I know a bit php, ajax & javascript (and jQuery) - but since I am a bit of a begginer I need a little help to connect the dots for this easy task:
I want to let the user enter sometext (lets say: "I saw the sun, the sun is so nice and shiny") and I want to send this text to the server and then replace the word "sun" with "moon") - send it back to the user and write to him: "I saw the moon, the moon is so nice and shiny").
I just need help with understanding:
How do I send the text to the server using php.
How do I manipulate it and sends it back to the user.
I am sure there is a tutorial for it - I just didn't know how to look for it so I am asking here - to get a good start.
<?php
if (isset($_POST['text'])) //only execute if some text were sent to this script
{
echo str_replace('sun','moon',$_POST['text']); //manipulate the text and print it
die(); // stop script execution
}
?>
<html>
<head>
<script>
$(document).ready(function(){//when document finished to load
$('#send').click(function(){//when user clicked on send button
var text = $('#text').val();//get the text from input field
$.post('',{text:text},function(data){ // send the text to the current script as a post variable called text
alert(data); // when we received the response display it in alert window
});
});
});
</script>
</head>
<body>
<input type="text" id="text" />
<input type="button" value="send" id="send" />
</body>
</html>
no ajax needed
<?php
if ($_POST){
if ($_POST['name']){
echo str_replace('sun', 'moon', $_POST['name']);
}
}
?>
<form method="post" action="">
<input type="text" name="text" />
<input type="submit" />
</form>
if you want ajax, do this
<?php
if ($_POST){
if ($_POST['name']){
echo str_replace('sun', 'moon', $_POST['name']);
}
}else{
?>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js" type="text/javascript" charset="utf-8"></script>
<script>
$("form").submit(function(){
$.post('<?php echo $_SERVER['PHP_SELF']; ?>', {name:$("#name").val()},
function(html){
$("body").append(html);
}
);
return false;
</script>
<body>
<form method="post" action="">
<input type="text" id="name" name="name" />
<input type="submit" />
</form>
</body>
<?php
}
?>
Google is your friend. I searched for "ajax for beginners jquery php". There are hundreds, thousands of resources to walk you through this very thing. Here's one tutorial that I found:
http://www.devirtuoso.com/2009/07/beginners-guide-to-using-ajax-with-jquery/
Try to google for ajax and jquery tutorials. Basically it would work something like this:
Get user text input and call a javascript function that will send that text to process.php page
//input page
function getAjax(text){
var $data = "sentText="+ text +";
$.ajax({
type: "GET",
dataType: 'text',
url: "process.php",
data: $data,
success: function(output){
$('#responseDiv').html(output); //output is the text echoed from the php page
}
});
}
on the process.php page you take that text from a $_GET variable
//process.php
$sentText = $_GET['sentText'];
echo $sentText.' - some new text added';
Related
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.
<?php
'<form method="post" action="postnotice.php");>
<p> <label for="idCode">ID Code (required): </label>
<input type="text" name="idCode" id="idCode"></p>
<p> <input type="submit" value="Post Notice"></p>
</form>'
?>
Alright, so that's part of my php form - very simple. For my second form (postnotice.php):
<?php
//Some other code containing the password..etc for the connection to the database.
$conn = #mysqli_connect($sql_host,$sql_user,$sql_pass,$sql_db);
if (!$conn) {
echo "<font color='red'>Database connection failure</font><br>";
}else{
//Code to verify my form data/add it to the database.
}
?>
I was wondering if you guys know of a simple way - I'm still quite new to php - that I could use to perhaps hide the form and replace it with a simple text "Attempting to connect to database" until the form hears back from the database and proceeds to the next page where I have other code to show the result of the query and verification of "idCode" validity. Or even database connection failure. I feel it wrong to leave a user sitting there unsure if his/her button click was successful while it tries to connect to the database, or waits for time out.
Thanks for any ideas in advance,
Luke.
Edit: To clarify what I'm after here was a php solution - without the use of ajax or javascript (I've seen methods using these already online, so I'm trying to look for additional routes)
what you need to do is give form a div and then simply submit the form through ajax and then hide the div and show the message after you get the data from server.
<div id = "form_div">
'<form method="post" id = "form" action="postnotice.php";>
<p> <label for="idCode">ID Code (required): </label>
<input type="text" name="idCode" id="idCode"></p>
<p> <input type="submit" value="Post Notice"></p>
</form>'
?>
</div>
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<script>
$(function () {
$('form').on('submit', function (e) {
$.ajax({
type: 'post',
url: 'postnotice.php',
data: $('form').serialize(),
success: function (data) {
if(data == 'success'){
echo "success message";
//hide the div
$('#form_div').hide(); //or $('#form').hide();
}
}
});
e.preventDefault();
});
});
</script>
postnotice.php
$idCode = $_POST['idCode'];
// then do whatever you want to do, for example if you want to insert it into db
if(saveSuccessfulIntoDb){
echo 'success';
}
try using AJAX. this will allow you to wait for a response from the server and you can choose what you want to do based on the reponse you got.
http://www.w3schools.com/ajax/default.ASP
AJAX with jQuery:
https://api.jquery.com/jQuery.ajax/
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();
$.ajax({
url: 'www.xyz.com/index.php?' + formData,
type: 'POST',
data:{
},
success: function(data){},
error: function(data){},
})
I have this code.
<html>
<head>
<script src="http://code.jquery.com/jquery-latest.js"></script>
</head>
<body>
<form action = "" method = "POST" id = "form">
<img src = "circle.gif" id="loading" style="display:none; "/>
<input type="text" name = "text2">
<input type="submit" name="submit2" value="Send">
</form>
<?
if (isset($_POST['submit2'])){
echo $_POST['text2'];
}
?>
<script>
$('#form').submit(function(e) {
$('#loading').show();
return false;
});
</script>
</body>
</html>
I want to store in my db the value written in the textbox using PHP, and while it's being saved, I want to show a gif using jQuery, once the page is loaded, this gif should be removed.
Then, If I don't comment nothing, gif appears when submit button is submitted but echo fails.
If I comment the jQuery script, PHP echoes the vale written.
If I comment the PHP script, gif is shown but no echo of course...
How could I do what i'm asking. I know that my full script does until only showing the gif, but this without this I can't continue.
You can achieve your desired behaviour, but you need to do it by submitting an AJAX request to the server and then handling the return value. So basically you'd add this ajax request to the click or submit event of the form, and handle the behaviour and request via javascript.
Perhaps something like this:
<html>
<head>
<script src="http://code.jquery.com/jquery-latest.js"></script>
</head>
<body>
<form action = "formSubmitHandler.php" method = "POST" id = "form">
<img src = "circle.gif" id="loading" style="display:none; "/>
<input type="text" name = "text2">
<input type="submit" name="submit2" value="Send">
</form>
<script>
$(document).ready(function(){
$('#form').submit(function(){
// Show the loading icon before the processing begins
$('#loading').show();
// Send the query/form details to the server
jQuery.ajax({
data: $(this).serialize(),
url: this.action,
type: this.method,
success: function(results) {
// Now that the processing has finished, you
// can hide the loading icon
$('#loading').hide();
// perhaps display some other message etc
}
})
return false;
});
});
</script>
</body>
</html>
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>