i have a simple form where one one field is there which is random javascript file whose name ranges from 1 to 500000 so i have used rand function to populate its field without any issue
i want form to be automatically submitted after loading but i dont want pages to be refreshed and even i dont want field value to be changed after form is submitted. my code is very clear but dont know why it is not working
index.html
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.4/jquery.min.js"></script>
<title>testing</title>
</head>
<body>
<script type="text/javascript" >
$(function() {
$("form1#form1").submit(function() {
var searchBox = $("#searchBox").val();
var dataString = 'searchBox='+ searchBox ;
if(searchBox=='')
{
$('.success').fadeOut(200).hide();
$('.error').fadeOut(200).show();
}
else
{
$.ajax({
type: "POST",
url: "test34.php",
data: dataString,
success: function(){
$('.success').fadeIn(200).show();
$('.error').fadeOut(200).hide();
}
});
}
return false;
});
});
</script>
<form method="post" name="form1" id="form1" enctype="multipart/form-data">
<input type="text" class="status" name="searchBox" id="searchBox" value="<?= "".rand(1,2889889).".js"; ?>">
<input class="button" type="submit" value="Submit" /></form>
</body>
</html>
test34.php
<?php
include('connect.php');
$title23 = $_POST['searchBox'];
mysql_query("insert into test (title) values('$title23')");
echo $title23;
?>
but it is not submitting the form automatically please advise
Your selector should probably look like this $("form#form1") instead of $("form1#form1")
After you've bound your submit method to the form, try triggering the submit method by doing this.
return false;
});
$("form1#form1").trigger('submit'); // Here
});
Related
I've got a form (id="my_form_id") and inside the form I have Dropdown list populated using PHP and a query from a Mysql, this mysql query pick the column named grupo and populated the dropdown list, the mysql table field named nombregrupo is put and updated (onchange) in a "input text" named nombregrupo using the Java script function "showname". I've got a text (named "nombregrupon") for typing and changing field nombregrupo.
I want to send the information selected from drop down list (field grupo) and typed in the text (named "nombregrupon") to ajax jquery function, and this ajax function calls a php file (postdata.php) for processing a returning data back to ajax.
I can pass the information entered in the text (named "nombregrupon") to ajax function but I cannot pass the information from dropdown list (grupo) to ajax function, this is the whole code:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
<script type="text/JavaScript">
<!--
function MM_goToURL() { //v3.0
var i, args=MM_goToURL.arguments; document.MM_returnValue = false;
for (i=0; i<(args.length-1); i+=2) eval(args[i]+".location='"+args[i+1]+"'");
}
</script>
//Java script function for updating input text nombregrupo
<script language="JavaScript">
function showname(what)
{
what.form.nombregrupo.value=what.options[what.selectedIndex].title
}
window.onload=function() {
showname(document.form1.grupo)
}
</script>
//Ajax Jquery for send data variables grupo and nombregrupon to php
<script src="//code.jquery.com/jquery-1.11.3.min.js"></script>
<script>
$(document).ready(function(){
$('#my_form_id').on('submit', function(e){
//Stop the form from submitting itself to the server.
e.preventDefault();
var nombregrupon = $('#nombregrupon').val();
var grupo= $('#grupo').val();
$.ajax({
type: "POST",
url: 'postdata.php',
data: { grupo: grupo, nombregrupon: nombregrupon },
success: function(data){
//Send Alert
alert(data);
//Show data returne from php postdata file
$('#result').html(data);
}
});
});
});
</script>
</head>
<body>
<form id="my_form_id" name="form1">
<?php
$con = mysql_connect("localhost","root","mysq1passw0rd");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("llamadas", $con);
//PHPMysql Query for fill dropdown list
$query1 = "SELECT * FROM grupostroncales ".
"WHERE grupo!='$grupox' ORDER BY grupo";
$result1 = mysql_query($query1) or die(mysql_error());
//On change call java script funtion for update Input text nombregrupo
echo "<select name=\"grupo\" onchange=\"showname(this)\">";
while($row1=mysql_fetch_array($result1)){
echo "<option value=\"$row1[grupo]\" title=\"$row1[nombregrupo]\">$row1[grupo]</option>";
}
echo "</select>";// Closing of list box
?>
//Input Text for showing group Name
<input name="nombregrupo" type="text" id="nombregrupo"/>
//Input Text for entering new group Name
<input name="nombregrupon" type="text" id="nombregrupon" />
<input name="submit" type="submit" id="submit" value="Acept" />
<div id="result"></div>
</form>
</body>
</html>
You have a few options:
You can add the id to the select element, as you forgot that
var grupo=$('#grupo').val()
echo "<select name=\"grupo\" onchange=\"showname(this)\">"; <-- no id
//----------------
echo "<select id=\"grupo\" name=\"grupo\" onchange=\"showname(this)\">";
or You can use the name
var grupo=$('select[name="grupo"]').val()
or You can serialze the whole form
$.ajax({
data : $('#my_form_id').serialize(),
...
});
Personally I would just serialize the form
Thanks again, I modified the code addind "select id" and It worked.
Please, Now I'm trying to pass the whole form variables using serialize but I doesn't work, this is my code:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
<script type="text/JavaScript">
<!--
function MM_goToURL() { //v3.0
var i, args=MM_goToURL.arguments; document.MM_returnValue = false;
for (i=0; i<(args.length-1); i+=2) eval(args[i]+".location='"+args[i+1]+"'");
}
//Fir
</script>
<script language="JavaScript">
function showname(what)
{
what.form.nombregrupo.value=what.options[what.selectedIndex].title
}
window.onload=function() {
showname(document.form1.grupo)
}
</script>
<script src="//code.jquery.com/jquery-1.11.3.min.js"></script>
<script>
$(document).ready(function(){
var Formvar = $('#my_form_id');
$("#submit").click(function(){
$.ajax({
type: "POST",
url:Formvar.attr("action"),
//url: 'ActualizarGrupoTroncalb.php',
data:Formvar.serialize(),
success: function(data){
//Send Alert
alert(data);
//Show data from mysql query
$('#result').html(data);
}
});
});
});
</script>
</head>
<body>
<form action="ActualizarGrupoTroncalb.php" id="my_form_id" name="form1">
<?php
$con = mysql_connect("localhost","root","mysq1passw0rd");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("llamadas", $con);
//Hago query1 selecciono el Departamento a buscar
$query1 = "SELECT * FROM grupostroncales ".
"WHERE grupo!='$grupox' ORDER BY grupo";
$result1 = mysql_query($query1) or die(mysql_error());
echo "<select id=\"grupo\" name=\"grupo\" onchange=\"showname(this)\">";
while($row1=mysql_fetch_array($result1)){//Array or records stored in $nt
echo "<option value=\"$row1[grupo]\" title=\"$row1[nombregrupo]\">$row1[grupo]</option>";
}
echo "</select>";// Closing of list box
?>
<input name="nombregrupo" type="text" id="nombregrupo"/>
<input name="nombregrupon" type="text" id="nombregrupon" />
<input name="submit" type="submit" id="submit" value="Acept" />
<div id="result"></div>
</form>
</body>
</html>
#lexer, in regards to your second question, you may want to try .serializeArray() instead of serialize().
.serialize() returns a string of values formatted like a query string and that may not be what you are expecting.
FYI - I would have commented to the post but I cannot comment yet.
I changed my code and variables are serialized to php file: ActualizarGrupoTroncalb.php and recieved for php file because I can make a mysql query using variables, but returning data from php to ajax is not show in: <div id="resp"></div>
This my code:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
<script type="text/JavaScript">
<!--
function MM_goToURL() { //v3.0
var i, args=MM_goToURL.arguments; document.MM_returnValue = false;
for (i=0; i<(args.length-1); i+=2) eval(args[i]+".location='"+args[i+1]+"'");
}
</script>
<script language="JavaScript">
function showname(what)
{
what.form.nombregrupo.value=what.options[what.selectedIndex].title
}
window.onload=function() {
showname(document.form1.grupo)
}
</script>
<script src="//code.jquery.com/jquery-1.11.2.min.js"></script>
<script>
$(document).on('ready',function(){
$('#submit').click(function(){
var url = "ActualizarGrupoTroncalb.php";
$.ajax({
type: "POST",
url: url,
data: $("#formulario").serialize(),
success: function(data)
{
$('#resp').html(data);
}
});
});
});
</script>
</head>
<body>
<form method="post" id="formulario">
<?php
$con = mysql_connect("localhost","root","mysq1passw0rd");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("llamadas", $con);
//Hago query1 selecciono el Departamento a buscar
$query1 = "SELECT * FROM grupostroncales ".
"WHERE grupo!='$grupox' ORDER BY grupo";
$result1 = mysql_query($query1) or die(mysql_error());
echo "<select id=\"grupo\" name=\"grupo\" onchange=\"showname(this)\">";
while($row1=mysql_fetch_array($result1)){//Array or records stored in $nt
echo "<option value=\"$row1[grupo]\" title=\"$row1[nombregrupo]\">$row1[grupo]</option>";
}
echo "</select>";// Closing of list box
?>
<input name="nombregrupo" type="text" id="nombregrupo"/>
<input name="nombregrupon" type="text" id="nombregrupon" />
<input name="submit" type="submit" id="submit" value="Acept" />
</form>
<div id="resp"></div>
</body>
</html>
I am working on trying to post to ajax file with jquery and showing result on page posted from. Post gets posted but I get no reply from ajax. Am i missing something here?
form file
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Test Ajax</title>
<script src="js/jquery-1.10.2.js"></script>
</head>
<body>
<form id="myform" />
<input type="text" id="youTyped" value=""/>
<input type="submit" value="Submit">
</form>
<div id="answer"></div>
<script type="text/javascript">
$('#myform').submit(function(){
var youTyped = $("#youTyped").val();
var data = {youTyped:youTyped};
$.ajax({
type: "POST",
url: "scripts/ajaxTest.php",
data: data,
success: function(response){
$("#answer").html(response);
}
});
});
</script>
</body>
</html>
ajax file
<?php
if(isset($_POST['youTyped'])){
$youTyped = $_POST['youTyped'];
}
echo $youTyped;
?>
Well, first of all, you didn't write your form correctly, the elements you want to send are not inside the form.
Second, your js function is not avoiding the form for being sent, so you're reloading the page when submitting the form.
So, to correct it, and get the answer, just change the code to:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Test Ajax</title>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8/jquery.min.js"></script>
</head>
<body>
<form id="myform">
<input type="text" id="youTyped" value=""/>
<input type="submit" value="Submit">
<div id="answer"></div>
</form>
<script type="text/javascript">
$('#myform').submit(function(e){
// Avoid send the form as a normal request
e.preventDefault();
var youTyped = $("#youTyped").val();
var data = {youTyped:youTyped};
$.ajax({
type: "POST",
url: "scripts/ajaxTest.php",
data: data,
success: function(response){
$("#answer").html(response);
}
});
});
</script>
</body>
</html>
One last thing. To test it, when you're sending and receiving AJAX, if you're using Chrome you can use the development tools to know if you're sending correctly the requests. Just press F12, and then go to Network tab. If you're using correctly AJAX, you'll see how new lines are added when you press the submit button:
You need to return false on form submit. The form is being submited before ajax request. Also add a closing tag for form.
<script type="text/javascript">
$('#myform').submit(function () {
var youTyped = $("#youTyped").val();
var data = {
youTyped: youTyped
};
$.ajax({
type: "POST",
url: "scripts/ajaxTest.php",
data: data,
success: function (response) {
$("#answer").html(response);
}
});
return false;
});
</script>
Try if this code is working:
<body>
<input type="text" id="youTyped" value=""/>
<input type="submit" id="submit" value="Submit">
<div id="answer"></div>
<script type="text/javascript">
$(document).ready(function(){
$('#submit').click(function(){
var youTyped = $("#youTyped").val();
$.post('scripts/ajaxTest.php',{"youTyped" : youTyped},function(response)
{
$("#answer").html(response);
});
});
});
</script>
</body>
</html>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Test Ajax</title>
<script src="js/jquery.min.js"></script>
<script type="text/javascript">
$('#myform').submit(function(event){
event.preventDefault();
var youTyped = $("#youTyped").val();
var data = {youTyped:youTyped};
$.ajax({
type: "POST",
url: "ajaxTest.php",
data: data,
success: successMsg,
error: errorMsg
});
});
function successMsg(response){
alert(response);
$("#answer").html(response);
}
function errorMsg(){
alert("Invalid Ajax Call");
}
</script>
</head>
<body>
<form id="myform" />
<input type="text" id="youTyped" value=""/>
<input type="submit" value="Submit">
<div id="answer"></div>
</body>
</html>
As I toyed with the idea, thought that this would make a good addition/contribution to the question.
I added and modified a few things.
One of which being:
If nothing was typed into the form input field, it would return "You typed: Nothing".
PHP:
<?php
if(!empty($_POST['youTyped'])){
$youTyped = $_POST['youTyped'];
echo "You typed: " .$youTyped;
}
else if (empty($_POST['youTyped'])){
echo "You typed: Nothing";
}
?>
I have here a page in which I am trying to update a users email now the div refreshes on submit like it is ment to but it is not updating the database and I ain't for the life of me know why.
My layout is one page click a menu link and it loads the page into the content div (below does) now I am wanting to post a form and it reload in that div and update the mysql database but for some reason it don't want to update the database.
Anyone got any suggestions into why it's not updating the database? have i made a small error somewhere in the php or is it due to my page set up on loading this way?
Thanks in advance. :)
<?php
session_start();
include_once("./src/connect.php");
include_once("./src/functions.php");
//Update email
if (isset($_POST['update'])){
$email = makesafe($_POST['email']);
$result = mysql_query("UPDATE Accounts SET email='$email' WHERE `id`= '{$fetchAccount['id']}'")
or die(mysql_error());
echo "<font color='lime'>Updated.</font>";
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Home</title>
<link rel="stylesheet" href="./static/css/LoggedIn/Index.css" type="text/css" />
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script>
<script type="text/javascript">
jQuery(document).ready(function($) {
$("#Submit").click(function() {
var url = "Profile.php"; // the script where you handle the form input.
$.ajax({
type: "POST",
url: url,
data: $("#myForm").serialize(), // serializes the form's elements.
success: function(html){ $("#right").html(html); }
});
return false; // avoid to execute the actual submit of the form.
});
});
</script>
</head>
<body>
<div id="right">
<form method="post" action="Profile.php" id="myForm">
E-mail: <input type="text" value="<?=$fetchAccount['email']?>" name="email"><br>
<input type="submit" value="Edit" name="update" id="Submit">
</form>
</div>
</body>
</html>
Heres some changes, read code comments, though the main thing wrong with your code is ajax will get the whole page not just echo "<font color='lime'>Updated.</font>"; part.
<?php
session_start();
include_once("./src/connect.php");
include_once("./src/functions.php");
//Update email only if request is from ajax request
if(isset($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) === 'xmlhttprequest'){
//check its POST, and email is real, then do update or exit a error message back
if($_SERVER['REQUEST_METHOD']=='POST' && !empty($_POST['email']) && filter_var($_POST['email'], FILTER_VALIDATE_EMAIL)){
header('Content-Type: text/html');
//check its an email
if(!filter_var($_POST['email'], FILTER_VALIDATE_EMAIL)){
exit('<span style="color:red;">Invalid email.</span>');
}
//Do update
mysql_query("UPDATE Accounts
SET email='".mysql_real_escape_string($_POST['email'])."'
WHERE id= ".mysql_real_escape_string($fetchAccount['id'])) or die('<span style="color:red;">Error updating email. '.mysql_error().'</span>');
exit('<span style="color:lime;">Updated.</span>');
}
}
?>
<!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">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Home</title>
<link rel="stylesheet" href="./static/css/LoggedIn/Index.css" type="text/css" />
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"/></script>
<script type="text/javascript">
$(function(){
$("#Submit").click(function(e) {
$.ajax({
type: "POST",
url: "Profile.php",
data: $("#myForm").serialize(), // serializes the form's elements.
success: function(html){ $("#right").html(html); }
});
e.preventDefault();
});
});
</script>
</head>
<body>
<div id="right">
<form method="post" action="Profile.php" id="myForm">
E-mail: <input type="text" value="<?=htmlspecialchars($fetchAccount['email'])?>" name="email"/><br/>
<input type="submit" value="Edit" name="update" id="Submit"/>
</form>
</div>
</body>
</html>
Hope it helps
I have an HTML file that has a form with two fields. These fields' value should be posted to a PHP and this PHP should be fetched from the HTML using JQuery. This is what I implemented.
My HTML file:
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script src="//ajax.aspnetcdn.com/ajax/jQuery/jquery-1.9.1.min.js"></script>
<script>
$(document).ready(function(){
$("button").click(function(){
$("#first").load("result_jquery.php");
});
});
</script>
</head>
<body>
<div id="first"></div>
<div>
<form method="POST" id="myForm">
Name: <input type="text" name="name"/><br/>
Number: <input type="text" name="number"/><br/>
<button>submit</button>
</form>
</div>
</body>
This is my result_jquery.php
<?php
$n = $_POST["name"];
echo "hello ".$n;
?>
When I click the submit button, the hello is getting printed. But the name is not getting printed. Can you please help me with this. I don't know where I am going wrong.
I think that the use of the button element is the worry and the code that i will put now it is working properly as you need so try this and tell me the result :)
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script src="//ajax.aspnetcdn.com/ajax/jQuery/jquery-1.9.1.min.js"></script>
<script>
$(document).ready(function(){
$("#button").click(function(){
var n = $('[name="namee"]').val();
var nb = $('[name="number"]').val();
$("#first").load("result_jquery.php",{'namee':n,'number':nb},function(data){});
});
});
</script>
</head>
<body>
<div id="first"></div>
<div>
<form method="POST" id="myForm">
Name: <input type="text" name="namee"/><br/>
Number: <input type="text" name="number"/><br/>
<input type="button" value="Submit" id="button" />
</form>
</div>
</body>
</html>
copy this code:
<script type="text/javascript">
$(document).ready(function() {
$("#send").click(function() {
$.ajax({
type: "POST",
data : "name="+$( '#name' ).val(),
url: "result_jquery.php",
success: function(msg) {
$('#first').html(msg);
}
});
});
});
</script>
change this in form
<form method="POST" id="myForm">
Name: <input type="text" id="name" name="name"/><br/>
Number: <input type="text" id="number" name="number"/><br/>
<input type="button" id="send" value="Submit">
</form>
just try that and tell me the result :)
var n = $('[name="name"]').val();
var nb = $('[name="number"]').val();
$('#error').load("result_jquery.php", {'name':n,'number':nb},function(data){});
Note try to change the element name for the name field from "name" to "namee" and apply changes as needed look like this :
var n = $('[name="namee"]').val();
var nb = $('[name="number"]').val();
$('#error').load("result_jquery.php", {'namee':n,'number':nb},function(data){});
and the result_jquery.php file :
<?php
$n = $_POST["name"];
echo "hello ".$n;
?>
From the jQuery documentation on load:
This method is the simplest way to fetch data from the server. It is
roughly equivalent to $.get(url, data, success) except that it is a
method rather than global function and it has an implicit callback
function. When a successful response is detected (i.e. when textStatus
is "success" or "notmodified"), .load() sets the HTML contents of the
matched element to the returned data. This means that most uses of the
method can be quite simple:
You are performing a HTTP GET with that method, and not a POST.
My suggestion would be if you want to send an AJAX request to your server with information in it, get used to using the long form jQuery AJAX:
$.ajax({
data: 'url=encoded&query=string&of=data&or=object',
url: 'path/to/server/script.php',
success: function( output ) {
// Handle response here
}
});
For more info, see jQuery documentation: http://api.jquery.com/jQuery.ajax/
I want to enter different rooms with on button. How is this possible in Javascript. For example, there are three rooms, "Kitchen, toilet and bedroom". How can I use JS to enter any of these rooms depending on my choice. so if i enter "kitchen" in the input box its gonna take me to kitchen.php, if I enter toilet...the same button is going to take me to toilet.php etc.
This is the HTML input,
<form method="post">
<input style=""name="Text1" type="text"><br>
<input name="move" style="height: 23px" type="submit" value="Move">
</form>
Just use a select field jsfiddle demo:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta content="text/html; charset=utf-8" http-equiv="Content-Type" />
<title>Untitled 1</title>
<script type="text/javascript">
function submitForm() {
var myform = document.getElementById('myform');
var mytext = document.getElementById('room');
myroom = mytext.value.toLowerCase();
if (myroom == 'kitchen') {
myform.action = 'kitchen.php';
myform.submit();
} else if (myroom == 'toilet') {
myform.action = 'toilet.php';
myform.submit();
} else if (myroom == 'bedroom') {
myform.action = 'bedroom.php';
myform.submit();
} else return false;
}
window.onload = function(){
document.getElementById('move').onclick = submitForm;
}
</script>
</head>
<body>
<form id="myform" name="myform" method="post">
<input type="text" id="room" name="room" />
<button id="move" name="move" style="height: 23px">Move</button>
</form>
</body>
</html>
On the php side create three files to test to see if this works, toilet.php, kitchen.php, and bedroom.php with the following code in all three files. make sure the file names are lower cased:
<?php
echo $_POST['room'];
?>
Basically, based on the option that is selected, the JavaScript would change the form's action url and submit. If none is selected it'll return false and not submit. The submitForm function is attached to the move button with an onclick event.