Get value of HTML element within PHP on same page? - php

In the past, whenever I needed to get the value of an html element, I would always submit a form to load a different page. Something like:
page1.php
<form name="form1" action="page2.php" method="post">
<input type="hidden" name="input1" value="value1" />
</form>
page2.php
<?php
$data = $_POST['input1'];
?>
My question is, is it possible to get the value of 'input1' on the same page (page1.php) without requiring clicking a submit button?

With jQuery:
<?php
if (isset($_POST['data'])) {
$data = $_POST['data'];
print( "data is: $data" );
return;
}
?>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<script type="text/javascript" src="jquery-1.6.3.js"></script>
</head>
<body>
<div>Response from server: <span id="response"></span></div>
<script type="text/javascript">
$.post('test.php',{'data': "value1"}, function (data) {
$('#response').text(data);
});
</script>
</body>
</html>
Non-jQuery IE doesn't have the XMLHttpRequest object so here is a shim for IE:
Yes, using ajax / javascript:
var formData = new FormData();
formData.append("input1", "value1");
var xhr = new XMLHttpRequest();
xhr.open("POST", "page1.php");
xhr.send(formData);
if (typeof XMLHttpRequest == "undefined") XMLHttpRequest = function() {
try {
return new ActiveXObject("Msxml2.XMLHTTP.6.0");
}
catch (e) {}
try {
return new ActiveXObject("Msxml2.XMLHTTP.3.0");
}
catch (e) {}
try {
return new ActiveXObject("Microsoft.XMLHTTP");
}
catch (e) {}
//Microsoft.XMLHTTP points to Msxml2.XMLHTTP and is redundant
throw new Error("This browser does not support XMLHttpRequest.");
};

What for? Do you want to submit it to another file without needing to actually "submit" the form?
You use JavaScript for that.
document.form1.input1.value

Related

return value on ajax comes back with 0

I am trying to run a conversion in Ajax. I believe from what i found online, I have most everything correct. However when I use the calculate button, It returns 0 in my results div, instead of answer. I think the issue is my numeric value isn't getting properly pulled from the text box. I need to do it like this, so changing html input types isn't an option. I am extremely new to Ajax, and don't quite know how this works. Any help would be great.
my code:
<!DOCTYPE html>
<head>
<title>Ajax Money Conversion</title>
<script>
var http = createRequestObject();
function createRequestObject() {
var ro;
var browser = navigator.appName;
if(browser == "Microsoft Internet Explorer"){
ro = new ActiveXObject("Microsoft.XMLHTTP");
}else{
ro = new XMLHttpRequest();
}
return ro;
}
function moneyConversion(argLB) {
http.open('get', 'Conversion.php?pound=' +argLB);
http.onreadystatechange = handleResponse;
http.send(null);
}
function handleResponse() {
if(http.readyState == 4){
result = http.responseText.split(",");
document.getElementById("results").innerHTML = result[0];
}
}
</script>
</head>
<body>
<div>
<form name="myForm" action="#">
<h1>Enter amount of Dollars You Want To Convert to Pounds</h1>
<input type="text" name="txtCurrency" />
<input type="button" name="calcBtn" value="Calculate" id="calcBtn" onclick="moneyConversion()" />
</form>
</div>
<h1>Total</h1>
<div id="results">
</div>
</body>
</html>
my function conversion.php
<?php
$dollars=$_GET["pound"];
$conversion=($dollars * .6302);
print ("$conversion");``
?>
onclick="moneyConversion()"
should be
onclick="moneyConversion(this.form.elements.txtCurrency.value);"
otherwise you passing empty string to php

Basic jquery & PHP session.upload_progress.name

I am trying to get the file progress working with the new session.upload_progress.name functionality in PHP 5.4.
So far my code is this:
<?
session_start();
?>
<!DOCTYPE html>
<head>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<script type='text/javascript'>
$(document).ready(function () {
$("#jimbo").submit(function () {
setInterval(function() {
$.ajax({
url: "ajx.php",
success: function (data) {
$("#feedback").html(data + Math.random(999));
}
});
//$("#feedback").html("hello " + Math.random(999));
},500);
//return false;
});
});
</script>
</head>
<body>
<h1>Upload</h1>
<br/>
<form action="test.php" method="POST" enctype="multipart/form-data" id='jimbo'>
<input type="hidden" name="<?=ini_get('session.upload_progress.name'); ?>" value="myupload" />
<input type="file" name="file1" />
<input type="submit" id='submitme' />
</form>
<div id="feedback">Hello</div>
</body>
</html>
And then the ajx.php file:
<? session_start(); ?>
<pre>
<?
echo "SESSIONVAR<br/>";
var_dump($_SESSION);
?>
</pre>
Now. When I click the submit button (after selecting a file), The file starts uploading, but the setinterval doesnt start. However, If I have the return false; in there, I get the setInterval results, but the file doesnt start uploading. If I submit the file without returning false, and in a seperate window view the contents of ajx.php, I can see that the variable is working fine and updating. So how do I get the #feedback div to update once the form has been clicked?
note the session array is populated, the problem here is with the jquery and nothing else.
You can achieve similar functionality using javascript XMLHttpRequest objects 'upload' property. It has a couple of events you can hook into, 'progress' is one of them.
here's a sample I've used. It will add a row with the progression (in %) to a <div class="progression"> for each file from a <input type="file"> field:
function startUpload() {
var fileInput = document.getElementById("file1");
$('.progression').show();
for(var i = 0;i<fileInput.files.length;i++) {
doFileUpload(fileInput.files[i]);
}
}
function doFileUpload(file) {
var xhr = new XMLHttpRequest();
var data = new FormData();
var $progress = $('<div class=\"progress\"><p>' + file.name + ':</p><span>0</span>%</div>');
$('div.progression').append($progress);
data.append("file", file);
data.append("album", $("#album").val());
xhr.upload.onprogress = function(e) {
var percentComplete = (e.loaded / e.total) * 100;
$progress.find('span').text(Math.ceil(percentComplete));
};
xhr.onload = function() {
if (xhr.status == 200) {
var result = JSON.parse(xhr.responseText);
if(result.success == "true") {
console.log("Great success!");
}
else {
console.log("Error! Upload failed");
}
};
xhr.onerror = function() {
console.log("Error! Upload failed.");
};
xhr.open("POST", "/_admin/_inc/upload.php", true);
xhr.send(data);
}

changing content in a div using php, html ajax

i am trying to make the content in the div 'test' change according to the link i click and to go where the original text is
<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
<title>An XHTML 1.0 Strict standard template</title>
<meta http-equiv="content-type" content="text/html; charset=utf-8" />
<meta http-equiv="Content-Style-Type" content="text/css" />
<style type="text/css">
</style>
<script type="text/javascript">
function ajaxFunction(id, url){
var xmlHttp;
try {// Firefox, Opera 8.0+, Safari
xmlHttp = new XMLHttpRequest();
} catch (e) {// Internet Explorer
try {
xmlHttp = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
try {
xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
} catch (e) {
alert("Your browser does not support AJAX!");
return false;
}
}
}
xmlHttp.onreadystatechange = function(){
if (xmlHttp.readyState == 4) {
//Get the response from the server and extract the section that comes in the body section of the second html page avoid inserting the header part of the second page in your first page's element
var respText = xmlHttp.responseText.split('<body>');
elem.innerHTML = respText[1].split('</body>')[0];
}
}
var elem = document.getElementById(id);
if (!elem) {
alert('The element with the passed ID doesn\'t exists in your page');
return;
}
xmlHttp.open("GET", url, true);
xmlHttp.send(null);
}
</script>
</head>
<body>
<div id="test">THIS SHOULD CHANGE</div>
link1
a href="#" value="Make Ajax Call" id="ajax" onclick="ajaxFunction('test','example2.html'); return false;">link2</a>
</body>
</html>
the div is i am trying to change is this
<div id="test">THIS SHOULD CHANGE</div>
and the links im using are these
link1
link2
any ideas on this? as this is giving me a head ache
i am using the links in a sepreat php file named menu and includeing it with
<?php include 'menu.php'; ?>
would this make a diffrence?
if i rename one the html files i get a error saying it cant be found but other than that nothing shows when there correct
!!!update!!!
ok if i change the link text from
link1
to this
<form>
<input type="button" value="Make Ajax Call" id="ajax" onclick="ajaxFunction('test','example.html');"/>
</form>
it works fine but looks really bad so is there a better way to change the link?
If you are trying to load external content into a div depending on what link was clicked? Then you could try something like the following. It is basically a striped down version of what you have and works fine for me.
<script type="text/javascript">
function loadiv(temp){
var req = new XMLHttpRequest();
req.open("GET", temp, false);
req.send(null);
var page = req.responseText;
document.getElementById("test").innerHTML = page;
};
</script>
link1
link2
<div id="test"></div>
Or use jQuery
<script type="text/javascript">
$(function(){
$('a').on('click', function(e){
e.preventDefault();
var page_url = $(this).prop('href');
$('#test').load(page_url);
});
});
</script>
Load Form 1
Load Form 2
<div id="test"></div>
I was wrong with my advice, sequence doesn't matter when accessing the outer scope.
function z() {
var a=1;
function x() {
console.log(a);
console.log(b);
}
var b=2;
x();
}
z();
I still recommend using Firefox with firebug addon or your favorite browser with a similar feature: check the javascript console for errors. You cn also check the ajax response there.

how can i change the output of JavaScript to go to directly to another page instead of an alert window?

<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Validate Zip Code</title>
<script type="text/javascript">
function IsValidZipCode(zip) {
var isValid = /20132/.test(zip);
if (isValid)
alert('Valid ZipCode');
else {
alert('yep')
}
}
</script>
</head>
<body>
<form>
<input id="txtZip" name="zip" type="text" /><br />
<input id="Button1" type="submit" value="Check My Zipcode"
onclick="IsValidZipCode(this.form.zip.value)" />
</form>
</body>
</html>
I need to use this to allow the user to go either to a page that says sorry we cannot service your area or to another page that says yes we can service your area based on wheter their zipcode is listed.
also how can i add more than one zip code in the isValid = line?
Setting window.location.href = "your url here"; answers the first part of your question.
"also how can i add more than one zip code in the isValid = line?"
If you want to stick with a regex test you can use the regex or |:
var isValid = /^(20132|20133|20200|90210|etc)$/.test(zip);
Note that I've also added ^ and $ to match the beginning and end of the entered string - the way you had it you'd also get matches if the user entered a longer string containing that code, e.g., "abc20132xyz" would match.
I'd be more inclined to do this validation server-side though.
if (isValid)
document.location.href="validzipcode.html";
else {
document.location.href="yep.html";
}
function IsValidZipCode(zip) {
var isValid = /20132/.test(zip);
if (isValid)
document.location.href="valid_zip.html";
else {
document.location.href="not_valid_zip.html";
}
}
<script type="text/javascript">
function IsValidZipCode(zip) {
var isValid = /20132/.test(zip);
if (isValid)
window.location = baseurl."/valid.php"
else {
window.location = baseurl."/invalid.php"
}
}
</script>
You can redirect it in javascript with following code , that can be put in your example instead of alert()
window.location.href = "http://stackoverflow.com";
You can try ajax for this:
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Validate Zip Code</title>
<script language="javascript" type="text/javascript">
<!--
//Browser Support Code
function IsValidZipCode(zip){
var ajaxRequest; // The variable that makes Ajax possible!
try{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
} catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
try{
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
} catch (e){
// Something went wrong
alert("Your browser broke!");
return false;
}
}
}
//alert(ajaxRequest);
// Create a function that will receive data sent from the server
ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4){
//var ajaxDisplay = document.getElementById('ajaxDiv');
//ajaxDisplay.innerHTML = ajaxRequest.responseText;
alert(ajaxRequest.responseText);
if(ajaxRequest.responseText=="")
{
alert("sorry we cannot service your area");
}
else{
window.location = "Page.php?zip="+zip+"";
}
}
}
//var age = document.getElementById('age').value;
//var wpm = document.getElementById('wpm').value;
//var sex = document.getElementById('sex').value;
var queryString = "?zip=" + zip;
ajaxRequest.open("GET", "zip_check.php" + queryString, true);
ajaxRequest.send(null);
}
//-->
</script>
</head>
<body>
<form name="form1" method="post">
<input id="txtZip" name="zip" type="text" /><br />
<input type="button" id="Button1" value="Check My Zipcode"
onclick="IsValidZipCode(document.form1.zip.value)" />
</form>
</body>
</html>

submit a form via Ajax and update a result div

I was using a self submitting form to process the data but I now need to process it separately so now I need to submit a form, return the results and place it in a div. It seems using AJAX is a good way to do this to have the data return to the original page where the form is. I have had a look at alot of examples and I don't really understand how to do it or really how its working.
Say I wanted to send this form data from index.php to my process page twitterprocess.php what do I need to do and get it to return to display the data processed.
<form method="POST" action="twitterprocess.php">
Hashtag:<input type="text" name="hashtag" /><br />
<input type="submit" value="Submit hashtag!" />
</form>
This is what I have been using to display the results.
<?php foreach($results as $result) {
$tweet_time = strtotime($result->created_at);?>
<div>
<div class="tweet"> <?php echo displayTweet($result->text),"\r\n"; ?>
<div class="user"><?php echo "<strong>Posted </strong>" . date('j/n/y H:i:s ',$tweet_time) ?><strong> By </strong><a rel="nofollow" href="http://twitter.com/<?php echo $result->from_user ?>"><?php echo $result->from_user ?></a></div>
</div>
<br />
<? } ?>
I'm new to AJAX but any guidance would be greatly appreciated
*When you use AJAX the output generated on other page is the result for this page.
*Now when you want to post data and retrieve results through the use of AJAX then in form part of your html don't use type="submit" for button, but simply go for type="button".
*action attribute should be left blank as you are going to trigger the action through your AJAX code.
*Well rest all your solution in the code snippet below:
Below is the HTML code along with AJAX
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>Simple Form Handling Through AJAX</title>
<script type="text/javascript">
function loadXmlDoc(fname, lname){
var xmlhttp;
if (window.XMLHttpRequest){
xmlhttp = new XMLHttpRequest();
}
else{
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function(){
if (xmlhttp.readyState == 4 && xmlhttp.status == 200){
document.getElementById("ajaxify").innerHTML = xmlhttp.responseText;
}
}
xmlhttp.open("POST", "demo_ajax3.php", true);
xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xmlhttp.send("fname=" + fname + "&" + "lname=" + lname);
}
</script>
</head>
<body>
<p>
<span id="ajaxify"> </span>
</p>
<form id="frm" action="#">
<input type="text" name="fn" />
<input type="text" name="ln" />
<input type="button" name="submit" value="submit" onclick="loadXmlDoc(fn.value, ln.value)" />
</form>
</body>
</html>
Below is the PHP code that is used in above code
<?php
$fname = $_POST["fname"];
$lname = $_POST["lname"];
echo "Hello " . $fname . " " . $lname;
?>
Assign some id to your submit button, i'd use id="submit" and some id for your text field (i use id="text");
Client-side js:
$("#submit").click(function () {
var postData = new Object(); //for complex-form
postData.hashTag = $("#text").val();
$.ajax({
type: 'POST', //or 'GET' if you need
contentType: "application/json; charset=UTF-8", //i use json here
dataType: "json",
url: "some_url",
data: JSON.stringify(postData), //or smth like param1=...&param2=... etc... if you don't want json
success: function (response) {
//handle response here, do all page updates or show error message due to server-side validation
},
error: function () {
//handle http errors here
}
});
return false; //we don't want browser to do submit
});
So, if user has js enabled = your code will do ajax request, otherwise - regular post request will be made;
On a server-side you have to handle ajax and regular submit different to make it work correct in both cases. I'm not good in php so can't do any advise here
You can use jQuery, for example,
function doPost(formdata){
var url="/twitterprocess.php";
var senddata={'data':formdata};
$.post(url,senddata,function(receiveddata){
dosomethingwithreceiveddata(receiveddata);
}
your php will get senddata in JSON form. You can process and send appropriate response. That response can be handled by dosomethingwithreceiveddata.
I find the Ajax Form plugin a good tool for the job.
http://www.malsup.com/jquery/form/#tab4
A basic code example could be:
$(document).ready(function() { // On Document Ready
var options = {
target: '#output1', // ID of the DOM elment where you want to show the results
success: showResponse
};
// bind form using 'ajaxForm'
$('#myForm1').ajaxForm(options);
});
// the callback function
function showResponse(responseText, statusText, xhr, $form) {
alert('status: ' + statusText + '\n\nresponseText: \n' + responseText +
'\n\nThe output div should have already been updated with the responseText.');
}
All your PHP file have to do is echo the html (or text) back that you want to show in your DIV after the form has been submitted.
If you do not want to use jquery try this in pure javascript
function SendData(Arg) {
xmlhttp=null;
var uri = "/twitterprocess.php";
if(window.XMLHttpRequest) {
xmlhttp=new XMLHttpRequest();
} else if(window.ActiveXObject) {
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
if(xmlhttp!=null) {
xmlhttp.onreadystatechange = function() {
if(xmlhttp.readyState==4) {
if(xmlhttp.status==200) {
var xmlDoc = xmlhttp.responseXML;
var DateNode=xmlDoc.getElementsByTagName('Date')[0].firstChild.nodeValue;
var Xml2String;
if(xmlDoc.xml) {
Xml2String = xmlDoc.xml
} else {
Xml2String = new XMLSerializer().serializeToString(xmlDoc);
}
document.getElementById("CellData").value=Xml2String;
} else {
alert("statusText: " + xmlhttp.statusText + "\nHTTP status code: " + xmlhttp.status);
}
}
}
}

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