I want to make a form where you write a number send it with Jquery and is shown by PHP without refreshing the page ,by far i have this code but if preventDefault() is placed it will post but the content is not shown,also if i remove it, the content is shown but the page will pe refreshed.
Javascript:
$('#testut').unbind('submit').bind('submit',function(ev) {
ev.preventDefault();
$.ajax({
type: 'post',
url: 'incerc.php',
data: $('form').serialize(),
success: function () {
$('.success').fadeIn(500);
}
});
});
}
HTML:
Test
<br />
<?php
if(isset($_POST['test'])) {
$test = $_POST['test'];
$numar = $_POST['numar'];
echo 'Order name:'.$test.'<br />Number:'.$numar.'<br />';
} else {
echo 'No <br />';
}
?>
<div id='testut'>
<form id='test' method='post'>
<input type='text' name='numar'><br />
<input type='submit' name='test' onclick ='check()' value='Place'><br />
<span class='success' style='display:none'>Order was successfully placed.</span>
</form>
</div>
If I understand you correctly, you're making a dynamic ajax request and expecting your if(isset($_POST['test'])) to run on success. That won't work. That PHP code is static and will only be run to build the page. What you'll need to do is dynamically build the results after receiving a response from the POST operation.
The form response should return JSON, i.e.: { orderName: 'test name', numar: 1321321 }
Then your success function would look like (roughly):
function( data ){
$('.success')
.html( 'Order name:' + data.orderName + '<br />Number:' + data.numar + '<br />' )
.fadeIn(500);
}
Don't use unbind submit and bind submit, eventually it is of no use, 'cause the default action (form submitting on submit button click) is not been prevented in this way.
If you really trying to get data from another php file (incerc.php) to your page, you need to prevent your page from been reloaded by preventDefault() method inside your event handler:
$('#test').on('submit', function(ev){
ev.preventDefault();
var form = $(this);
$.post('incerc.php', form.serialize())
.done(function(data){
$("#form_result").html('Number:' + data + '<br />');
})
.fail(function(){
$("#form_result").html('No <br />');
});
$('.success').fadeIn(500);
});
And without page reload you can only insert response from $.post ajax call to some DOM element (#form_result in my example) of your page, to show it. Here is the HTML for my example:
<div id="form_result"></div>
<div id='testut'>
<form id='test' method='post'>
<input type='text' name='numar'><br />
<input type='submit' name='test' value='Place'><br />
<span class='success' style='display:none'>Order was successfully placed.</span>
</form>
</div>
Then you don't need to pass $_POST['test'] - value of your submit button - in ajax call to the incerc.php, 'cause it can be simply found in the DOM.
However, if you are trying to get data from the same page of yours, you will need the right PHP implementation of ajax response in the beginning:
if(isset($_POST['test']))
{
$test = $_POST['test'];
$numar = $_POST['numar'];
echo 'Order name:'.$test.'<br />Number:'.$numar.'<br />';
die();
}
...
I wanted to execute a query after the POST in php,the example was just for test, so i managed to do this(on example).
Javascript:
function check(){
$("#test").submit(function(e)
{ var numar = $('#numar').val();
var test = $('#testim').val();
var dataString = 'numar='+ numar + '&test=' + test;
$.ajax(
{
url : 'incerc.php',
type: "POST",
data : dataString,
success:function(data, textStatus, jqXHR)
{
$('.success').fadeIn(500);
$('body').load('incerc.php?'+dataString);
},
error: function(jqXHR, textStatus, errorThrown)
{
$('.success').html("Nu a mers");
}
});
e.preventDefault();
e.unbind();
});
}
HTML:
<?php if(isset($_GET['test'])){
$test = $_GET['numar'];
$numar = $_GET['numar'];
echo 'Order name:'.$test.'<br />Number:'.$numar.'<br />';}
else{echo 'No <br />';} ?>
<div id='testut'>
<form name='test' id='test' method='POST'>
<input type='text' id='numar' name='numar'><br />
<input type='submit' id='testim' name='test' onclick ='check()' value='Place'><br />
<span class='success' style='display:none'>Order was successfully placed.</span>
</form>
So now the body will load when the data is sent sucessfully , and will get the result correctly ,also in the adress bar it will still show 'incerc.php' not 'incerc.php?'+dataString.
Related
Tell me please, there is a form for sending data to the database. Without a script it works fine, but nothing happens with the script. In the console — Form Data has all the data, and the 200th code arrives, but is not added to the database.
PHP:
<?php
$data = $_POST;
if (isset($data['add'])) {
$posts = R::dispense('posts');
$posts->head = $data['head'];
$posts->desc = $data['desc'];
R::store($posts);
}
?>
HTML:
<form method="POST" id="FormID">
<input type="text" name="head" required />
<input type="text" name="desc" required />
<button type="submit" name="add">Добавить</button>
JS:
<script>
$("#FormID").submit(function(e)
{
var form = $(this);
var url = form.attr('action');
e.preventDefault();
$.ajax({
type: "POST",
url: url,
data: $("#FormID").serialize(),
success: function(data)
{
c = "hello";
$('#FormStatus').text(c);
}
});
});
</script>
You said:
if (isset($data['add'])) {
So the code only does anything if add in the data.
<button type="submit" name="add">Добавить</button>
add is a submit button. It will be included in the data when you submit the form.
data: $("#FormID").serialize(),
You aren't submitting the form. jQuery serialize does not include submit buttons because they aren't successful controls when you aren't submitting the form.
Use some other mechanism to determine if there is data to process (such as the presence of head and desc.
You have forget the action for your form
Why don't simply use $data['name'] instead of R::dispense?
If you what to do a POST request why don't you use $.post()?
What you need is these:
PHP Code:
<?php
$data = $_POST;
if (isset($data['add'])) {
if(isset($data['head']) AND !empty($data['head']) AND isset($data['desc']) AND !empty($data['desc'])) {
$head = htmlspecialchars($data['head']);
$desc = htmlspecialchars($data['desc']);
echo "Hello from server";
}
else {
echo "Please fill the form";
}
}
?>
HTML:
<form method="POST" id="FormID" action="path_to_php_file.php">
<input type="text" name="head" required />
<input type="text" name="desc" required />
<button type="submit" name="add">Добавить</button>
</form>
JS:
<script>
$("#FormID").submit(function(e)
{
e.preventDefault();
var form = $(this),
url = form.attr('action');
var data = {};
// To have post paramaters like
// { 'head' : 'Head value', 'desc' : 'Desc value' }
$.each(form.serializeArray(), function(i, field) {
data[field.name] = field.value;
});
$.post(url, data, function(Responses) {
// Will write "Hello from server" in your console
console.log(Responses);
});
});
</script>
Please, can somebody publish a mistakes corrected and tested code for my problem?
Program does - 22.php has the form. When the user enter and click Submit button, the result should be taken from 23.php and displayed in div on 22.php
I already tried solutions below and none of them solve my problem;
1) I changed to: $("#testform").submit(function(event){
2) I included "return false;" at the end to prevent it to actually submit the form and reload the page.
3) clear my browser cache
I can see what happen the program with my computer;
1) I do not get error message after I click submit.
2) I can see the tab of the page reloads quickly and the entered text fields are cleared.
3) No error message or result shows.
<html>
<head>
<title>My first PHP page</title>
<script type="text/javascript"
src="http://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<script>
$(document).ready(function() {
$("#btn").click(function(event){
event.preventDefault();
var myname = $("#name").val();
var myage = $("#age").val();
yourData ='myname='+myname+'&myage='+myage;
$.ajax({
type:'POST',
data:yourData,//Without serialized
url: '23.php',
success:function(data) {
if(data){
$('#testform')[0].reset();//reset the form
$('#result').val(data);
alert('Submitted');
}else{
return false;
}
};
});
});
});
</script>
</head>
<body>
<form method="post" id="testform">
Name:
<input type="text" name="name" id="name" />Age:
<input type="text" name="age" id="age" />
<input type="submit" name="submit" id="btn" />
</form>
<div id='result'></div>
</body>
</html>
<?php
if ( isset($_POST['name']) ) { // was the form submitted?
echo "Welcome ". $_POST["name"] . "<br>";
echo "You are ". $_POST["age"] . "years old<br>";
}
?>
you don't need to change your php code
try submit form with submit event ...
$("#testform").submit(function(event){
use `dataType:json`; in your ajax ..
yourData =$(this).serialize();
Your php
<?php
if ( isset($_POST['name']) ) { // was the form submitted?
$data['name']= 'welcome '.$name;
$data ['age']= 'you are '.$age;
print_r(json_encode($data));exit;
}
?>
Now In Your Success function
var message = data.name + ' ' + data.age;
$('#result').html(message );
You are sending myname and checking name(isset($_POST['name']) in php.
don't use .value() use .html() for data rendering. and console log the data and see whats request and response using firebug.
Can you try this one?
To be changed
var yourData ='name='+myname+'&age='+myage; // php is expecting name and age
and
$('#result').html(data); // here html()
the code becomes
$(document).ready(function() {
$("#btn").click(function(event){
event.preventDefault();
var myname = $("#name").val();
var myage = $("#age").val();
var yourData ='name='+myname+'&age='+myage; // php is expecting name and age
$.ajax({
type:'POST',
data:yourData,//Without serialized
url: '23.php',
success:function(data) {
if(data){
$('#testform')[0].reset();//reset the form
$('#result').html(data); // here html()
alert('Submitted');
}else{
return false;
}
}
});
});
});
Try formatting your post data like this inside your ajax function.
$.ajax({
type:'POST',
data : {
myname: myname
myage: myage
}
...
}
EDIT
Try removing the ; in
return false;
}
};
to
return false;
}
}
You can change at both end ajax and php:
#PHP:
You can check for correct posted data which is myname and myage not name and age.
<?php
if ( isset($_POST['myname'])) { // was the form submitted?
echo "Welcome ". $_POST["myname"] . "<br>";
echo "You are ". $_POST["myage"] . "years old<br>";
}
?>
or #Ajax:
yourData ='name='+myname+'&age='+myage;
//--------^^^^^^----------^^^^----change these to work without changing php
Just noticed the #result is an div element. So, you can't use .val() but use .html(), .append() etc:
$('#result').html(data);
I have a form with an input field for a userID. Based on the entered UID I want to load data on the same page related to that userID when the user clicks btnLoad. The data is stored in a MySQL database. I tried several approaches, but I can't manage to make it work. The problem is not fetching the data from the database, but getting the value from the input field into my php script to use in my statement/query.
What I did so far:
I have a form with input field txtTest and a button btnLoad to trigger an ajax call that launches the php script and pass the value of txtTest.
I have a div on the same page in which the result of the php script will be echoed.
When I click the button, nothing happens...
Test.html
<script type="text/javascript" src="http://code.jquery.com/jquery-1.4.2.min.js"></script>
<script type="text/javascript" src="http://ajax.microsoft.com/ajax/jquery.validate/1.7/jquery.validate.min.js"></script>
<script>
//AJAX CALL
function fireAjax(){
$.ajax({
url:"testpassvariable.php",
type:"POST",
data:{userID:$("#txtTest").val(),},
success: function (response){
$('#testDiv').html(response);
}
});
}
</script>
</head>
<body>
<form name="testForm" id="testForm" action="" method="post" enctype="application/x-www-form-urlencoded">
<input type="text" name="txtTest" id="txtTest"/>
<input type="button" id="btnLoad" name="btnLoad" onclick="fireAjax();"
<input type="submit" name="SubmitButton" id="SubmitButton" value="TEST"/>
</form>
<div id="testDiv" name="testDiv">
</div>
</body>
The submit button is to insert updated data into the DB. I know I have to add the "action". But I leave it out at this point to focus on my current problem.
testpassvariable.php
<?php
$player = $_POST['userID'];
echo $player;
?>
For the purpose of this script (testing if I can pass a value to php and return it in the current page), I left all script related to fetching data from the DB out.
As the documentation says 'A page can't be manipulated safely until the document is ready.' Try this:
<script>
$(document).ready(function(){
//AJAX CALL
function fireAjax(){
$.ajax({
url:"testpassvariable.php",
type:"POST",
data:{userID:$("#txtTest").val(),},
success: function (response){
$('#testDiv').html(response);
}
});
}
});
</script>
You need to correct two things:
1) Need to add $(document).ready().
When you include jQuery in your page, it automatically traverses through all HTML elements (forms, form elements, images, etc...) and binds them.
So that we can fire any event of them further.
If you do not include $(document).ready(), this traversing will not be done, thus no events will be fired.
Corrected Code:
<script>
$(document).ready(function(){
//AJAX CALL
function fireAjax(){
$.ajax({
url:"testpassvariable.php",
type:"POST",
data:{userID:$("#txtTest").val(),},
success: function (response){
$('#testDiv').html(response);
}
});
}
});
</script>
$(document).ready() can also be written as:
$(function(){
// Your code
});
2) The button's HTML is improper:
Change:
<input type="button" id="btnLoad" name="btnLoad" onclick="fireAjax();"
To:
<input type="button" id="btnLoad" name="btnLoad" onclick="fireAjax();"/>
$.ajax({
url: "testpassvariable.php",
type: "POST",
data: {
userID: $("#txtTest").val(),
},
dataType: text, //<-add
success: function (response) {
$('#testDiv').html(response);
}
});
add dataType:text, you should be ok.
You need to specify the response from the php page since you are returning a string you should expect a string. Adding dataType: text tells ajax that you are expecting text response from php
This is very basic but should see you through.
Change
<input type="button" id="btnLoad" name="btnLoad" onclick="fireAjax();"/>
Change AJAX to pass JSON Array.
data = $(this).serialize() + "&" + $.param(data);
$.ajax({
type: "POST",
dataType: "json",
url: "action.php",
data: data,
....
// action.php
header('Content-type: application/json; charset=utf-8');
echo json_encode(array(
'a' => $b[5]
));
//Connect to DB
$db = mysql_connect("localhst","user","pass") or die("Database Error");
mysql_select_db("db_name",$db);
//Get ID from request
$id = isset($_GET['id']) ? (int)$_GET['id'] : 0;
//Check id is valid
if($id > 0)
{
//Query the DB
$resource = mysql_query("SELECT * FROM table WHERE id = " . $id);
if($resource === false)
{
die("Database Error");
}
if(mysql_num_rows($resource) == 0)
{
die("No User Exists");
}
$user = mysql_fetch_assoc($resource);
echo "Hello User, your number is" . $user['number'];
}
try this:- for more info go here
$(document).ready(function(){
$("#btnLoad").click(function(){
$.post({"testpassvariable.php",{{'userID':$("#txtTest").val()},function(response){
$('#testDiv').html(response);
}
});
});
});
and i think that the error is here:-(you wrote it like this)
data:{userID:$("#txtTest").val(),}
but it should be like this:-
data:{userID:$("#txtTest").val()}
happy coding :-)
UPDATE:
This is the error:
412 (Precondition Failed)
I am trying to call a php script from ajax, I currently have the below ajax, which when the button in the form (also below) is clicked will call a php script passing it the form data, which will then be submitted to the database.
However, it is not working; and what's more I am just getting a blank error back, so I do not even know what is going wrong.
Could someon please point me in the right direction?
Thanks.
HTML form:
<form name="report-form" id="report-form" action="" method="POST">
<textarea id="reason-box" type="text" name="reason-box" cols="40" rows="5" maxlength="160" onkeypress=""></textarea>
<input id="reportedID" name="reportedID" type="text" />
<!--<input id="report-submit" value="" name="submit" type="submit" onclick="submitReport()"/> -->
<button id="report-submit" name="submit" onclick="submitReport()"></button>
</form>
AJax call:
function submitReport()
{
var ID=$('#reportedID').val();
var reason=$('#reason-box').val();
var formData = "ID="+ID+"&reason="+reason;
alert(formData);
//This code will run when the user submits a report.
$.ajax(
{
url: 'submit_report.php',
type: "POST",
data: formData,
success: function(data)
{
alert("Report Submitted!");
},
error: function(xhr,err)
{
alert(err.message);
alert("responseText: "+ xhr.responseText);
}
});
}
Now I have already tested the php script, and that works fine, the problem started when I added the ajax call so I know it is something to do with the ajax not the php.
This should correct the problem with submitting:
Your jQuery Ajax call won't succeed because the POST data isn't supplied in the correct format.
If the ajax should succeed the form is also posted resulting in a 405 error.
<button id="report-submit" name="submit" onclick="submitReport(event)"></button>
function submitReport(event)
{
event.preventDefault();
....... // your code
}
Now the default action of your form will be prevented (resulting in a 405 error). And only the ajax request is submitted.
In the button element we pass the event object on to the function. We use event.preventDefault() to make sure the button doesn't run it's default action, which is submitting the form.
You could also prevent this by deleting the form element as a wrapper, but maybe you want to use other features (like validation) on the form.
Form data in a jQuery ajax request needs to be an object called data:
var formData = {"ID" : ID, "reason" : reason};
jQuery will reform this to a correct query string for the submit.
I would do it like this:
<form name="report-form" id="report-form" action="" method="POST">
<textarea id="reason-box" type="text" name="reason-box" cols="40" rows="5" maxlength="160"></textarea>
<input id="reportedID" name="reportedID" type="text" />
<button id="report-submit" type="submit" name="submit" value="submit"></button>
</form>
<script type="text/javascript">
jQuery("document").ready(function(){
var $ = jQuery
$("form").submit(function(){
var data = "";
data = $(this).serialize() + "&" + $.param(data);
$.ajax({
type: "POST",
url: "submit_report.php",
data: data,
success: function(data)
{
alert("Report Submitted!");
},
error: function(xhr,err)
{
alert(err.message);
alert("responseText: "+ xhr.responseText);
}
});
return false;
});
});
</script>
and then use $reason=$_POST['reason-box']; and $ID=$_POST['reportedID']; inside your PHP script
this is optional to choose the form for submitting data or you can do it without the HTML form this is what i do
<textarea id="reasonbox" type="text" name="reason-box" cols="40" rows="5" maxlength="160" onkeypress=""></textarea>
<input id="reportedID" name="reportedID" type="text" />
<button id="report-submit" ></button>
and the using folloing javascript and jquery style
<script type="text/javascript">
$(function() {
$("#report-submit").click(function(){
try
{
$.post("your php page address goes here like /mypage.php",
{
//in this area you put the data that is going to server like line below
'reasonbox':$("#reason-box").val().trim(),
'reportedID':$("#reportedID").val().trim()
}, function(data){
data=data.trim();
//this is data is sent back from server you can send back data that you want
//like message or json array
});
}
catch(ex)
{
alert(ex);
}
});
});
</script>
I hope it helps
I followed a tutorial to adapt the code. Here I am trying trying to auto-populate my form fields with AJAX when an 'ID' value is provided. I am new to Jquery and can't get to work this code.
Edit 1 : While testing the code, Jquery isn't preventing the form to submit and sending the AJAX request.
HTML form
<form id="form-ajax" action="form-ajax.php">
<label>ID:</label><input type="text" name="ID" /><br />
<label>Name:</label><input type="text" name="Name" /><br />
<label>Address:</label><input type="text" name="Address" /><br />
<label>Phone:</label><input type="text" name="Phone" /><br />
<label>Email:</label><input type="email" name="Email" /><br />
<input type="submit" value="fill from db" />
</form>
I tried changing Jquery code but still I couldn't get it to work. I think Jquery is creating a problem here. But I am unable to find the error or buggy code. Please it would be be very helpful if you put me in right direction.
Edit 2 : I tried using
return false;
instead of
event.preventDefault();
to prevent the form from submitting but still it isn't working. Any idea what I am doing wrong here ?
Jquery
jQuery(function($) {
// hook the submit action on the form
$("#form-ajax").submit(function(event) {
// stop the form submitting
event.preventDefault();
// grab the ID and send AJAX request if not (empty / only whitespace)
var IDval = this.elements.ID.value;
if (/\S/.test(IDval)) {
// using the ajax() method directly
$.ajax({
type : "GET",
url : ajax.php,
cache : false,
dataType : "json",
data : { ID : IDval },
success : process_response,
error: function(xhr) { alert("AJAX request failed: " + xhr.status); }
});
}
else {
alert("No ID supplied");
}
};
function process_response(response) {
var frm = $("#form-ajax");
var i;
console.dir(response); // for debug
for (i in response) {
frm.find('[name="' + i + '"]').val(response[i]);
}
}
});
Ajax.php
if (isset($_GET['action'])) {
if ($_GET['action'] == 'fetch') {
// tell the browser what's coming
header('Content-type: application/json');
// open database connection
$db = new PDO('mysql:dbname=test;host:localhost;', 'xyz', 'xyz');
// use prepared statements!
$query = $db->prepare('select * from form_ajax where ID = ?');
$query->execute(array($_GET['ID']));
$row = $query->fetch(PDO::FETCH_OBJ);
// send the data encoded as JSON
echo json_encode($row);
exit;
}
}
I don't see where you're parsing your json response into a javascript object (hash). This jQuery method should help. It also looks like you're not posting your form using jquery, but rather trying to make a get request. To properly submit the form using jquery, use something like this:
$.post( "form-ajax.php", $( "#form-ajax" ).serialize() );
Also, have you tried adding id attributes to your form elements?
<input type="text" id="name" name="name"/>
It would be easier to later reach them with
var element = $('#'+element_id);
If this is not a solution, can you post the json that is coming back from your request?
Replace the submit input with button:
<button type="button" id="submit">
Note the type="button".
It's mandatory to prevent form submition
Javascript:
$(document).ready(function() {
$("#submit").on("click", function(e) {
$.ajax({type:"get",
url: "ajax.php",
data: $("#form-ajax").serialize(),
dataType: "json",
success: process_response,
error: function(xhr) { alert("AJAX request failed: " + xhr.status); }
});
});
});