XML Request not working properly - php

I have an XML request in my javascript file that is not transferring my variable correctly to PHP and I cannot figure out why. Could I possibly be missing a reference library of some sort?
Here is the function in question. I do know that the str variable has what I want inside. I am leaving old trial code in comments just in case you want to see what I have tried.
function PHP_Con(str)
{
var xmlhttp;
if (str=="")
{
document.getElementById("txtHint").innerHTML="";
return;
}
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
}
}
if(!xmlhttp)
{
alert("Fehler");
}
else
{
/*$.ajax({
type:"POST",
url:"dbConnection.php",
data:{"test" : "str"},
success:function()
{
alert("success");
}
});*/
alert(str);
xmlhttp.open("GET","dbConnection.php?test="+str,true);
// Requestheader senden
xmlhttp.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
// Request senden
xmlhttp.send();
/*var json_string= JSON.stringify(str); // convert it into a json string.
$.post('dbConnection.php' , {test : json_string },function(){
alert("success");
}); */
}
}
And here is my corresponding PHP code:
$test = intval($_POST['test']);
echo $test;
Also, I know about isset but I am trying to get the variable to show up. I'd rather get an error when it is not there, if that makes since...
Thank you very much. :) If there are easier ways to do this, I would be interested in that as well.
But as of now I feel like I do need to have a Form that calls a JS function, then my JS has variables retrieved from user input, and then those variables go to PHP so that they can be checked against my database... it all seems like the right order to me, but I admit to not having a very clear view of how JS variables show up in PHP and also of how PHP can respond in a way that is reflected through HTML code... for example I would like to "echo" back to a span in a p with the results of the comparison (user input against the database contents) and I have no idea of how that is done...

You are sending a GET request so you have to use $_GET instead of $_POST
$test = intval($_GET['test']);
echo $test;
a POST request would look like
xmlhttp.open("POST","dbConnection.php",true);
// Requestheader senden
xmlhttp.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
// Request senden
xmlhttp.send("test="+encodeURIComponent(str));

Related

AJAX no data in responsetext

Right up front...I am very new to using Ajax.
I'm working on a web site where I want the results of one Select object to determine the options in the second Select object(from a database query). I'm using PHP and it appears that the only way to do this is to use Ajax. I've written a short html page to test my Ajax knowledge and it seems to work just find on Firefox but not on Chrome or IE. I've done a lot of research and found all sorts of folks with similar problems but no real solution.
I'm making the XMLHTTPRequest call to a local file in the same folder even so I should not be experiencing any cross-domain problems. Any help would be greatly appreciated.
Here's my Javascript function that gets called when the Select box is changed:
...
function getData(str)
{
var xmlhttp;
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.open("GET","ajax_info.php?color=",true);
xmlhttp.setRequestHeader("Content-Type", "text/xml");
xmlhttp.send();
alert(xmlhttp.responseText);
}
********ajax_info.php
+++++++++++++++++++++
//this is the php file that runs in response to the xmlhttprequest. It just generates a string of number at this time.
<?php
$str = "";
$i = 0;
for($i; $i<1000; $i++)
{
$str = $str.$i."-";
}
echo $str;
?>
You need to attach an event handler to your xmlhttp object to catch the onreadystatechange event. Note that when you alert your value, the asynchronous ajax call has just fired and has not finished yet (you are not checking for that anyway):
xmlhttp.onreadystatechange=function() {
if (xmlhttp.readyState==4 && xmlhttp.status==200) {
alert(xmlhttp.responseText);
}
}
xmlhttp.open("GET","ajax_info.php?color=",true);
xmlhttp.setRequestHeader("Content-Type", "text/xml");
xmlhttp.send();
Well in that case you should try jQuery. It will be lot easier for you to make ajax request.
Here is an example for your problem
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<script>
// FOR GET REQUEST
$.get("ajax_info.php",{color:'value'},function(data) {
alert(data); // RETRIEVE THE RESULT
});
</script>

multi pages in AJAX

I just built a web site by using this script:
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.min.js"></script>
<script>
function loadpage(page)
{
document.getElementById("pageContent").innerHTML="Yükleniyor...";
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("pageContent").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET",page,true);
xmlhttp.send();
}
</script>
It can load any page thanks to AJAX. But yet there is a question: when I load any page containing any HTML form, when i click "submit", it leaves the main page, I mean I can't send form variables by AJAX. the only thing I need is to pass form variables by using "href" and the loadpage() function I mentioned above.
How can I do get form input's values and send to another PHP file?
you can use jQuery.
$(document).ready(function(){
$("#div_load").load("page.html");
});
whit this code you can open any page (Ex: page.html) in any div(Ex:div whit id=div_load).
and for sending data use it:
$(".class_div").click(function(){
$.post("ajax.php",
{
name:"naser",
age:"23"
},
function(data,status){
// do something when done
});
});
As you are using jQuery, you can do:
$('form').submit(funciton() {
var data = $(this).serialize();
// Call Ajax
return false;
});
I advice you to read about:
http://api.jquery.com/category/ajax/ and http://api.jquery.com/serialize/.

jquery div not showing more them on user profile

I have been working on this code for a while and I am finally stumped and cannot figure out what the heck to do to get this issue fixed.
I have a jquery code that works beautifully for the get profile, but when i need to return the values in a div, it only shows the first profile of the user, but if a user posts more then once on the blog, it will not show the profile information. I have tried to append more information for each profile div to be different, but its still not working.
Here is the jQuery code for the GET user profile and return response.
function showUser(str)
{
var profileDiv = document.getElementById("profile_"+ str);
if (str=="")
{
return;
}
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
profileDiv.innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","getuser.php?q="+str,true);
xmlhttp.send();
}
Also here is the PHP script that i am using to pass the information
"<div id=\"info\" onmouseover=\"showUser(" .$blogrow['id_user'].")\"><imgalign= \"left\" style=\"vertical-align:top;\" class=\"imgb\" height=\"41\" width=\"41\" src=\"profilepics/".$blogrow['pic']."\" />".$blogrow['author']." <br>".$blogrow['timestamp']."<br></div><br>";
echo "</div>";
here is the div part as well that stores the information
echo "<div id=\"txtHint\"><div id=\"profile_".$blogrow['id_user']."\"></div></div>";
The problem relies on your HTML markup, which is invalid. Element ids must be unique in a HTML page but I see a lot of repeated ids such as #info, #theDiv, #txtHint and #profile_X
A quick fix for your problem would be to change all those and any other repeating ids to a class and then use the ajax code provided by #Rohan Kumar but using a class selector to append the content to every mention of the user in the page
function showUser(str)
{
$.ajax({
url:'getuser.php',
data:{q:str},
type:'GET',
success:function(data){
$(".profile_"+ str).html(data);
}
});
}
This is definitely not the most efficient or elegant solution but I think it would work. If you were to try and improve your code I would suggest binding all divs of class .info to a mouseenter handler, using data-attributes to get the user id and maybe maintaining a list of the profiles retrieved so you don't end up making redundant calls to your php
Using $.ajax it will more simple like,
function showUser(str)
{
$.ajax({
url:'getuser.php',
data:{q:str},
type:'GET',
success:function(data){
$("#profile_"+ str).html(data);
}
});
}
But, before this you need to add any version of jQuery

Calling a PHP page with MySQL query from Javascript function then returning results to another javascript function

I am refactoring some code. I have a PHP page that contains a MySQL query and stores the result in a PHP variable $my_result. This result is then echoed to a Flash SWF during embedding with SWFObject.
I now want to call this PHP page that makes the query from a javascript function like so - one change I have made to the PHP is that instead of storing the result in a variable $my_result I am echoing the result.
Javascript function to call the PHP page and make the database query
function getNewUploads() {
if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else {// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function() {
if (xmlhttp.readyState==4 && xmlhttp.status==200) {
alert(xmlhttp.responseText); // shows the correct output in an alert box
return xmlhttp.responseText;
}
}
xmlhttp.open("GET","dBaseConnect_uploadStep2.php",true);
xmlhttp.send();
}
Then I have the embedding of the Flash which I only want to happen when a certain tab is clicked on a page and is handled by jabvascript -
function show_tab(tab_id) {
$(tab_id).show();
if(tab_id == "#tab_2") {
var data_string = getNewUploads(); // CALLING THE FUNCTION TO CALL THE PHP QUERY
alert(data_string); // shows undefined in the alert box
var so = new SWFObject(".....");
so.addVariable("theDataString", data_string);
so.write("flashcontent2");
}
}
So it seems that the getNewUploads() function does not return the result from the PHP page.
Can anyone shed some light on my mistake please. Thanks
The call is asynchronous. After you call send, the call begins in the background. In the meantime, getNewUploads returns. Later, the function you've assigned is called with the answer. When you do return xmlhttp.responseText, you're returning from this anonymous function (assigned to onreadystatechange), not from getNewUploads (which is already done).
You can use a callback instead. Something like:
function getNewUploads(callback) {
if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else {// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function() {
if (xmlhttp.readyState==4 && xmlhttp.status==200) {
callback(xmlhttp.responseText);
}
}
xmlhttp.open("GET","dBaseConnect_uploadStep2.php",true);
xmlhttp.send();
}
function show_tab(tab_id) {
$(tab_id).show();
if(tab_id == "#tab_2") {
getNewUploads(function(data_string) {
alert(data_string);
var so = new SWFObject(".....");
so.addVariable("theDataString", data_string);
so.write("flashcontent2");
}); // CALLING THE FUNCTION TO CALL THE PHP QUERY
}
}
We define getNewUploads to take a single argument, callback. Then, in the success function, we call the callback with a single argument, the response text. In show_tab, we pass an anonymous function that takes a single parameter (the response text) to getNewUploads, as the callback parameter.
The true in xmlhttp.open("GET","dBaseConnect_uploadStep2.php", true); makes it asynchronous and hence the error. Change it to:
function getNewUploads() {
//initialize xmlhttp
xmlhttp.open("GET", "dBaseConnect_uploadStep2.php", false);
xmlhttp.send();
if(xmlhttp.status == 200)
return xmlhttp.responseText;
else
return "Oops :(";
}

Can't Get A Simple Ajax Response

I've implemented more complex AJAX before with javascript and PHP, but for some reason this refuses to work. This is copied almost directly from the W3 example.
var xmlhttp;
function changeLoc(str)
{
xmlhttp=GetXmlHttpObject();
if (xmlhttp==null)
{
alert ("Browser does not support HTTP Request");
return;
}
var url="action.php";
url=url+"?q="+str;
url=url+"&sid="+Math.random();
xmlhttp.onreadystatechange=stateChanged;
xmlhttp.open("GET",url,true);
xmlhttp.send(null);
}
function stateChanged()
{
if (xmlhttp.readyState==4)
{
alert(xmlhttp.responseText);
}
}
function GetXmlHttpObject()
{
if (window.XMLHttpRequest)
{
// code for IE7+, Firefox, Chrome, Opera, Safari
return new XMLHttpRequest();
}
if (window.ActiveXObject)
{
// code for IE6, IE5
return new ActiveXObject("Microsoft.XMLHTTP");
}
return null;
}
And the simple action.php
<?php
echo 'here';
?>
The function changeLoc is called from a link on the html page. It gets into the readyState = 4 condition , but the alert is blank. I know it's something really simple, but I can't find it.
Thank you.
Use firebug to see if there are any issues(like there are any 404 etc). Also its better to choose a javascript framework like jQuery for AJAX.
After seeing some of the suggestions I saw that the call to my php file was never occurring through firebug. The action that was occurring was when I navigated back to my index through the links on my page that call the Ajax functionality. Once i removed the href from my links, the php worked!
Very tricksy, but now I know better
Thank you for your help.

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