I am trying to post the element information that jQuery pulls, when a user clicks on table cell, to a new page that will use that information (an id in this case) in a sql query. i.e., the user clicks a cell and the job he/she clicks has an id of 25, that is to be passed to my php page that queries the database for the job with that id and then populates the page with said information. The user can then alter the information from the query and submit it to update the database table. I have the id from the click function and a success alert tells me that the info was posted. The problem is that when the page is opened it states that the posted name index is undefined.
Here is my script to get the information:
<script>
$(document).ready(function()
{
$("table.jobs tbody td#job").click(function()
{
var $this = $(this);
var col = $this.text();
var LeftCellText = $this.prev().text();
if (col == '')
alert("Please pick another column");
else
$.ajax(
{
type:"POST",
url:"../php/jobUpdate.php",
data:"name=" + LeftCellText,
success: function()
{
window.location = "../php/jobUpdate.php";
}
});
});
});
</script>
and here is the simple php page it is sending to:
$name = $_POST['name'];
echo $name;
I am new to jQuery, and I cannot figure out why this is not working?
When you use ajax, the second page ../php/jobUpdate.php processes the data sent by the first page, and returns a value (or even a huge string of html, if you want).
The first page receives the new data in the ajax routine's success function and can then update the current page. The updating part happens in the success: function, so you're on the right track.
But in your success function, you are redirecting the user to the 2nd page -- after already being there and processing the data. Redirecting them is probably not what you want to do.
Try replacing this:
success: function()
{
window.location = "../php/jobUpdate.php";
}
with this:
success: function(data)
{
alert(data);
}
If you want to see how to update the first page with the data received via ajax, try adding an empty DIV to your html, like this:
<div id="somestuff"></div>
Then, in the success: function of the ajax routine, do this:
$('#somestuff').html(data);
(Note that the term "data" can be any name at all, it only needs to match the name used in the function param. For example:
success: function(whatzup) {
alert(whatzup);
}
From your comment to my previous post, it seems that you don't need ajax at all. You just need a form in your HTML:
<form id="MyForm" action="../php/jobUpdate.php" method="POST">
<input type="hidden" id="jobID" name="yourJobID">
</form>
Note that forms are invisible until you put something visible inside them.
You can have select controls (dropdowns) in there, or all form elements can be invisible by using hidden input fields (like the HTML just above), which you can populate using jQuery. Code to do that would look something like this:
<script>
$(document).ready(function() {
$("table.jobs tbody td#job").click(function() {
var $this = $(this);
var col = $this.text();
var LeftCellText = $this.prev().text();
//Set value of hidden field in form. This is how
//data will be passed to jobUpdate.php, via its `name` param
$('#jobID').val(LeftCellText);
if (col == '')
alert("Please pick another column");
else
$('#myForm').submit();
});
});
</script>
If you add more values to your form to send over to jobUpdate.php, just ensure that each element has a name, such as <input type="text" name="selectedJobType"> (this element, type="text", would be visible on screen).
In the jobUpdate.php file, you would get these values thus:
$SJT = $_POST['selectedJobType'];
$id = $_POST["yourJobID"];
//Use these vars in MySQL search, then display some HTML based on results
Note that the key referenced in the $_POST[] array (selectedJobType / yourJobID) is always identical to the name specified in the HTML tag. These names are case sensitive (and spelling counts, too).
Hope this isn't TMI, just wish to cover all the bases.
On your success function causing the window to reload will delete any of the variables passed in via .ajax.
What you can try is returning the data and use it in the existing page.
success: function(msg) {
$('#someDiv').append(msg);
}
The reason the index is not defined is because you are using a string in the data-argument, however, that is actually an array-like object. :)
data: { name: col }
that should be the line you need to change. Otherwise I have not seen any problems. Also if I can give you a little idea, I wouldn't use POST actually. In fact, I'd use GET. I can not confirm if that is saver or not, but using $_SERVER["HTTP_REFFERER"] you can check from where that request is coming to determine if you want to let it pass or not.
The way I would suggest is, that you sent the ID in a GET-request and have the PHP code return the data using json_decode(). Now in jQuery, you can use $.getJSON(url, function(data){}) - which is, for one, shorter and a bit faster.
Since you probably will crop the URL yourself here, make sure that you use a function like intVal() in JS to make sure you are sending an intenger instead of a malicious string :)
Related
I have a dropdown selector on a page that allows a user to select a template type (for example, "human" or "dog").
Based on what template is selected, different fields will need to populate below the dropdown (for example, text fields for "parents names" or a dropdown list for "breed") that are unique to each template.
I will have a button that the user will click once the data fields are put in that will output data to an "output div" section of the same page when clicked (no POSTing data as it's not being saved). The output will have different output logic based on the selected template (for example, "I'm a human named X" or "I'm a dog, my breed is Y").
My real program will be more complex and each template will have a php class that stores all of the logic. Since I will be dealing with both php objects and variables gathered by jquery, what's the best way to let them interact?
For 1., I know I can do something easy like -
var selected_template = $('#my-template-dropdown :selected').text();
if (selected_template == 'Human'){
$('#my-fields').html('<?php echo HumanTemplate::render_fields(); ?>');
}
which is easy enough, but for 2. I need to pass variables from jquery to php, then return output back to jquery.
I would like some advice on the easiest way to do this before I start down the wrong path.
HTML
Allow the user to select the template type:
<form>
<select id="my-template-dropdown" name='template'>
<option value="dogs">Dogs</option>
<option value="humans">Humans</option>
</select>
</form>
<div id="my-fields"><div>
<div id="output"><div>
jQuery
Any time the user changes the template selection, request new content to display via AJAX, and insert it on the current page so the page does not have to refresh:
$('#my-template-dropdown').on('change', function() {
var template = $(this).val();
$.ajax({
url: 'http://your-site/path/to/' + template,
success: function(resp) {
$('#my-fields').html(resp);
}
});
});
PHP
http://your-site/path/to/template simply generates the HTML you want to display for that template, eg (just an example, don't know if this is suitable for your app):
if ($template == 'humans') {
echo HumanTemplate::render_fields();
} else if ($template == 'dogs') {
echo DogTemplate::render_fields();
}
For part 2, assuming all the logic you refer to is in the template rendered by PHP, you could then handle it with jQuery. This is pretty crude, you probably need something more sophisticated (eg a full template which you swap variables into?), but you get the idea:
$('#output').on('click', 'button', function(e) {
e.preventDefault();
// fields in your template which the user will fill
var species = $('#species').val(),
title = $('#title').val();
// Probably better to have this text as a template in your source
$('#output').html("I'm a " + species + ' named ' + title);
});
NOTE the gotcha in the event handler. Event handlers will only attach to elements that exist at the time the handler is defined. Since the content is injected after page load, an event handler like $('#button).on('click', function() {... would have no effect when clicking a button inserted via AJAX. The syntax here attaches to the parent #output div, which does exist at page load, and filters for clicks on a button. See the jQuery event delegation docs for more info.
Another option would be to POST the submitted data to some PHP controller, which generates and returns the output. This way all your logic is in the one place. For example, here the user's click will query the same PHP file which generated the initial template, this time including the values the user has entered. It could then generate the required output and return it, to be inserted on the page. You'd need to update the PHP so it can determine which of these cases it is handling (eg hidden field?); alternatively if you wanted to keep those separate you could hit another PHP file all together.
$('#output').on('click', 'button', function(e) {
var template = $('#my-template-dropdown').val(),
$form = $('form'),
data = $form.serialize(); // Values from all fields user has entered
$.ajax({
url: 'http://your-site/path/to/' + template,
data: data,
success: function(resp) {
$('#output').html(resp);
}
});
});
The best way to pass data from jQuery to PHP, is by using AJAX.
Mozilla has an excellent guide on getting started, that i recommend you follow.
An example of how you can achieve what you are requesting, is by trying the following:
var selected_template = $('#my-template-dropdown :selected').text();
var ajaxurl = 'ajax.php',
data = {'select_template': selected_template };
$.post(ajaxurl, data, function (response) {
console.log(response);
});
On the PHP end (Ajax.php in my example) It could look something like this
if(isset($_POST['select_template'])) {
// do something with the input from jQuery
$selected_template = $_POST['select_template'];
// return the result back to the client
echo $seleted_template;
}
?>
$selected_template will be sent back to the client, and response in the AJAX function will be whatever the server returned. So the console.log(response) should display whatever was being sent to the server
You can have a look to the function wp_localize_script.
This function make available PHP datas to JS files on the page load through the wp_enqueue_scripts action.
This will not work like an Ajax request and only populate data for a specific handle on page load. But you can mix this method with ajax in the same script.
Hope it helps even it doesn't seems to fit to your case.
As your class not fires on page load, you can use the action wp_ajax_{custom _action} and wp_ajax_nopriv_{custom_action} . For example, that's usually used to populate multiple dropdown, each time an event is trigger by the user, a php function returns result the js script.
I'd like to update a form, more exactly I wish to display an extra scroll menu, depending on the user's choice in a first scroll menu.
I have a page, mypage.php page on which there is a form. Here it is :
<form method="post" action="wedontcare.php" enctype="multipart/form-data">
<label for="base">3 - Select a DB </label><br />
<?php
include 'functions.php';
offers_choice_DB();
if (isset($_POST['base_name_choice'])){
offers_choice_table($_POST['base_name_choice']);
}
I have a separated file "functions.php" where are declared all the functions I use here. offers_choice_DB() displays a scroll menu where I can select a database (actually, this function performs a MySQL query and echoes the result in a scroll menu). If the user selects a database, then $_POST['base_name_choice'] exists. And it does, because when i only work with PHP/HTML, all is doing fine.
My purpose is to allow the user to select a database, then for this database I'd like to display a second scroll menu that displays some tables from this DB. This scroll menu will only be displayed if a POST value has been set. The offers_choice_table($_POST['base_name_choice']) function takes this value as an argument, then echoes the HTML for the scroll menu, containing the tables. Here we are !
Oh, and the submit button is not important here, because I want to have my second scroll menu displayed before the user clicks on the submit button, so we just disregard the target page, ok ?
Before, everything was OK : I used tests, conditions (isset...) but it was not dynamic, I had to call other pages, ...etc. And now I want, as you guessed, to use jQuery to refresh mypage.php as soon as the user selects a database so that an extra menu appears.
I started to listen to a change in my scroll menu, but then I don't know what to do to refresh my page with a POST parameter containing the selected database. Anyway, here is my code :
<script type="text/javascript">
$( '#base_name_choice' ).change(function() {
var val = $(this).val(); //here I retrieve the DB name
alert( val ); //I notify myself to check the value : it works
$.ajax({
type: 'POST', //I could have chosen $.post...
data: 'base_name_choice='+val, //I try to set the value properly
datatype: 'html', //we handle HTML, isn't it ?
success: function(){
alert( "call ok" ); //displays if the call is made
//and here...???? I don't know whatto do
}
})
});
</script>
Here it is...any help will be appreciated ! Thanks :)
Regards
The issue here is that your page is rendered first (html + php preprocessing). That means that once your page is rendered, you won't be able to make direct php method calls such as offers_choice_table or change the $_POST parameters.
How you normally do this, is by making an AJAX call from your javascript to a PHP script/method which than generates the second menu based on the parameter that the user chose.
So, you don't need this part:
if (isset($_POST['base_name_choice'])){
offers_choice_table($_POST['base_name_choice']);
}
because you will call "offers_choice_table" method with an ajax call.
You make the ajax call to a url which will return the second menu
$.ajax({
type: "GET",
url: "generate_second_menu.php",
data: { base_name_choice: val},
success: function(data){
alert( "call ok" ); //displays if the call is made
// here you can append the data that is returned from your php script which generates the second menu
$('form').append(data);
}
})
You should use GET instead of POST, in your new PHP file:
if (isset($_GET['base_name_choice'])) {
offers_choice_table($_GET['base_name_choice']);
}
You SHOULD check if the variable has a value set, especially if your function is expecting one. You can use whatever functions you want in this file, it is like any other PHP file you would write. Include any files that you want.
You should make sure to avoid SQL injection when using a GET or POST value in a query: How can I prevent SQL injection in PHP?
In the .ajax() call, there is a callback function success, this runs if the server response is good (i.e. not a 404 or 500). There first parameter in that function is what the server returned. In your case, it would be the HTML you echoed out. You can use jQuery to append the response to an element:
$.ajax({
type: "GET",
url: "generate_second_menu.php",
data: { base_name_choice: val},
success: function(data) {
// it looks like you're only returning the options of the select, so we'll build out the select and add them
var $newSelect = $('<select></select>');
$newSelect.append(data);
// and then add the select to the form (you may want to make this selector for specific)
$('form').append($newSelect);
}
});
I am echoing out a list of content from my database and have a star icon next to each item. I tried to make it so that when the user clicks the star it will find that content in my database and change the 'popular' value in that field (a 'like' button). I have been trying to do this by having each star wrapped in a tag and then making the id of each form 'formWithId'x ...... x being the id of the content that is being displayed.
Here it is in code:
echo "<form method = 'post' class ='formId".$row['id']."' id = 'likePostWithId".$row['id']."'>";
//all inputs are hidden and send the 'id' value of the content so the php file will know which field of the table to like in the mysql database
echo "<input type ='text' name = 'email' style = 'display: none;' value= '".$row['email']."' >
<input type ='text' name = 'id' style = 'display: none;' value= '".$row['id']."' >
//this is the star icon
<i class = 'ss-icon' id = '".$row['id']."'>⋆ </i>";
echo "</form>";
So, now what i need is to be able to send the form by using jquery validate.... but the problem i am having is that I have always used jquery validation (using the submit handler) to submit specific forms whose id's are predefined and dont change based on the id of the content sent from the database.
Here is my example that i am trying to
$(document).ready(function(){
$(".ss-icon").click(function(){
var id= $(this).attr('id');
$("#likePostWithId"+id).submit();
});
});
$(function(){
$(".ss-icon").click(function(){
//this doesn't work...
$("#likePostWithId").validate({
submitHandler:function(data){
//do stuff
}
)};
)};
see where $("#likePostWithId")... i dont know how to make it choose the one with the correct id (example $("#likePostWithId"+id)) would interpret as #likePostWithId6 or $likePostWithId7 where the number is the id of the mysql content
anyway i have no clue if this will be clear to any of you... I don't really know how to ask this question.. But basically i need to send an id (the id of the content from mysql) to php form without reloading page
any help is appreciated.
thanks
You may achieve this using jQuery's Ajax. Simply give each star icon a unique id (this id should have some relation with the MySQL data, like index key). Pass this argument (the id) using jQuery's Ajax function to a php script which takes the data (unique for each MySQL entry, like index) which changes it's popularity field entry. On success function, you may give some animation or color change to star icon informing that the data is successfully posted. just give it a try.
try this
$(function(){
$(".ss-icon").click(function(){
var id= $(this).attr('id');
$("#likePostWithId" + id ).validate({ <--- here
submitHandler:function(data){
$("#likePostWithId"+id).submit();
}
)};
)};
this however submits the form... you can use ajax to submit the form wihtout reloading
Maybe you can use $.post() to solve your problem. Then you don't have to mess around with forms:
$(document).ready(function(){
$(".ss-icon").click(function(){
$.post('yourPHPUrl', { contentID: $(this).attr('id') }, successFunction);
});
});
function successFunction(data) {
if (data.success) { //just a sample, you can give a JSON-Object back to the success function
$('#' + data.ID).css() ... // do something with the clicked star
}
}
Your code:
$(function(){
$(".ss-icon").click(function(){
//this doesn't work...
$("#likePostWithId").validate({
submitHandler:function(data){
//do stuff
}
)};
}); // <-- this was missing
)};
You were missing a set of brackets but you still shouldn't implement it like that. You are merely re-initializing the plugin on every click. (The plugin already has the click event handler built-in and it's captured automatically.)
.validate() must be called once within the DOM ready to initialize the plugin. As per documentation, the submitHandler callback function is "the right place to submit a form via Ajax after it validated."
$(function(){
$("#likePostWithId").validate({ // initialize plugin
// other options & rules,
submitHandler: function(form){
// do ajax
return false; // blocks form redirection since you already did ajax
}
)};
)};
Play with a demo: http://jsfiddle.net/XnfDc/
On my website I have an update form that when submitted alters my database.
What I want to do however is have all the elements that contain the old values,
AJAX
// UPDATE
$(".bill-upd-submit").click(function() {
var elem = $(this);
$.post("update_bill.php", elem.parent(".bill-upd").serialize(), function(data) {
//INSERT INTO OLD VALUES
});
});
PHP
In my PHP I've tried the following with no luck, has anybody done this before or can give me advice on where I may be going wrong?
echo "elem.closest('li').find('.cost').html('');";
You do not need to have the server provide the jQuery code. You can just type it into the callback function, where you have your //INSERT INTO OLD VALUES-comment.
$.post("update_bill.php", elem.parent(".bill-upd").serialize(), function(data) {
//We will use the "data"-variable that was passed to this function by jQuery.
elem.closest('li').find('.cost').html(data);
});
In the above code, the "data"-variable contains that which was given back by the server. So for that solution, make sure that "update_bill.php" will return the HTML that you want to insert into 'li .cost'.
Is it possible to get an element id into a PHP variable?
Let's say I have a number of element with IDs:
<span id="1" class="myElement"></span>
<span id="2" class="myElement"></span>
How do I get this into a PHP variable in order to submit a query. I suppose I would have to resubmit the page, which is OK. I would like to use POST. Can I do something like:
<script language="JavaScript">
$(document).ready(function(){
$(".myElement").click(function() {
$.post("'.$_SERVER['REQUEST_URI'].'", { id: $(this).attr("id") });
});
});
</script>
I need to pass $(this).attr('id') into $newID in order to run
SELECT * from t1 WHERE id = $newID
jQuery is a very powerful tool and I would like to figure out a way to combine its power with server-side code.
Thanks.
This is like your question: ajax post with jQuery
If you want this all in one file (posting to active file) here is what you would need in general:
<?php
// Place this at the top of your file
if (isset($_POST['id'])) {
$newID = $_POST['id']; // You need to sanitize this before using in a query
// Perform some db queries, etc here
// Format a desired response (text, html, etc)
$response = 'Format a response here';
// This will return your formatted response to the $.post() call in jQuery
return print_r($response);
}
?>
<script type='text/javascript'>
$(document).ready(function() {
$('.myElement').click(function() {
$.post(location.href, { id: $(this).attr('id') }, function(response) {
// Inserts your chosen response into the page in 'response-content' DIV
$('#response-content').html(response); // Can also use .text(), .append(), etc
});
});
});
</script>
<span id="1" class="myElement"></span>
<span id="2" class="myElement"></span>
<div id='response-content'></div>
From here you can customize the queries and response and what you would like to do with the response.
You have two "good" choices in my mind.
The first is to initiate a post request every time the ordering changes. You might be changing the ordering using jQuery UI sortable. Most libraries that support dragging and dropping also allow you to put an event callback on the drop simply within the initialization function.
In this even callback, you'd initiate the $.post as you have written it in your code (although I would urge you to look up the actual documentation on the matter to make sure you're POSTing to the correct location).
The second strategy is to piggyback on a form submission action. If you're using the jQuery Form Plugin to handle your form submissions, they allow you to indicate a before serialize callback where you can simply add into your form a field that specifies the ordering of the elements.
In both cases, you'd need to write your own function that actually serializes the element IDs. Something like the following would do just fine (totally untested; may contain syntax errors):
var order = [];
$('span.myElement').each(function(){
// N.B., "this" here is a DOM element, not a jQuery container
order.push(this.id);
});
return order.join(',');
You're quite right, something along those lines would work. Here's an example:
(btw, using $.post or $.get doesn't resubmit the page but sends an AJAX request that can call a callback function once the server returns, which is pretty neat)
<script language="JavaScript">
$(document).ready(function(){
$(".myElement").click(function() {
$.post(document.location, { id: $(this).attr("id") },
function (data) {
// say data will be some new HTML the server sends our way
// update some component on the page with contents representing the element with that new id
$('div#someContentSpace').html(data);
});
});
});
</script>
Your approach looks perfectly fine to me, but jQuery does not have a $_SERVER variable like PHP does. The url you would want to provide would be window.location (I believe an empty string will also work, or you can just specify the url on your own). You seem to be sending the ID just fine, though, so this will work.
If you want the page to react to this change, you can add a callback function to $.post(). You can do a variety of things.
$.post(window.location, {id: this.id}, function (data) {
//one
location.reload();
//two
$("#responsedata").html(data);
//three
$("#responsedata").load("affected_page.php #output");
});
I think number 2 is the most elegent. It does not require a page reload. Have your server side php script echo whatever data you want back (json, html, whatever), and it will be put in data above for jQuery to handle however you wish.
By the way, on the server side running the query, don't forget to sanitize the $id and put it in quotes. You don't want someone SQL Injecting you.