My asp code is not read completely and it doesn't allow me to make it work.
I am totally new to ASP, but I searched a lot of sources for the task I need: to get the file names from a server directory
I call .asp file with an XMLHttpRequest the following way:
function readDir() {
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("demo").innerHTML = this.responseText;
console.log(this);
}
}
xhttp.open('get', 'loader/asp/getFiles.asp', true);
xhttp.send();
}
and the getFiles.asp contains simple output:
<%
Response.Write "First statement"
Response.Write "Second statement"
%>
As a response from the server i get the this content, instead of the result or at least the full source code.
Can somebody give an advice regarding this?
Thanks.
Related
I have a simple code in a Javascript file that calls a PHP file using Ajax. At the moment it's a simple call and I don't need to pass any variables.
I'm using IIS and the folders structure is the following:
ROOT > sendMail.php
ROOT > JS > chat.js
The call starts from chat.js and it's the following:
function sendEmail(params){
var xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("txtHint").innerHTML = this.responseText;
}
};
xmlhttp.open("POST", "sendMail.php?q=" + params, true);
xmlhttp.send();
}
But I receive "404 Not Found". Someone can help me?
Thanks in advance!
I am finding a way to load multiple files at a time in my html document using vanilla AJAX ( I don't want any dependecy, the way I will get using jQuery ). Here is a piece of code I grabbed from W3schools' AJAX Documentation. :
function loadDoc() {
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.querySelector("#tst").innerHTML = this.responseText;
}
}
xhttp.open("GET", "ajax_info.txt", true);
xhttp.send();
}
I am pretty new to AJAX. So, I have a confusion. The above function works fine, but I have multiple files to load and I don't want to call this function multiple times, to load the files. Let me describe my desired output. I am imagining the syntax of the updated function will be :
loadDoc(["one.txt", "two.txt", "three.txt"])
So, we observe that the input will be an array. And now the ouput will follow this format :
content of "one.txt" + break + content of "two.txt" + break + content of "three.txt" and so on....
If the content of "one.txt" is "Hello World !", content of "two.txt" is "I love", and content of "three.txt" is "Javascript", the output will be :
Hello World !
I love
Javascript
So, can you please help me update my code ?
Pass the list of files to load to a PHP/ASP/whatever page which assembles the text for you.
function loadDoc(files) {
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.querySelector("#tst").innerHTML = this.responseText;
}
}
xhttp.open("GET", "ajax_info.php?files="+files, true);
xhttp.send();
}
The files parameter could be a comma delimited list, which the page uses to get the contents and assemble it into a single return string.
I'm recently working on a website project. Therefor I have a website.php with all html code, a function.php and saveArray.js . In website.php I'm printing a html table with a button at the bottom. Through the button click I'm getting to the saveArray.js, where I save all the table data in an array.
With this code
var arrString = JSON.stringify(tableData);
var request = new XMLHttpRequest();
request.open('post', 'function.php', true);
request.setRequestHeader('Content-Type', 'application/x-www-form-
urlencoded');
request.send('daten=' + arrString);
I post the JS array to function.php. In function.php I do something with the array and in an if statement I want to show a modal.
The modal itself works, but I want to show it on website.php page. Which doesn't happends, because I'm currently on function.php .
How can I solve this ?
EDIT: In my array is an ID and I want to check if this ID is already in my database or not. Depending on this result I want to show the modal and upload the data if necessary. All the checking is happening in function.php
I suppose you want to inject the string returned (the modal PHP code) by your function in function.php in your current page ('website.php').
To do this, you'll have to inject the response given by the XMLHttpRequest when the request is finished.
Let's suppose we want to add all the contents within
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("demo").innerHTML =
this.responseText;
}
};
See, You are not handling the response of the request.So handle the response.and restuern the status of the request from function.php and if data is saved the open the model. You need not go to the function.php page. See the code
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
// this is response of the request //now check it
//Suppose you returned " data saved" as response from function.php
if(this.responseText='data saved'){
//Open model here
}
}
};
xhttp.open("POST", "function.php", true);
xhttp.send();
I have an anchor link with no destination, but it does have an onClick event:
<li><a href onClick='deletePost()'> Delete </a> </li>
I understand that I cannot directly execure PHP code blocks in JavaScript due to the nature of PHP and it being a server side language, so I have to utilize AJAX to do so.
When the delete link is clicked, I need it to execute this query (del_post.php)
<?php include("connect.php");
$delete_query = mysqli_query ($connect, "DELETE FROM user_thoughts WHERE id = 'id' ");
?>
I have tried to understand AJAX using similar past questions, but due to being relatively new, I cannot completely grasp it's language. Here is what I have tried:
function deletePost() {
xmlhttp=new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200){
xmlhttp.open("GET", "del_post.php", false);
xmlhttp.send();
}
}
}
But clicking the link just changes the URL to http://localhost/.
I believe the (main) problem is your empty "href" attribute. Remove that, or change it to href="#" or old school href="javascript:void()" (just remove it, imo).
It's been a while since I used XMLHttpRequest and not something like jQuery's .ajax, but I think you need to do it like so (mostly you need to .open/send before you watch for the state change):
var xmlHttpReq = new XMLHttpRequest();
if (xmlHttpReq) {
xmlHttpReq.open('GET', 'your-uri-here.php', true/false);
xmlHttpReq.onreadystatechange = function () {
if (xmlHttpReq.readyState == 4 && xmlHttpReq.status == 200) {
console.log('success! delete the post out of the DOM or some other response');
}
else {
console.log('there was a problem');
}
}
xmlHttpReq.send();
}
Can you please provide your : del_post.php file?
Normally you can show a text or alert in a
<div id="yourname"></div>
by using callback in an AJAX request :
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById("yourname").innerHTML = xmlhttp.responseText;
}
This response is coming from your PHP file for example :
function remove_record(ARG){
if ($condition==true)
echo "TRUE";
else
echo "FALSE";
}
You should remove href attribute from anchor tag and style the element with CSS.
Also, your script should look like this:
<script>
function deletePost() {
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (xhttp.readyState == 4 && xhttp.status == 200) {
// Do something if Ajax request was successful
}
};
xhttp.open("GET", "del_post.php", true);
xhttp.send();
}
</script>
You are trying to make the http request inside the callback.
You just need to move it outside:
function deletePost() {
var xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
alert(xmlhttp.responseText);
}
}
xmlhttp.open("GET", "del_post.php", false);
xmlhttp.send();
}
Removing the href attribute will prevent the refresh. I believe that is valid in HTML5.
Ok... I'm just a hobbyist, so please forgive me any inaccuracies in the typing but this works: A format I use for an ajax call in an <a> element is:
<a href="javascript:" onclick="functionThatReallyCallsAjax()">
So that I have more flexibility(in case I need to check something before I send the ajax). Now, for an ajax call you need:
What file to call
What to do with the response from the file you called
What to do if an I/O error happens
So we have this function - not mine, leeched amongst thousands from somewhere - probably here :) - and probably well known, my apologies to the author, he is a genius: This is what you call for the ajax thing, where 'url' is the file you want to 'ajax', 'success' is the name of the function that deals with results and error is the name of the function that deals with IO errors.
function doAjaxThing(url, success, error) {
var req = false;
try{
// most browsers
req = new XMLHttpRequest();
} catch (e){
// IE
try{
req = new ActiveXObject("Msxml2.XMLHTTP");
} catch(e) {
// try an older version
try{
req = new ActiveXObject("Microsoft.XMLHTTP");
} catch(e) {
return false;
}
}
}
if (!req) return false;
if (typeof success != 'function') success = function () {};
if (typeof error!= 'function') error = function () {};
req.onreadystatechange = function(){
if(req.readyState == 4) {
return req.status === 200 ?
success(req.responseText) : error(req.status);
}
}
req.open("GET", url, true);
req.send(null);
return req;
}
You will naturally need to include the success+error functions:
function dealWithResponse(textFromURL)
{
//textFromURL is whatever, say, a PHP you called in the URL would 'echo'
}
function ohNo()
{
//stuff like URL not found, etc.
alert("I/O error");
}
And now that you're armed with that, this is how you compose the real call inside the function you called at the <a>:
function functionThatReallyCallsAjax()
{
//there are probably many scenarios but by having this extra function,
//you can perform any processing you might need before the call
doAjaxThing("serverFile.php",dealWithResponse,ohNo);
}
One scenario might be when you need to pass a variable to the PHP you didn't have before. In this case, the call would become:
doAjaxThing("serverFile.php?parameter1=dogsRock",dealWithResponse,ohNo);
And now not only you have PHP sending stuff to JS, you have JS sending to PHP too. Weeeee...
Final words: ajax is not a language, its a javascript 'trick'. You don't need to fully understand what the first 'doAjaxThing' function does to use this, just make sure you are calling it properly. It will automatically 'call' the 'deal WithResponse' function once the response from the server arrives. Notice that you can continue doing your business (asynchronous - process not time-tied) till the response arrives - which is when the 'deal WithResponse' gets triggered -, as opposed to having a page stop and wait (synchronous - time tied) until a response arrives. That is the magic of ajax (Asynchronous JAvascript and Xml).
In your case you want to add the echo("success") - or error! - in the PHP, so that the function 'dealWithResponse' knows what to do based on that info.
That's all I know about ajax. Hope this helps :)
I'm new to coding and now I'm trying to write a code that gets the text data form a file and replace the present text with the new one. I'm using AJAX to do the task but the problem is first I'm getting the error message and then expected answer The error message is what I have included in the code to display when there is error. Even though i'm getting the desired answer I want to know why the error message is displayed. Here is my code
<!DOCTYPE html>
<html>
<head>
<script src="http://ajax.aspnetcdn.com/ajax/jquery/jquery-1.9.0.js"></script>
<script src="http://ajax.aspnetcdn.com/ajax/jquery/jquery-1.9.0.min.js"></script>
<script>
function loadXML() {
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("p1").innerHTML = xmlhttp.responseText;
}
else {
alert(" there is a error in your code");
}
}
xmlhttp.open("GET","robots.txt", true);
xmlhttp.send(null);
}
</script>
</head>
<body>
<p id="p1">This is Text</p>
<button id="b1" onclick="loadXML()">Click me </button>
</body>
</html>
The problem is your if-block in onreadystatechange. During the request and response, xmlhttp.readyState changes multiple times and onreadystatechange is called every time this happens.
If you do it like this, it may work:
xmlhttp.onreadystatechange = function () {
if (xmlhttp.readyState == 4 ) {
if( xmlhttp.status == 200 ) {
document.getElementById("p1").innerHTML = xmlhttp.responseText;
} else {
alert(" there is a error in your code");
}
}
It is simpler, however, to use the other event-methods like onload and onerror, as described here.
1- include 1 jQuery file jquery-1.9.0.js same as jquery-1.9.0.min.js but jquery-1.9.0.min.js is minified file
2- what is the Error message from javaconsole log ?
3- you are using jQuery so why you are working with XMLHttpRequest when jQuery $.get already using best of HttpRequest for all browsers
you can replace your JS code with this
<script type="text/javascript">
$( document ).ready(function() {
$.get('robots.txt',function(data){ $('#p1').html(data) });
});
<script>
Are you using a webserver ? Ajax doesn't work from local file url : 'file://...'
alert(" there is a error in your code");
Actually there is no error in your code. xmlhttp.onreadystatechange is called in different states of the request. Thats why you check for xmlhttp.readyState == 4. Learn more about the different stages here: http://www.w3schools.com/ajax/ajax_xmlhttprequest_onreadystatechange.asp
So your code is running fine, you just alert an error message, that is not an error at all.