I have the following snippet.
var myData = {
video: $("input[name='video[]']:checked").serialize(),
sinopse: $("#sinopse").val(),
dia: $("#dia").val(),
quem: $("#quem").val()
};
jQuery.ajax({
type: "POST", // HTTP method POST or GET
url: "response.php", //Where to make Ajax calls
dataType:"text", // Data type, HTML, json etc.
data:myData, //Form variables
success:function(response){
$("#responds").append(response);
$(".video").val(''); //empty text field on successful
$("#sinopse").val('');
$("#dia").val('');
$("#quem").val('');
$("#FormSubmit").show(); //show submit button
$("#LoadingImage").hide(); //hide loading image
},
all variables except "video" are getting saved in database. What can be wrong here? (maybe the problem is with jquery)
refer this link:- serialize
You can get selected checkboxes values by
$("input[type='checkbox']:checked").serialize();
Related
I can't display my json data in the modal form. But data is coming which I can see from browser inspect. I have tried this code, can anyone one please help me.One is select form another is input field.
$.ajax({
url:"catch.php",
method:"GET",
data:{user_id:user_id},
dataType:"json",
success:function(data)
{ response = JSON.parse(data);
$('#exampleModalLong').modal('show');
$("#ven option[value="+data.vendor+"]").attr("selected","selected");
$('#cost').val(data.repair_cost);
}
})
});
Basically what I'm trying to do is post comments to a page without having to refresh the entire page. Just the Comments DIV so it looks like it posted and refreshed smoothly.
The form submits to the same page it's on. Everything I've found shows me how to refresh content constantly using intervals. I just want the comments DIV to refresh when someone posts a comment.
I can't find the correct ajax code to do this the way I want.
Here is my code:
var submit_button = $('#submit_button');
submit_button.click(function() {
var commentSubmitted= $('commentSubmitted').val();
var update_div = $('#update_div');
$.ajax({
type: 'POST',
url: '/blog/',
data: data,
success:function(html){
update_div.html(html);
}
});
});
in the same PHP file, I have the post to DB:
if($_POST[commentSubmitted])
{
$query="INSERT INTO comments (commentSubmitted) VALUES ('$commentSubmitted')";
mysql_query($query);
}
The HTML is for the form:
<form id="flow" method='post' action='/blog/'>
<textarea name='commentSubmitted' ></textarea>
<input type='submit' value='Post'/>
The DIV containing all comments looks like so:
<DIV id='AllComments'>
// comments displayed here
</DIV>
So after submitting the form, I would like the 'AllComments' DIV to reload.
The best would be to use jQuery to make the ajax call to the server and retrieve the data you want.
You have two ways of retrieving the data. Either retrieve the additional comments to show in a json array and handle it with javascript, or create the html on the server side and append/replace the html in the comments section.
Using Json
$.ajax({
url: "the_ajax_url_here",
type: "post",
data: {paramTitle: "paramValue", paramTitle1: "paramValue1"},
dataType: "json"
success: function(response) {
// handle the response
}
});
Retrieving Html
$.ajax({
url: "the_ajax_url_here",
type: "post",
data: {paramTitle: "paramValue", paramTitle1: "paramValue1"},
dataType: "html"
success: function(response) {
// set the html of comments section to the newly retrieved html
$("comments_section_selector").html(response);
}
});
What I would do is retrieve the newly added comment in a json array and then using javascript append it to the comments section.
edit:
After seeing your code I have some comments that might help you.
I would personally prefer the code that handles the ajax request in a separate file.
In that file you can store the new comment and create the html to display that comment.
Then in the success function just append the new html to the comment section like so:
success: function(response) {
$('#AllComments').append(response);
}
You can also make new comment appear on top using prepend
$('#AllComments').prepend(response);
Simple as that hope you are upto it
submit_button.click(function() {
var commentSubmitted= $('commentSubmitted').val();
var update_div = $('#update_div');
$.ajax({
type: 'POST',
url: '/blog/',
data: data,
success:function(html){
update_div.html(html);
}
});
});
Then you go to insert data
if($_POST[commentSubmitted])
{
$query="INSERT INTO comments (commentSubmitted) VALUES ('$commentSubmitted')";
mysql_query($query);
//After Inserting data retrieve back all the comments from db
$sql = "select * from comments";//any query and execute it
$query = mysql_query($sql);
while($data = mysql_fetch_array($query)){
echo $data["comments"];//Echo your commenets here
}
exit;
}
Thats it
I have a div, which shows Rate me section. i.e. Five stars. When the user click on it, the rating value gets prompted.
<div class="rateit bigstars" id="rateit99" data-rateit-starwidth="32" data-rateit-starheight="32" onclick="alert($('#rateit99').rateit('value'))">
</div>
In here rather than an alert, I want to make an ajax call, which sends(post data) Rated value to data.php file.
How can it be done?
I don't understand what's on your HTML code, but i can understand what you are trying to achieve.
First clean your HTML code to make it easier to read for the purpose.
Here is an example of how you can receive the data and pass it using ajax
// Use the id of your trigger element
$('#submit').click(function() {
// Make your datastring (you can get other values with var a = $('#id').val();
var dataString = 'value1='+ value1 + '&value2=' + value2 ;
$.ajax({
type: "POST",
url: "data.php",
data: dataString,
cache: false,
success: function(data)
{
// handle the result
}
});
});
I have been trying for the last couple of days to serialize a form in jquery and pass it to php. For some reason it just doesn't work... Question, What steps do you take to serialize a form in Jquery... I get the id of the form pass it into $('#fomrID').serialize()...
And it still doesn't work. When I debugg with firebug All I get is an empty string... Do you use the name or the Id of the form. Do you use e.disableDefalut or whatever that is? I can't figure out why I can't serialize my form please help
$('#profilepicbutton').change(function(){
alert("Boy I hate PHP");
var formElement = $('#profilepicinput').attr('value');
dataString = $("#registerpt3").serialize();
$.ajax({
type: "POST",
url: "register3.php",
data: dataString, //"profilePic="+formElement,
success: function(data){
alert( "Data Saved: " + data );
$.ajax({
type: "POST",
url: "retrievePic.php",
data: "",
success: function(data){
alert(data);
var image = new Image();
$(image).load(function(){
console.log("we have uploaded this image");
}).attr('src', 'images/');
alert("");
},
error: function(msg){
alert("");
}
});
},
error:function(msq){
alert("" + msg);
}
}
);
});
<form id="registerpt3" name="registerpt3" enctype="multipart/form-data" action="register3.php" onSubmit="" method="post">
<input name="profilepicinput" type="file" id="profilepicbutton" />
</form>
According to http://api.jquery.com/serialize
For a form element's value to be included in the serialized string, the element must have a name attribute. Values from checkboxes and radio buttons (inputs of type "radio" or "checkbox") are included only if they are checked. Data from file select elements is not serialized.
Also, simply calling serialize won't actually upload the picture for you. If you want to do an asynchronous picture upload (or any file for that matter) I'd suggest looking into something like Uploadify: http://www.uploadify.com/
I'm trying to take values from a dropdown two boxes and send them to a PHP file which will draw an appropriate field from a mySQL database depending on the combination chosen and display it in a div without refreshing the page using AJAX. I have the second part sorted, but I'm stuck on the first part.
Here is the HTML: http://jsfiddle.net/SYrpC/
Here is my Javascript code in the head of the main document:
var mode = $('#mode');
function get() {$.post ('data.php', {name: form.him.value, the_key: #mode.val()},
function(output) {$('#dare').html(output).show();
});
}
My PHP (for testing purposes) is:
$the_key = $_POST['the_key'];
echo $the_key;
After I have it in PHP as a variable I can manipulate it, but I'm having trouble getting it there. Where am I going wrong? Thanks for your replies!
You need a callback function as well to have the server response to the POST.
$.post('ajax/test.html', function(data) {
$('.result').html(data);
});
This snippet will post to ajax/test.html and the anonymous function will be called upon its reply with the parameter data having the response. It then in this anonymous function sets the class with result to have the value of the server response.
Help ? Let me know and we can work through this if you need more information.
Additionally, $.post in jQuery is a short form of
$.ajax({
type: 'POST',
url: url,
data: data,
success: success
dataType: dataType
});
your jquery selectors are wrong:
html:
<select id="mode">
jquery selector:
$("#mode").val();
html:
<select name="player">
jquery selector:
$("select[name=player]").val();
You want to add a callback to your ajax request, its not too hard to do, here ill even give you an example:
$.ajax({
url: "http://stackoverflow.com/users/flair/353790.json", //Location of file
dataType: "josn",//Type of data file holds, text,html,xml,json,jsonp
success : function(json_data) //What to do when the request is complete
{
//use json_data how you wish to.;
},
error : function(_XMLHttpRequest,textStatus, errorThrown)
{
//You fail
},
beforeSend : function(_XMLHttpRequest)
{
//Real custom options here.
}
});
Most of the above callbacks are optional, and in your case i would do the following:
$.ajax({
url: "data.php",
dataType: "text",
data : {name: ('#myform .myinput').val(),the_key: $('#mode').val()},
success : function(value)
{
alert('data.php sent back: ' + value);
}
});
the ones you should always set are url,success and data if needed, please read The Documentation for more information.