Hi iam learning jquery with php. I created very small php and jquery code to getting value but it's not working. I check console but not giving any information i have referred jquery api documentation same process i followed but no use. How can i solve this.
<script>
$(document).ready(function(){
$("#submit").click(function(){
var term= $("#sterm").val();
$.post('oopsdata.php', {nwterm: term}, function(data){
("#container").html(data);
});
});
});
</script>
<body>
<form class="" action="" method="post">
<input type="text" name="" value="" id="sterm">
<input type="submit" name="" value="Search term" id="submit">
</form>
<div id="container">olddata</div>
PHP Code
<?php
$newterm = $_POST['nwterm'];
if($newterm == 'bio'){
echo "Request received This is the new data"
} else {
echo "no data received;"
}
?>
There are few issues with your code, such as:
You need to prevent your form from being submitted in the first place. Use jQuery's .preventDefault()
Missing $ before ("#container").html(data);
Based on the above two points, your jQuery code should be like this:
$(document).ready(function(){
$("#submit").click(function(event){
event.preventDefault();
var term= $("#sterm").val();
$.post('oopsdata.php', {nwterm: term}, function(data){
$("#container").html(data);
});
});
});
There are two syntax errors in your PHP code. Missing ; in both your echo statements.
So based on the above point, your PHP code should be like this:
<?php
$newterm = $_POST['nwterm'];
if($newterm == 'bio'){
echo "Request recieved This is the new data";
}else{
echo "no data recieved";
}
?>
Related
I posted a smaller to this yesterday and was shown how to do this but it doesn't work and the user never got back to me and I have been working on the same problem for hours.
I am trying to post a checkbox array from jQuery to php, when I run my code nothing seems to happen and when I try var_dump($_POST) this is all I get
Using this question as a reference, it seems that jQuery doesn't handle arrays too well. You can use to snippet from the accepted answer and it should work just fine.
serialize().replace(/%5B%5D/g, '[]')
Change the submit to a button or better use the form's submit event
why data-type html?
Your php does not seem to react to the serialised data but returns a button...
try my code here: http://plungjan.name/SO/sport.php
I am not unravelling the check box array - that is up to you
<?PHP
if (isset($_POST['saved'])) {
echo "saved"; exit(0);
}
else if (isset($_POST['Submit'])) {
echo var_dump($_POST["sport"]); exit(0);
}
?>
<!DOCTYPE html>
<html>
<head>
<title>Sports quiz</title>
<script src="https://code.jquery.com/jquery-3.2.1.min.js"></script>
<script>
$(function() {
$('#myForm').on("submit", function(ev) {
ev.preventDefault(); // cancel submit
var $form = $(this);
if ($("[type=checkbox]:checked").length ==0) {
alert("Please check one or more");
return false;
}
var formData = $form.serializeArray();
formData.push({name:"Submit",value:"submit"}); // note I changed the name from submit to Submit
$.post('sport.php',formData, function(data) {
console.log("Data",data);
if (confirm('You want to save \n' + data + ' as your sport?')) {
formData = $form.serializeArray();
formData.push({name:"saved",value:"saved"});
$.post('sport.php',formData,function(data) {
console.log("Saved Data",data);
});
}
});
});
});
</script>
</head>
<body>
<form id="myForm">
<input type="checkbox" name="sport[]" value="Football">Football<br>
<input type="checkbox" name="sport[]" value="Rugby">Rugby<br>
<input type="checkbox" name="sport[]" value="Golf">Golf<br>
<input type="checkbox" name="sport[]" value="Basketball">Basketball<br>
<br> <input type="submit" class="btn btn-info" name="Submit" value="submit">
</form>
</body>
</html>
Updated: It still not work after I add "#".
I am new to ajax. I am practicing to send value to php script ,and get result back.
Right now, I met one issue which I can not show my result in my html page.
I tried serves answers online, but I still can not fix this issue.
My index.html take value from the form and send form information to getResult.php.
My getResult.php will do calculation and echo result.
How do I display result into index.html?
Hers is html code
index.html
<html>
<body>
<form name="simIntCal" id="simIntCal" method="post"
>
<p id="Amount" >Amount(USD)</p>
<input id="amount_value" type="text" name="amount_value">
<p id="annual_rate" >Annual Rate of Interest
(%)</p>
<input id="rate_value" type="text" name="rate_value">
<p id="time_years" >Time (years)</p>
<input id="time_value" type="text" name="time">
<input id="calculate" type="submit" value="Calculate">
</form>
<p id="amount_inteCal" >The Amount (Acount
+ Interest) is</p>
<input id="result" type="text">
</body>
</html>
ajax script :
<script>
$('#simIntCal').on('submit', function (e) {
e.preventDefault();
$.ajax({
type: 'post',
url: 'getResult.php',
data: $('#simIntCal').serialize(),
success: function (result) {
$("#result").text(result);// display result from getResult.php
alert('success');
}
});
});
</script>
getResult.php
<?php
if ($_SERVER ["REQUEST_METHOD"] == "POST") {
//do some calculation
$result=10;//set result to 10 for testing
echo $result;
}
?>
You are missing the '#' in front of your css selector for result.
$("result").text(result);// display result from cal.php
Should be
$("#result").text(result);// display result from cal.php
index.php
----php start---------
if(isset($_POST['name'])){
echo 'Thank you, '.$_POST['name']; exit();
}
----php end ---------
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.0/jquery.min.js"></script>
<script>
function test(){
var formDATA = {'name': $('#input_name').val()}
$.ajax({
type: 'POST',
url: 'index.php',
data: formDATA,
success: function(response){
$('#result').html();
}
});
}
</script>
<input id="input_name" type="text" value="">
<button onclick="test();">Test Ajax</button>
<div id="result"></div>
Try something simple, this is a very basic version of ajax and php all in one page. Since the button triggers the function you don't even need a form (doesn't mean you shouldn't use one). But i left it simple so you could follow everything.
Sorry when i added php open and closing tags it didn't show up as code.
Also don't forget to include your jquery resources.
In your html file where you want the result to display, you probably want to be using a div.
Currently your code is using an input field:
<input id="result" type="text">
And what you probably want is something like this:
<div id="result"></div>
Unless you were intending to have the result show up in an input field inside your form, and in that case, the input field isn't actually inside your form code.
Because of my web style, i don't want to use input & textarea and get information by using $_POST[] and i need to get information that is in DIV element.
For example , I want to get information in this :
<div class="mine" name"myname">
this is information that i want to get and put into database by PHP !
</div>
and :
$_POST[myname];
But i can't do it with $_POST , How can i do it ??
And if this method can't do this , do you know any other method to get information from DIV like this ?
you can call a onsubmit function and make a hidden field at the time of form submission like this
HTML
need to give a id to your form id="my_form"
<form action="submit.php" method="post" id="my_form">
<div class="mine" name"myname">
this is information that i want to get and put into database by PHP !
</div>
<input type="submit" value="submit" name="submit" />
</form>
Jquery call on submit the form
$(document).ready(function(){
$("#my_form").on("submit", function () {
var hvalue = $('.mine').text();
$(this).append("<input type='hidden' name='myname' value=' " + hvalue + " '/>");
});
});
PHP : submit.php
echo $_POST['myname'];
You can use this method. First, with javascript get content of <div>
Code:
<script type="text/javascript">
var MyDiv1 = Document.getElementById('DIV1');
</script>
<body>
<div id="DIV1">
//Some content goes here.
</div>
</body>
And with ajax send this var to page with get or post method.
You would need some JavaScript to make that work, e.g. using jQuery:
$.post('http://example.org/script.php', {
myname: $('.mine').text()
});
It submits text found inside your <div> to a script of your choosing.
You can use following structure;
JS:
$(document).ready(function() {
$("#send").on("click", function() {
$.ajax({
url: "your_url",
method: "POST",
data: "myname=" + $(".mine").text(),
success: function(response) {
//handle response
}
})
})
})
HTML:
<div class="mine" name"myname">
this is information that i want to get and put into database by PHP !
</div>
<input type="button" name="send" id="send" value="Send"/>
You can see a simulation here: http://jsfiddle.net/cubuzoa/2scaJ/
Do this in jquery
$('.mine').text();
and post data using ajax.
Put the content of DIV in a variable like below:
var x = document.getElementById('idname').innerHTML;
I Have an requirement to pass form data to php using ajax and implement it in php to calculate the sum , division and other arithmetic methods I am a new to ajax calls trying to learn but getting many doubts....
It would be great help if some one helps me out with this
index.html
<script type="text/javascript">
$(document).ready(function(){
$("#submit_btn").click(function() {
$.ajax({
url: 'count.php',
data: data,
type: 'POST',
processData: false,
contentType: false,
success: function (data) {
alert('data');
}
})
});
</script>
</head>
<form name="contact" id="form" method="post" action="">
<label for="FNO">Enter First no:</label>
<input type="text" name="FNO" id="FNO" value="" />
label for="SNO">SNO:</label>
<input type="text" name="SNO" id="SNO" value="" />
<input type="submit" name="submit" class="button" id="submit_btn" value="Send" />
</form>
In count.php i want to implement
<?php
$FNO = ($_POST['FNO']);
$SNO=($_post['SNO']);
$output=$FNO+$SNO;
echo $output;
?>
(i want to display output in count.php page not in the first page index.html)
Thanks for your help in advance.
You can use a simple .post with AJAX. Take a look at the following code to be able to acheive this:
$('#form').submit(function() {
alert($(this).serialize()); // check to show that all form data is being submitted
$.post("count.php",$(this).serialize(),function(data){
alert(data); //check to show that the calculation was successful
});
return false; // return false to stop the page submitting. You could have the form action set to the same PHP page so if people dont have JS on they can still use the form
});
This sends all of your form variables to count.php in a serialized array. This code works if you want to display your results on the index.html.
I saw at the very bottom of your question that you want to show the count on count.php. Well you probably know that you can simply put count.php into your form action page and this wouldn't require AJAX. If you really want to use jQuery to submit your form you can do the following but you'll need to specify a value in the action field of your form:
$("#submit_btn").click(function() {
$("#form").submit();
});
I have modified your PHP code as you made some mistakes there. For the javscript code, i have written completely new code for you.
Index.html
<html>
<head>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>
</head>
<body>
<form name="contact" id="contactForm" method="post" action="count.php">
<label for="FNO">Enter First no:</label>
<input type="text" name="FNO" id="FNO" value="" />
<label for="SNO">SNO:</label>
<input type="text" name="SNO" id="SNO" value="" />
<input type="submit" name="submit" class="button" id="submit_btn" value="Send" />
</form>
<!-- The following div will use to display data from server -->
<div id="result"></div>
</body>
<script>
/* attach a submit handler to the form */
$("#contactForm").submit(function(event) {
/* stop form from submitting normally */
event.preventDefault();
/* get some values from elements on the page: */
var $form = $( this ),
//Get the first value
value1 = $form.find( 'input[name="SNO"]' ).val(),
//get second value
value2 = $form.find( 'input[name="FNO"]' ).val(),
//get the url. action="count.php"
url = $form.attr( 'action' );
/* Send the data using post */
var posting = $.post( url, { SNO: value1, FNO: value2 } );
/* Put the results in a div */
posting.done(function( data ) {
$( "#result" ).empty().append( data );
});
});
</script>
</html>
count.php
<?php
$FNO = $_POST['FNO'];
$SNO= $_POST['SNO'];
$output = $FNO + $SNO;
echo $output;
?>
There are a few things wrong with your code; from details to actual errors.
If we take a look at the Javascript then it just does not work. You use the variable data without ever setting it. You need to open the browser's Javascript console to see errors. Google it.
Also, the javascript is more complicated than is necessary. Ajax requests are kind-of special, whereas in this example you just need to set two POST variables. The jQuery.post() method will do that for you with less code:
<script type="text/javascript">
$(document).ready(function(){
$("#form").on("submit", function () {
$.post("/count.php", $(this).serialize(), function (data) {
alert(data);
}, "text");
return false;
});
});
</script>
As for the HTML, it is okay, but I would suggest that naming (i.e. name="") the input fields using actual and simple words, as opposed to abbreviations, will serve you better in the long run.
<form method="post" action="/count.php" id="form">
<label for="number1">Enter First no:</label>
<input type="number" name="number1" id="number1">
<label for="number2">Enter Second no:</label>
<input type="number" name="number2" id="number2">
<input type="submit" value="Calculate">
</form>
The PHP, as with the Javascript, just does not work. PHP, like most programming languages, are very picky about variables names. In other words, $_POST and $_post are not the same variable! In PHP you need to use $_POST to access POST variables.
Also, you should never trust data that you have no control over, which basically means anything that comes from the outside. Your PHP code, while it probably would not do much harm (aside from showing where the file is located on the file system, if errors are enabled), should sanitize and validate the POST variables. This can be done using the filter_input function.
<?php
$number1 = filter_input(INPUT_POST, 'number1', FILTER_SANITIZE_NUMBER_INT);
$number2 = filter_input(INPUT_POST, 'number2', FILTER_SANITIZE_NUMBER_INT);
if ( ! ctype_digit($number1) || ! ctype_digit($number2)) {
echo 'Error';
} else {
echo ($number1 + $number2);
}
Overall, I would say that you need to be more careful about how you write your code. Small errors, such as in your code, can cause everything to collapse. Figure out how to detect errors (in jQuery you need to use a console, in PHP you need to turn on error messages, and in HTML you need to use a validator).
You can do like below to pass form data in ajax call.
var formData = $('#client-form').serialize();
$.ajax({
url: 'www.xyz.com/index.php?' + formData,
type: 'POST',
data:{
},
success: function(data){},
error: function(data){},
})
this a simple example in how to submit form using the Jquery form plugins and retrieving data using html format
html Code
<html>
<head>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.7/jquery.js"></script>
<script src="http://malsup.github.com/jquery.form.js"></script>
<script>
// prepare the form when the DOM is ready
$(document).ready(function() {
// bind form using ajaxForm
$('#htmlForm').ajaxForm({
// target identifies the element(s) to update with the server response
target: '#htmlExampleTarget',
// success identifies the function to invoke when the server response
// has been received; here we apply a fade-in effect to the new content
success: function() {
$('#htmlExampleTarget').fadeIn('slow');
}
});
});
</script>
</head>
<body>
<form id="htmlForm" action="post.php" method="post">
Message: <input type="text" name="message" value="Hello HTML" />
<input type="submit" value="Echo as HTML" />
</form>
<div id="htmlExampleTarget"></div>
</body>
</html>
PHP Code
<?php
echo '<div style="background-color:#ffa; padding:20px">' . $_POST['message'] . '</div>';
?>
this just work fine
what i need to know if what if i need to Serialize the form fields so how to pass this option through the JS function
also i want show a loading message while form processed
how should i do that too
thank you
To serailize and post that to a php page, you need only jQuery in your page. no other plugin needed
$("#htmlForm").submit(function(){
var serializedData= $("#htmlForm").serialize();
$.post("post.php", { dat: serializedData}, function(data) {
//do whatever with the response here
});
});
If you want to show a loading message, you can do that before you start the post call.
Assuming you have div with id "divProgress" present in your page
HTML
<div id="divProgress" style="display:none;"></div>
Script
$(function(){
$("#htmlForm").submit(function(){
$("#divProgress").html("Please wait...").fadeIn(400,function(){
var serializedData= $("#htmlForm").serialize();
$.post("post.php", { dat: serializedData},function(data) {
//do whatever with the response here
});
});
});
});
The answer posted by Shyju should work just fine. I think the 'dat' should be given in quotes.
$.post("post.php", { 'dat': serializedData},function(data) {
...
}
OR simply,
$.post("post.php", serializedData, function(data) {
...
}
and access the data using $_POST in PHP.
NOTE: Sorry, I have not tested the code, but it should work.
Phery library does this behind the scenes for you, just create the form with and it will submit your inputs in form automatically. http://phery-php-ajax.net/
<?php
Phery::instance()->set(array(
'remote-function' => function($data){
return PheryResponse::factory('#htmlExampleTarget')->fadeIn('slow');
}
))->process();
?>
<?php echo Phery::form_for('remote-function', 'post.php', array('id' => ''); ?> //outputs <form data-remote="remote-function">
Message: <input type="text" name="message" value="Hello HTML" />
<input type="submit" value="Echo as HTML" />
</form>
<div id="htmlExampleTarget"></div>
</body>
</html>