PHP if (isset($_POST['uname'])) not doing anything? - php

Question

I have two pages. 1st page has two text forms like so:
<form name="NameForm">
Name: <input type = "text" name = "uname">
Location: <input type = "text" name = "ulocation">
<button type="button" onClick="MakeRequest()"">Save</button>
</form>
It pushes the information into page number two using javascript like so (note this is on the same page as the code above):
<script>
function MakeRequest()
{
// get values
var uname = document.NameForm.uname.value;
var ulocation = document.NameForm.ulocation.value;
// validation stuff here that detects browser
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("result").innerHTML=xmlhttp.responseText;
}
}
var url = "location.php?uname=" + uname + "&ulocation=" + ulocation;
xmlhttp.open("POST", url, true);
xmlhttp.send();
}
</script>
So the problem is this, the php scripts on the page that does all server communication is not reading and storing the variables from this post request into my database.(my other get methods that view items in the database works fine)
if (isset($_POST['uname']))
{
$name = $_POST['uname'];
$location = $_POST['ulocation']
}
then the query goes somehting like
//$table and the other undefined variables are the names of my table & columns
$query = "INSERT INTO $table ($tablename, $tablelocation) VALUES ('$name', '$location')";
Basically I'm trying to get that query to work. If i remove the If statement, it stores $name to the database but not $location.
EDIT:
I forgot to add
<div id="result">
</div>

Answer 1

You are sending a GET.
to send a POST try:
[edited] perform the functions that order
function XHR(){
if(typeof XMLHttpRequest !=="undefined"){
try{ return new XMLHttpRequest(); } catch (e){}
}
if(typeof ActiveXObject !=="undefined"){
try{ return new ActiveXObject("Msxml2.XMLHTTP"); }catch(e){}
try{ return new ActiveXObject("Microsoft.XMLHTTP"); }catch(e){}
}
return false;
}
function MakeRequest()
{
// get values
var uname = document.NameForm.uname.value;
var ulocation = document.NameForm.ulocation.value;
// validation stuff here that detects browser
var url = "location.php";
xmlhttp = XHR();//Fixed
xmlhttp.open("POST", url, true);
xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");//Fixed
xmlhttp.onreadystatechange=function(){
if (xmlhttp.readyState==4) {
if(xmlhttp.status==200){
document.getElementById("result").innerHTML = xmlhttp.responseText;
} else {
document.getElementById("result").innerHTML = "ERROR:"+xmlhttp.status;
}
}
};
xmlhttp.send("uname=" + uname + "&ulocation=" + ulocation);
}

Answer 2

<form name="NameForm">
Name: <input type = "text" name = "uname">
Location: <input type = "text" name = "ulocation">
<button type="button" onClick="MakeRequest()"">Save</button>
</form>
Your missing a form method. In your case you wish:
<form name='NameForm" method="POST">
If this does not resolve your issue, then download and use firebug for firefox or chrome console to debug javascript errors.
There will be no output of errors in JS to the text. You will need to use a debug console.
perform an insert via html form
I would modify your HTML to:
<form name="NameForm" method="POST">
Name: <input type = "text" name = "uname">
Location: <input type = "text" name = "ulocation">
<button type="button" onClick="MakeRequest()"">Save</button>
<input type='submit' name='SubmitForm' value='SUBMIT THIS FORM'>
</form>
Then my PHP code:
<?php
if(isset($_POST['SubmitForm'])){
$Name = $_POST['uname'];
$Location = $_POST['ulocation'];
// Perform validation for these inputs, check if empty, set correctly ETC
$query = "INSERT INTO $table ($tablename, $tablelocation) VALUES ('$name', '$location')";
}
then call your Javascript function inside your PHP script; or perform an ajax/jquery call to run the insert without the need of a submit button