ "left": "+=768px" }, "fast" );
}
});
return false;
});
});
you are trying to pass query string in data it should be json data.
Does your method edit client has all the parameters you are passing?
A simple way to test this is doing the following:
change this line to be like this
url: "/ajaxload/editclient.php" + "?" + dataString;
and remove this line
data: dataString
The correct way of doing it should be, create a javascript object and send it in the data like so:
var sendData ={
fnome: $('.fnome').val(),
fmorada: $('.fmorada').val(),
flocalidade: $('.flocalidade').val(),
fcodigopostal: $('.fcodigopostal').val(),
ftelemovel: $('.ftelemovel').val(),
femail: $('.femail').val(),
fnif: $('.fnif').val(),
fempresa: $('.fempresa').val()
}
$.ajax({
url: "/ajaxload/editclient.php",
dataType: 'json',
data: sendData,
success: function() {
$('.primeirosector').animate({ "left": "+=768px" }, "fast" );
}
});
Another thing shouldn't this be a post request?
Hope it helps
I am using a function that will submit with ajax and without the help of a button click. But I am currently undergoing two issues which with trial and error haven't found plausible solutions:
First is there any way I can disable the enter button click(this causes the whole page to refresh)?
JSFIDDLE basic example in how the JS function works
Second, It feels like I am going the roundabout way to display what has been posted. How can I change this part of the function $('#special').html('<p>' + $('#resultval', result).html() + '</p>'); to have it POST just inside a div called #special without the need of span or <p> #resultval?
Everytime i echo through php I have to do set it like this to display a result: <div id="special"><span id="resultval">This is the result.</span></div>
<script>
$(document).ready(function() {
var timer = null;
var dataString;
function submitForm(){
$.ajax({ type: "POST",
url: "posting.php",
data: dataString,
success: function(result){
$('#special').html('<p>' + $('#resultval', result).html() + '</p>');
}
});
return false;
}
$('#ytinput').on('keyup', function() {
clearTimeout(timer);
timer = setTimeout(submitForm, 050);
var name = $("#ytinput").val();
dataString = 'name='+ name;
});
});
</script>
$(document).ready(function() {
var timer = null;
var dataString;
function submitForm(event){// the function call on click or on submit onclick=submitForm(event);
event.preventDefault(); //to prevent enter key
$.ajax({ type: "POST",
url: "posting.php",
data: dataString,
success: function(result){
$('#special').text(result); //you can use text() or html() only
}
});
return false;
}
$('#ytinput').on('keyup', function() {
clearTimeout(timer);
timer = setTimeout(submitForm, 050);
var name = $("#ytinput").val();
dataString = 'name='+ name;
});
});