I have some experience in JAVA GUI programming and I want to achieve the same in a PHP form.
Situation: I want to have a php form with a submit button. When the button is pressed an ActionEvent should be called to update another part of the form.
How to implement such a feature with HTML,PHP,JAVASCRIPT ?
Load latest version of jQuery:
<script type="text/javascript" src="http://code.jquery.com/jquery-latest.min.js"></script>
HTML code:
<form>
<button type="button" class="formLoader">Click button</button>
<div id="formContentToLoad"></div>
</form>
jQuery code:
<script type="text/javascript">
$(function(){
$(".formLoader").click(function(){
$("#formContentToLoad").load("scriptToRun.php");
});
});
</script>
Whatever markup you need to update in the form, can be put into scriptToRun.php
Use jQuery
Javascript
$(document).ready(function() {
$(".myForm").submit(function() {
$.ajax({
type: "POST",
url: "myForm.php",
data: $(this).serialize(),
success: function(response) {
// todo...
alert(response);
}
})
})
});
Html
<form method="POST" class="myForm">
<input type="text" id="a_field" name="a_field" placeholder="a field" />
<input type="submit" value="Submit" />
</form>
PHP
<?php
if(isset($_POST)) {
$a_field = $_POST["a_field"];
// todo..
}
If you want to use PHP and HTML to submit a form try this:
HTML Form
<form action="" method="post">
<input type="text" placeholder="Enter Name" name="name" />
<input type="submit" name="sendFormBtn" />
</form>
PHP
<?php
if(isset($_POST["sendFormBtn"]{
$name = isset($_POST["name"]) ? $_POST["name"] : "Error Response Here";
//More Validation Here
}
Related
I have an html form that loads the main portion of a document, postload an ajax request goes off and gets an xml file that is parsed out to create 'sub' forms which can be updated/submitted. This is the form 'preload'
<html>
<head>
<script src="jquery.js">
<script src="jquery.forms.js">
<script>
$(document).ready(function () {
//Script to execute when form is loaded
loadOrder(unid);
});
</script>
</head>
<body>
<form id="mainform" name="main" method="post" action="whatever">
<input type="hidden" id="unid" name="unid" value="123" />
</form>
<div id="orderForms">
</div>
</body>
</html>
Here is the form post load :
<html>...
<div id="orderForms">
<form id="order_1" name="order" method="post" action="whatever">
<input type="hidden" id="pid_1" name="pid" value="123" />
<input type="hidden" id="unid_1" name="unid" value="456" />
</form>
<form id="order_2" name="order" method="post" action="whatever">
<input type="hidden" id="pid_2" name="pid" value="123" />
<input type="hidden" id="unid_2" name="unid" value="789" />
</form>
</div>
</body>
</html>
JS code:
function loadOrders(unid){
var rUrl = "url";
$.ajax({type: "GET", url: rUrl, async: true, cache: false, dataType: "xml", success: postLoadOrders}) ;
}
function postLoadOrders(xml){
nxtOrder = 1;
var html="";
$('order',xml).each(function() {
// parses the xml and generates the html to be inserted into the <div>
});
$("#orderForms").html(html);
}
This all works, the main form loads, the 'hidden' forms in the <div> are written in. The trouble happens when I put a button on the main form that does this...
function submitOrder(){
$("#pid_1").val('555');
$("#order_1").formSerialize();
$("#order_1").ajaxSubmit();
}
If I alert($("#pid_1").val()) prior to the .val('555') it shows the original value, when I alert after, it shows the new value, however it submits the original value, and if I open the html in firebug the value isn't showing as changing.
If I put a hidden field into the main form, that exists when the document loads and change its value, not only does the new value post, it also shows as being changed when examining the source.
Any ideas?
$('order',xml).each(function() {
});
this is not Object in JQuery
You can edit:
$('[name=order]',xml).each(function() {
});
I have two questions:
How can I create a second form in the same page and ajax handle the right form depending on which submit button got clicked?
Now let's suppose I have multiple forms in the same page. I want to change where the output goes depending on which submit button is clicked.
Source code:
Ajax:
<script type="text/javascript">
$(document).ready(function(){
$("#myform").validate({
submitHandler: function(form) {
// do other stuff for a valid form
$.post('post.php', $("#myform").serialize(), function(data) {
$('#results').html(data);
});
}
});
});
</script>
Html:
<form id="myform" action="" method="POST">
<input type="text" name="a"><input type="submit" value="submit">
</form>
<div id="results"></div> <!-- post.php output goes here.-->
<!-- until here everything is working fine -->
<!-- now if I want to add a second form I don't know what id I should use -->
I would use a class on your forms to bind to, and then each form can have an element that holds the output id which you select -
javascript
<script type="text/javascript">
$(document).ready(function(){
$(".myforms").validate({
submitHandler: function(form) {
// do other stuff for a valid form
$.post('post.php', $(form).serialize(), function(data) {
var output = $(form).data('output');
$("#"+output).html(data);
});
}
});
});
</script>
html
<form class="myforms" action="" method="POST" data-output="results1">
<input type="text" name="a"><input type="submit" value="submit">
</form>
<div id="results1"></div>
<form class="myforms" action="" method="POST" data-output="results2">
<input type="text" name="a"><input type="submit" value="submit">
</form>
<div id="results2"></div>
<form class="myforms" action="" method="POST" data-output="results3">
<input type="text" name="a"><input type="submit" value="submit">
</form>
<div id="results3"></div>
i have form with one input and one submit button.
<form method='POST' action='' enctype='multipart/form-data' id="form_search">
<input type='hidden' name="action" id="form_1" value='1' />
</span><input id="query" type="text" name="mol" value="">
<input type='submit' value='Search' name="Search" id="Search" />
on form submission form input data goes to php below code
if (isset($_POST['Search'])) {
$_SESSION["query"] = $_POST["mol"];
$_SESSION["action"] = $_POST["action"];
}
i want to avoid page refresh on form submission. i tried e.preventDefault() and return false;
methods in my java script but not working(this methods helping me from page refresh but does not allowing me to send data to php code)
please help me out of this problem, please suggest working ajax code for this problem.
Page refresh will delete you previous data so to reserve it you can use $.post() or $.ajax()
You can prevent page refreshing by adding one of these two things in event handler function
for pure js
return false;
for jquery you can use
e.preventDefault(); // e is passed to handler
Your complete code will be something like
using $.post() in js
function checkfunction(obj){
$.post("your_url.php",$(obj).serialize(),function(data){
alert("success");
});
return false;
}
html
<input type='submit' onclick="return checkfunction(this)" />
or same effect with onsubmit
<form onsubmit="return checkfunction(this)" method="post">
Without ajax you can simply add the checked attribute in PHP. So for example if your radio group has the name radio and one has value a, the other b:
<?php
$a_checked = $_POST['radio'] === 'a';
$b_checked = $_POST['radio'] === 'b';
?>
<input type="radio" name="radio" value="a"<?=($a_checked ? ' checked' : '')?>></input>
<input type="radio" name="radio" value="b"<?=($b_checked ? ' checked' : '')?>></input>
So when a user submits the form and you display it again, it will be like the user submitted it even the page refreshes.
<input type="radio" name="rbutton" id="r1">R1
<input type="radio" name="rbutton" id="r2">R2
<input type="button" id="go" value="SUBMIT" />
<div id="result"></div>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$('#go').click(function(){
var val1 = $('input:radio[name=rbutton]:checked').val();
var datastring = "partialName="+val1;
$.ajax({
url: "search.php",
type: "POST",
data: datastring,
success: function(data)
{
$("#result").html(data);
}
});
});
});
</script>
I have the following php files. One that captures a user's text from a simple form, and the other echos this text. The files are shown below..
file1.php:
<form action="output.php" method="post">
Paste text document: <br><br>
<textarea name="doc" rows="5" cols="50" value="">
</textarea><br><br>
<input type="submit" name="submit" value="submit">
<input type="reset" name="reset" value="Clear">
</form>
<br><br>
<div id="output_area">
</div>
</body>
</html>
output.php:
<html lang="en">
<head><title>Output</title></head>
<body>
<?php
$doc = $_POST['doc'];
echo $doc;
?>
</body>
</html>
I want the output to be displayed between the DIV tags, rather than to load and show the output on a new page. I want the text to be output in between the DIV tags on the same page. Maybe the best way is to use AJAX to display output.php within the DIV tags and output the text after the user has clicked on the "submit" button. How can I use AJAX to do this in the most basic way possible?
You can use JQuery ajax function:
JQuery:
$("form").on("submit", function(event){
event.preventDefault();
$.ajax({
type: "POST",
url: "output.php",
data: { doc: $('#doc').val()}
})
.done(function( data) {
$("#output_area").html(data);
});
});
Change html:
<textarea name="doc" id="doc" rows="5" cols="50" value=""></textarea>
You can use the function:
$(document).ready(function() {
$( "#output_area" ).load( "output.php" );
});
But remeber!
That the load function from jquery, loads everything on the page you are asking for, so dont use:
<html lang="en">
<head><title>Output</title></head>
<body>
</body>
</html>
Now the load(function), will put in the PHP you made in output.php
if you want to output the answer on a submit i would do it like:
First add:
Show doc
Then add jquery:
$(document).ready(function() {
$('.pressme').on('click', function(){
var yourdata = $(this).parent().find("textarea[name='doc']").val();
$.ajax({
type: "POST",
url: "output.php",
data: {doc: yourdata},
success: function(data) {
$('#output_area').html(data);
});
});
});
I added a <a>, instead of using the <input type="submit"> because the submit, will post the form, thats not what you need, after what i understand, you need to preview the textarea in a div.
Hope it helps!
If you want your $_POST variable to be shown on the same page, then
1.change form's action to it. In your case
action="file1.php" or leave it empty to reload the page on submit (not valid in HTML5)
2.insert echo part directly in your div
<div id="output_area">
<?php
echo $_POST['doc'];
?>
</div>
not in another file
You need not use ajax for this. You can do this by using javascript. It will improve your performance of your code.
Try This:
<script>
function show() {
document.getElementById("output_area").innerHTML=document.getElementById("doc").value;
return false;
}
</script>
<form action="#" method="post">
<textarea name="doc" id="doc" rows="5" cols="50" value="">
</textarea>
<input type="submit" name="submit" value="submit" onclick="return show();">
<input type="reset" name="reset" value="Clear">
</form>
<div id="output_area"></div>
</body>
</html>
I am trying to build a page with multiple forms submitting different values to another php script "another-script.php" using Javascript as shown below:
<html>
<head>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
</head>
<body>
<div id="simple-msg"></div>
<form name="ajaxform" id="ajaxform" action="another-script.php" method="POST">
<input type=hidden name=a value='value1'>
<input type="button" id="simple-post" value="Run Code" />
</form>
<form name="ajaxform" id="ajaxform" action="another-script.php" method="POST">
<input type=hidden name=a value='value2'>
<input type="button" id="simple-post" value="Run Code" />
</form>
<form name="ajaxform" id="ajaxform" action="another-script.php" method="POST">
<input type=hidden name=a value='value3'>
<input type="button" id="simple-post" value="Run Code" />
</form>
<script>
$(document).ready(function()
{
$("#simple-post").click(function()
{
$("#ajaxform").submit(function(e)
{
$("#simple-msg").html("<img src='/logo/progress_bar.gif'/>");
var postData = $(this).serializeArray();
var formURL = $(this).attr("action");
$.ajax(
{
url : formURL,
type: "POST",
data : postData,
success:function(data, textStatus, jqXHR)
{
$("#simple-msg").html('<pre>'+data+'</pre>');
},
});
e.preventDefault(); //STOP default action
e.unbind();
});
$("#ajaxform").submit(); //SUBMIT FORM
});
});
</script>
</body>
</html>
The forms should be able to submit different values by themselves but when using Javascript to submit to a different script "another-script.php", only the first form works. Would you please kindly advise what I should do to make each form submitting their own values.
Thank you very much!