I'm building a simple forum on which I have a user details page with two text fields, one for the user's biography and another for his interests.
When the user clicks on the save icon, a handler on the jquery is suposed to call an ajax call to update the database with the new value of the biography/interests but the ajax call isn't being called at all and I can't figure it out since I don't find any problems with the code and would apreciate if someone could take a look at it.
this is the textarea:
<textarea rows="4" cols="50" id="biography" readonly><?php if($info['bio'] == "") echo "Não existe informação para mostrar";
else echo $info['bio']; ?></textarea>
Here is the icon the user clicks on:
<li style="display:inline;" class="infoOps-li"><img class="info-icons" id="save1" src="assets/icons/save.png" alt=""></li>
this is the jequery with the ajax call:
$("#save1").click(function(){
var bio = $("#biography").val();
alert(bio); //this fires up
$.ajax({
url:"assets/phpScripts/userBioInterest.php", //the page containing php script
type: "post", //request type,
dataType: 'json',
data: {functionName: "bio", info:bio},
success:function(result){
alert(result.abc); //this doesn't fire
}
});
$("#biography").prop("readonly","true");
});
I know that the jquery handler is being called correctly because the first alert is executed. The alert of the ajax success function isn't, so I assume that the ajax call isn't being processed.
On the php file I have this:
function updateBio($bio)
{
$user = $_SESSION['userId'];
$bd = new database("localhost","root","","ips-connected");
$connection = $bd->getConnection();
if($bio == "")
{
echo json_encode(array("abc"=>'empty'));
exit();
}
if($stmt = mysqli_prepare($connection,"UPDATE users SET biografia = ? WHERE user_id = ?"))
{
mysqli_stmt_bind_param($stmt,'si',$bio,$user);
mysqli_stmt_execute($stmt);
mysqli_stmt_close($stmt);
echo json_encode(array("abc"=>'successfuly updated'));
}
$bd->closeConnection();
}
if(isset($_POST['functionName']))
{
$function = $_POST['functionName'];
echo $function;
if(isset($_POST['info']))
$info = $_POST['info'];
if($function == "bio")
{
updateBio($info);
}
else if($function == "interest")
{
updateInterests($info);
}
}
Can anyone shed some light on why isn't the ajax call being called?
Thank you
EDIT: changed "function" to "functionName" in json data object as suggested.
A possible problem is dued to a wrong parsing of the PHP output (for example due to a PHP error). You are reading the output as JSON, so if the output is not a JSON, success callback will not be triggered.
$("#save1").click(function(){
var bio = $("#biography").val();
alert(bio); //this fires up
$.ajax({
url:"assets/phpScripts/userBioInterest.php",
type: "post", //request type,
dataType: 'json',
data: {function: "bio", info:bio},
success:function(result){
alert(result.abc); //this doesn't fire
},
error: function(result){
alert("An error has occurred, check the console!");
console.log(result);
},
});
$("#biography").prop("readonly","true");
});
Try with this code, and check if an error is printed to the console.
You can use complete too, check here: http://api.jquery.com/jquery.ajax/
Related
At this moment I am using laravel. In this context I am having a form which is successfully submitted by using ajax to a controller. and that controller make it to the database. But the problem is as the ajax is doing its job the whole page remain unmoved / unchanged after the submission even the database is updated.
Now what I want
I want to give feedback to the user that your post is successfully submitted there. or what I want to do in further, I want to refresh the section in which the post is collected from the database as this post can be retrieved from there. But by using ajax only.
So there is no need to collect the whole page or refresh.
here is my form structure
`
{{ Form::open(array('route' => array('questions.store'), 'class' => 'form-horizontal' )) }}
blah blah blaaa .......
<script type="text/javascript">
$(".form-horizontal").submit(function(e){
$(this).unbind("submit")
$("#ask").attr("disabled", "disabled")
var that = $(this),
url = that.attr('action'),
type = that.attr('method'),
data = {};
that.find('[name]').each(function(index, value){
var that = $(this),
name = that.attr('name'),
value = that.val();
data[name] = value;
});
$.ajax({
url: url,
type: type,
data: data,
success: function(response){
console.log(response);
}
});
return false;
});
</script>
{{ Form::close() }}
`
As it is very much visible that the post is updated through a route & controller I want to have another dive and a success message at this script to be displayed after the success of posting. I am looking for some professional structure using what there is minimal need to have interaction with the server side and give user a better page viewing experience.
Thanks a lot for helping me in this research.
I am not sure if I understand you well, but if you want to notify the user about the result of an ajax-called db update you need to have
a route for the ajax save db call - it should point to a method that does the db update.
the db update method should return some value indicating the success/failure of update (for example OK or FAIL)
the only result of calling the method will be just plain text page with OK or FAIL as body
fetch the result by ajax and inform user accordingly (after form submit button)
check out the below code for ajax call itself (inside the form submit handler) to see what I mean
var db_ajax_handler = "URL_TO_YOUR_SITE_AND_ROUTE";
var $id = 1; //some id of post to update
var $content = "blablabla" //the cotent to update
$.ajax({
cache: false,
timeout: 10000,
type: 'POST',
tryCount : 0,
retryLimit : 3,
url: db_ajax_handler,
data: { content: $content, id: $id }, /* best to give a CSRF security token here as well */
beforeSend:function(){
},
success:function(data, textStatus, xhr){
if(data == "OK")
{
$('div.result').html('The new Question has been created');
}
else
{
$('div.result').html('Sorry, the new Question has not been created');
}
},
error : function(xhr, textStatus, errorThrown ) {
if (textStatus == 'timeout') {
this.tryCount++;
if (this.tryCount <= this.retryLimit) {
//try again
$.ajax(this);
return;
}
return;
}
if (xhr.status == 500) {
alert("Error 500: "+xhr.status+": "+xhr.statusText);
} else {
alert("Error: "+xhr.status+": "+xhr.statusText);
}
},
complete : function(xhr, textStatus) {
}
});
EDIT: as per comment, in step 2 (the method that is called with AJAX) replace
if($s)
{
return Redirect::route('questions.index') ->with('flash', 'The new Question has been created');
}
with
return ($s) ? Response::make("OK") : Response::make("FAIL");
EDIT 2:
To pass validation errors to the ajax-returned-results, you cannot use
return Response::make("FAIL")
->withInput()
->withErrors($s->errors());
as in your GIST. Instead you have to modify the suggested solution to work on JSON response instead of a plain text OK/FAIL. That way you can include the errors in the response and still benefit from the AJAX call (not having to refresh the page to retrieve the $errors from session). Check this post on the Laravel Forum for a working solution - you will get the idea and be able to fix your code.
I have an ajax call that sends data from a form to a php file that will then insert that data to the database. I put a call to die in said php file because I want to try something but it doesn't work.
addUserForm.php
<script>
$(document).ready(function () {
var $form = $('form');
$form.submit(function (event) {
event.preventDefault();
var formData = $form.serialize(),
url = $form.attr('action');
$.ajax({
type: "POST",
url: url,
data: formData,
success: function () {
//$("#div1").load("table.php");
alert('User Successfully Added');
document.getElementById("form1").reset();
}
});
});
});
</script>
Here is the php file:
addUser.php
<?php
include('sqlconnection.php');
die('here');
$firstname = $_POST['fname'];
$lastname = $_POST['lname'];
$middlename = $_POST['mname'];
$password = $_POST['pword'];
$username = $_POST['uname'];
$gender = $_POST['gender'];
$utype = $_POST['utype'];
$query = "INSERT INTO user (firstname,lastname,middlename,gender) VALUES ('$firstname','$lastname','$middlename','$gender')";
mysqli_query($con,$query);
$result = mysqli_query($con,"SELECT id FROM user WHERE firstname = '$firstname'");
$row = mysqli_fetch_assoc($result);
$uid=$row['id'];
$result = mysqli_query($con,"INSERT INTO accounts (u_id,username,password,account_type) VALUES ('$uid','$username',md5('$password'),'$utype');");
?>
Even when there is a die call in adduser.php it still alerts that the user was successfully added.
That's because die() only terminates/ends the PHP script. From an AJAX point of view the request was successful.
You should echo the info in the PHP page and then output the content of the response in your AJAX.
You could also set the response header in your PHP Script to something other than 200/OK, such as 401/Unauthorized or 400/Bad Request. Basically all 400 and 500 status codes indicate error.
Since the PHP code executes successfully even thou die(), the ajax will trigger the success and you will recevie the success message.
to stop your javascript at any point you can add return false;
In your success block
success: function () {
//$("#div1").load("table.php");
alert('User Successfully Added');
document.getElementById("form1").reset();
}
You can add these two line like this
success: function (data) {
/* These two lines*/
console.log(data);
return false;
//$("#div1").load("table.php");
alert('User Successfully Added');
document.getElementById("form1").reset();
}
So once you're done with your debugging you can remove those lines..
Die function doesn't stop javascript. It just stop PHP.
For exemple, if you add the die() function before inserting datas, datas will not be inserted but the success funciton will be executed and you will have alert.
If you want to execute the Error function, you have to add Throw exception or header 403 in the php file.
The jQuery success function just make sure the page is loaded. A die in PHP doesn't change that. You will have to check with returned data.
success
Type: Function( PlainObject data, String textStatus, jqXHR jqXHR )
A function to be called if the request succeeds. The function gets passed three arguments: >The data returned from the server, formatted according to the dataType parameter; a string >describing the status; and the jqXHR (in jQuery 1.4.x, XMLHttpRequest) object. As of jQuery >1.5, the success setting can accept an array of functions. Each function will be called in >turn. This is an Ajax Event.
ajax request php file, die ('here') is equivalent to echo 'here', return value, that successful implementation
Hi I’m quite new to jquery -ajax and I’d like some help please to join it with CI.
I have followed this tutorial on Submitting a Form with AJAX and I’d like to add this functionality to my CodeIgniter site. What I’d like to do is when the user submits the form, if there are any validation errors to show the individually on each input field (as in native ci process), or if this is not possible via validation_errors() function. If no errors occured to display a success message above the form.
Here's my code so far:
my view
// If validation succeeds then show a message like this, else show errors individually or in validation_errors() in a list
<div class="alert alert-success">Success!</div>
<?php echo validation_errors(); //show all errors that ajax returns here if not individualy ?>
<?php echo form_open('admin/product/add, array('class' => 'ajax-form')); ?>
<p>
<label for="product_name">Product *</label>
<input type="text" name="product_name" value="<?php echo set_value('product_name', $prod->product_name); ?>" />
<?php echo form_error('product_name'); ?>
</p>
<p>
<label for="brand">Brand</label>
<input type="text" name="brand" value="<?php echo set_value('brand', $prod->brand); ?>" />
<?php echo form_error('brand'); ?>
</p>
...
my controller
public function add($id){
// set validation rules in CI native
$rules = $this->product_model->rules;
$this->form_validation->set_rules($rules);
if ($this->form_validation->run() === true) {
// get post data and store them in db
$data = $this->input_posts(array('product_name', 'brand', 'category_id', 'description'));
$this->product_model->save($data, $id);
// no errors - data stored - inform the user with display success-div
} else {
// validation failed - inform the user by showing the errors
}
//load the view
$this->load->view('admin/products/add', $data);
}
and here’s the js script
$(document).ready(function () {
$('form.ajax-form').on('submit', function() {
var obj = $(this), // (*) references the current object/form each time
url = obj.attr('action'),
method = obj.attr('method'),
data = {};
obj.find('[name]').each(function(index, value) {
// console.log(value);
var obj = $(this),
name = obj.attr('name'),
value = obj.val();
data[name] = value;
});
$.ajax({
// see the (*)
url: url,
type: method,
data: data,
success: function(response) {
console.log(response); // how to output success or the errors instead??
}
});
return false; //disable refresh
});
});
How should I pass my validation results (either success or the post errors) throught the ajax request and display them on my view??
From some little research I did I've found that you can use a single controller, that holds both the native proccess and the ajax request (instead of using 2 controllers), but my main difficulty is, I don't understand how the results of the validation will pass through the js script and display them on my view?? Please note that I don't want to display anything on an alert box, instead show the results on a div or the errors individualy(if possible).
EDIT I did some changes to my application, here's the code so far:
the controller
public function manage($id = NULL){
$this->load->library('form_validation');
$data['categ'] = $this->category_model->with_parents();
//fetch a single product or create(initialize inputs empty) a new one
if (isset($id) === true) {
$data['prod'] = $this->product_model->get($id);
$data['attr'] = $this->attribute_model->get_by('product_id', $id, null, true);
} else {
$data['prod'] = $this->product_model->make_new();
$data['attr'] = $this->attribute_model_model->make_new();
}
if (isset($_POST['general_settings'])) {
if ($this->form_validation->run('product_rules') === true) {
// get post inputs and store them in database
$data = $this->product_model->input_posts(array('product_name', 'brand', 'category_id', 'general_description'));
$this->product_model->save($data, $id);
$status = true;
} else {
// validation failed
$status = validation_errors();
}
if ( $this->input->is_ajax_request() ) {
echo json_encode($status);
exit;
}
redirect('admin/product');
}
//if (isset($_POST['attributes_settings'])) { the same thing here }
// load the view
$this->load->view('admin/products/manage', $data);
}
and the js
success: function(response) {
//console.log(response);
if (data.status === true) {
$('#ajaxResults').addClass('alert alert-success').html(response);
} else {
$('#ajaxResults').addClass('alert alert-error').html(response);
};
}
But I'm having some issues
Although I get the error messages from validation_errors() as an alert-error when there are no errors I get the true in an alert-error too, insted of alert-success.
2.how should I return the success message too? eg. a message saying "Saves were done!".
Althought in a non-ajax-request the data are stored in the database, in case fo ajax the don't store. Any ideas What may be wrong???
HTML:
<div id="ajaxResults"></div>
Javascript ajax:
success: function(response) {
$('#ajaxResults').text(response);
}
this script you've wrote is only if the validation succeeds, right?
Wrong. The code in "success" gets executed any time you get a response back from the server (assuming the HTTP header is 200). Does your javascript knows if the server has any error for you? No.
You need your JavaScript to recognize if the validation failed or succeeded. You have many ways to do that. One of these could be sending the message to display followed by a 0 or 1.
So your PHP will looks like:
return "0 " . $errorMessage;
and
return "1 " . $successMessage;
and your javascript should then recognize, with if statement and substring, if the message starts with 0 or with 1.
Use this way i hope this will work for you
<script type='text/javascript'>
var base_url = '<?=base_url()?>';
function ajax_call()
{
var ids = $("#all_users").val();
$.ajax({
type:"POST",
url: base_url+"expense/home/get_expense",
data: "userid=" + ids,
success: function(result){
$("#your_div_id").html(result);
}
});
}
</script>
I want to pop up an alert box after checking whether some data is stored in the database. If stored, it will alert saved, else not saved.
This is my ajax function:
AjaxRequest.POST(
{
'url':'GroupsHandler.php'
,'onSuccess':function(creategroupajax){ alert('Saved!'); }
,'onError':function(creategroupajax){ alert('not saved');}
}
);
but now it show AjaxRequest is undefined.
How can I fix this?
This of course is possible using Ajax.
Consider the below sample code for the same.
Ajax call :
$.ajax({
url: 'ajax/example.php',
success: function(data) {
if(data == "success")
alert('Data saved.');
}
});
example.php's code
<?php
$bool_is_data_saved = false;
#Database processing logic here i.e
#$bool_is_data_saved is set here in the database processing logic
if($bool_is_data_saved) {
echo "success";
}
exit;
?>
function Ajax(data_location){
var xml;
try {
xml = new XMLHttpRequest();
} catch (err){
try {
xml = new ActiveXObject("Msxml2.XMLHTTP");
} catch (error){
try {
xml = new ActiveXObject("Microsoft.XMLHTTP");
} catch (error1){
//
}
}
}
xml.onreadystatechange = function(){
if(xml.readyState == 4 && xml.status == 200){
alert("data available");
}
}
xml.open("GET", data_location, true);
xml.send(null);
}
window.onload = function(){
Ajax("data_file_location");
}
You can create an addtitional table with date(time) of last update database and check if this date is later. You can use standard setInterval function for it.
This is possible using ajax. Use jQuery.ajax/pos/get to call the php script that saves the data or just checks if the data was saved previously (depends on how you need it exactly) and then use the succes/failure callbacks to handle its response and display an alert if you get the correct response.
Below code based on jQuery.
Try it
$.ajax({
type: 'POST',
url: 'http://kyleschaeffer.com/feed/',
data: { postVar1: 'theValue1', postVar2: 'theValue2' },
beforeSend:function(){
// this is where we append a loading image
$('#ajax-panel').html('<div class="loading"><img src="/images/loading.gif" alt="Loading..." /></div>');
},
success:function(data){
// successful request; do something with the data
$('#ajax-panel').empty();
$(data).find('item').each(function(i){
$('#ajax-panel').append('<h4>' + $(this).find('title').text() + '</h4><p>' + $(this).find('link').text() + '</p>');
});
},
error:function(){
// failed request; give feedback to user
$('#ajax-panel').html('<p class="error"><strong>Oops!</strong> Try that again in a few moments.</p>');
}
});
use the ajax to call the script and check values in the database through the script. If
data present echo success else not.lets look an example of it.
Assuming databasename = db
Assuming tablename = tb
Assuming tableColumn = data
Assuming server = localhost
Ajax:
$.ajax({
url: 'GroupsHandler.php',
success:function(data){
if(data=="saved")
{
alert("success");
}
}
});
Now in the myphpscript.php :
<?php
$Query = "select data from table";
$con = mysql_connect("localhost","user","pwd"); //connect to server
mysql_select_db("db", $con); //select the appropriate database
$data=mysql_query($Query); //process query and retrieve data
mysql_close($con); //close connection
if(!$empty(mysql_fetch_array($data))
{
echo "saved";
}
else
{
echo " not saved ";
}
?>
EDIT:
You must also include jquery file to make this type of ajax request.Include this at the top of your ajax call page.
<script src='ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js' type='text/javascript'></script>
First, decide whether to use POST or GET (I recommend POST) to pass AJAX data. Make a php file (ajax.php) such that it echos true or false after checking whether some data is stored in the database. You may test with a variable $your_variable = "some_data_to_check"; having a data inside and once you are finished, you may replace it with $your_variable = $_POST["ajaxdata"];.
Then in your page, set up AJAX using jQuery plugin like:
var your_data_variable = "data_to_send";
$.ajax({
type: "POST",
url: "ajax.php",
data: 'ajaxdata=' + your_data_variable,
success: function(result){
if(result == "true"){
alert("saved");
}else{
alert("not saved");
}
}
You may have a look at jQuery AJAX Tutorial, Example: Simplify Ajax development with jQuery.
I want to pop up an alert box after checking whether some data is stored in the database. If stored, it will alert saved, else not saved.
This is my ajax function:
AjaxRequest.POST(
{
'url':'GroupsHandler.php'
,'onSuccess':function(creategroupajax){ alert('Saved!'); }
,'onError':function(creategroupajax){ alert('not saved');}
}
);
but now it show AjaxRequest is undefined.
How can I fix this?
This of course is possible using Ajax.
Consider the below sample code for the same.
Ajax call :
$.ajax({
url: 'ajax/example.php',
success: function(data) {
if(data == "success")
alert('Data saved.');
}
});
example.php's code
<?php
$bool_is_data_saved = false;
#Database processing logic here i.e
#$bool_is_data_saved is set here in the database processing logic
if($bool_is_data_saved) {
echo "success";
}
exit;
?>
function Ajax(data_location){
var xml;
try {
xml = new XMLHttpRequest();
} catch (err){
try {
xml = new ActiveXObject("Msxml2.XMLHTTP");
} catch (error){
try {
xml = new ActiveXObject("Microsoft.XMLHTTP");
} catch (error1){
//
}
}
}
xml.onreadystatechange = function(){
if(xml.readyState == 4 && xml.status == 200){
alert("data available");
}
}
xml.open("GET", data_location, true);
xml.send(null);
}
window.onload = function(){
Ajax("data_file_location");
}
You can create an addtitional table with date(time) of last update database and check if this date is later. You can use standard setInterval function for it.
This is possible using ajax. Use jQuery.ajax/pos/get to call the php script that saves the data or just checks if the data was saved previously (depends on how you need it exactly) and then use the succes/failure callbacks to handle its response and display an alert if you get the correct response.
Below code based on jQuery.
Try it
$.ajax({
type: 'POST',
url: 'http://kyleschaeffer.com/feed/',
data: { postVar1: 'theValue1', postVar2: 'theValue2' },
beforeSend:function(){
// this is where we append a loading image
$('#ajax-panel').html('<div class="loading"><img src="/images/loading.gif" alt="Loading..." /></div>');
},
success:function(data){
// successful request; do something with the data
$('#ajax-panel').empty();
$(data).find('item').each(function(i){
$('#ajax-panel').append('<h4>' + $(this).find('title').text() + '</h4><p>' + $(this).find('link').text() + '</p>');
});
},
error:function(){
// failed request; give feedback to user
$('#ajax-panel').html('<p class="error"><strong>Oops!</strong> Try that again in a few moments.</p>');
}
});
use the ajax to call the script and check values in the database through the script. If
data present echo success else not.lets look an example of it.
Assuming databasename = db
Assuming tablename = tb
Assuming tableColumn = data
Assuming server = localhost
Ajax:
$.ajax({
url: 'GroupsHandler.php',
success:function(data){
if(data=="saved")
{
alert("success");
}
}
});
Now in the myphpscript.php :
<?php
$Query = "select data from table";
$con = mysql_connect("localhost","user","pwd"); //connect to server
mysql_select_db("db", $con); //select the appropriate database
$data=mysql_query($Query); //process query and retrieve data
mysql_close($con); //close connection
if(!$empty(mysql_fetch_array($data))
{
echo "saved";
}
else
{
echo " not saved ";
}
?>
EDIT:
You must also include jquery file to make this type of ajax request.Include this at the top of your ajax call page.
<script src='ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js' type='text/javascript'></script>
First, decide whether to use POST or GET (I recommend POST) to pass AJAX data. Make a php file (ajax.php) such that it echos true or false after checking whether some data is stored in the database. You may test with a variable $your_variable = "some_data_to_check"; having a data inside and once you are finished, you may replace it with $your_variable = $_POST["ajaxdata"];.
Then in your page, set up AJAX using jQuery plugin like:
var your_data_variable = "data_to_send";
$.ajax({
type: "POST",
url: "ajax.php",
data: 'ajaxdata=' + your_data_variable,
success: function(result){
if(result == "true"){
alert("saved");
}else{
alert("not saved");
}
}
You may have a look at jQuery AJAX Tutorial, Example: Simplify Ajax development with jQuery.