Ajax Request not returning PHP Loop Echo - javascript

The PHP is working fine when directly called in browser, but the ajax.responseText does not contain anything given out by the While Loop.
$result = $mysqli->query("SELECT * FROM `Dienstplan` WHERE MONTH(Datum) = '$Monat'");
while ($row = $result->fetch_assoc()) {
$Datum = $row['Datum'];
$Datum = strtotime($Datum);
$Datum = date('d.m.Y',$Datum);
$Tag = $row['Tag'];
$Dienst = $row['Dienst'];
//echo '<tr><td>'.$Tag.' '.$Datum.' '.$Dienst.'</td></tr>';
echo $Tag.' '.$Datum.' '.$Dienst.'';
}
echo "This is a Test!";
The ajax.responseText only contains the last echo after the While Loop. But when I call the PHP script directly in my Browser I can clearly see all the data given out by the loop. For some reason it's not contained in the Ajaxresponse.
Here is the Ajax Code:
var ajax;
ajax = new XMLHttpRequest();
ajax.open("GET","../php/FTL_Month_2.php",true);
ajax.send();
ajax.onreadystatechange=function(){
if (ajax.readyState==4 && ajax.status==200){
//alert(ajax.responseText);
document.getElementById("DataList").innerHTML=ajax.responseText;
}
}

As a best practice, you should get all the data you want to return, and store it in a variable. after that, out of the loop, echo this variable.

Related

Using AJAX to get an SQL WHERE query into javascript

I am trying to pass a javascript variable into an SQL WHERE query and I keep getting null in return.
On-click of a button, the buttonClick function is ran:
<script>
var var1;
function buttonClick(elem){
var1 = elem.src //this gets the url from the element
var path = var1.slice(48); //this cuts the url to img/art/9/1.jpg
ajax = theAjax(path);
ajax.done(processData);
ajax.fail(function(){alert("Failure");});
}
function theAjax(path){
return $.ajax({
url: 'info.php',
type: 'POST',
data: {path: path},
});
}
function processData(response_in){
var response = JSON.parse(response_in);
console.log(response);
}
</script>
Here is the code stored in the info.php file:
<?php
$path = $_POST['path'];
$result3 = mysqli_query("SELECT itemName from images WHERE imgPath='$path'");
$json = json_encode($result3);
echo $json
?>
As you can see, once I click the button, the buttonClick() function is ran and a variable stores the image path or src. That path variable is send to theAjax() function where it is passed to the info.php page. In the info.php page, the SQL WHERE query is ran and returned to the processData() function to be parsed and printed in the developer console. The value printed shows null.
Below is a picture of what I am trying to get from the database:
1.Check that path is correct or not? you can check inside jquery using console.log(path); or at PHP end by using print_r($_POST['path']);
2.Your Php code missed connection object as well as record fetching code.
<?php
if(isset($_POST['path'])){
$path = $_POST['path'];
$conn = mysqli_connect ('provide hostname here','provide username here','provide password here','provide dbname here') or die(mysqli_connect_error());
$result3 = mysqli_query($conn,"SELECT itemName from images WHERE imgPath='$path'");
$result = []; //create an array
while($row = mysqli_fetch_assoc($result3)) {
$result[] = $row; //assign records to array
}
$json = json_encode($result); //encode response
echo $json; //send response to ajax
}
?>
Note:- this PHP query code is wide-open for SQL INJECTION. So try to use prepared statements of mysqli_* Or PDO.
mysqli_query() required 1st parameter as connection object.
$result3 = mysqli_query($conn,"SELECT itemName from images WHERE imgPath='$path'"); // pass your connection object here
I think your issue is that you're trying to encode a database resource.
Try adjusting your PHP to look like the following:
<?php
$path = $_POST['path'];
$result3 = mysqli_query("SELECT itemName from images WHERE imgPath='$path'");
$return_data = [];
while($row = mysqli_fetch_assoc($result3)) {
$return_data[] = $row;
}
$json = json_encode($return_data);
echo $json
?>

JavaScript HttpRequest always return the same results

I have a HTML page JavaScript which send a GET request data to PHP file to return all datas saved in the database . PHP replies with a HTML-table - that works fine!
But: When if i click a button (which calls the same JavaScript function) to update my table in order to display the new data, i get the same result (and i have definitely new data on table).
If I call the PHP manually via the browser it'll show me the new results immediately and at this moment it is also working with JavaScript (but only once).
Here is a part of my code.
HTML/JS:
<button onclick="GetData()"></button>
<div id="test"></div>
<script>
function GetData(){
var xhttp = new XMLHttpRequest();
document.getElementById("test").innerHTML = "";
xhttp.onreadystatechange = function(){
if (xhttp.readyState == 4 && xhttp.status == 200){
document.getElementById("test").innerHTML = xhttp.responseText;
}
};
xhttp.open("GET", "../GetData.php", true);
xhttp.send();
}
</script>
PHP:
//DB details
$dbHost = 'localhost';
$dbUsername = 'lalalala';
$dbPassword = 'lalalalal';
$dbName = 'lalalala';
//Create connection and select DB
$db = new mysqli($dbHost, $dbUsername, $dbPassword, $dbName) or die ("UUUUPS");
$sql = "select name, beschreibung, image, video from data";
$result = $db->query($sql);
if ($result->num_rows > 0) {
$return = '<table class ="table table-hover"><thead><tr><th scope="col">Name</th><th scope="col">Beschreibung</th><th scope="col">Bilddatei</th><th scope="col">Video-Url</th></tr></thead><tbody>';
// output data of each row
while($row = $result->fetch_assoc()) {
$return .= "<tr><td>".$row["name"]."</td><td>".$row["beschreibung"]."</td><td><img src=data:image/png;base64,".base64_encode($row["image"])."/></td><td>".$row["video"]."</tr>";
}
$return .= "</tbody></table>";
$db->close();
echo $return;
} else {
echo "0 results";
}
Thank you for your help!
It seems your browser is caching your result, that's why you see data.
You can test it like this:
var random = Math.floor(Math.random() * 100);
xhttp.open("GET", "../GetData.php?"+random, true);
If this helps, look into expire headers in your PHP script. Also, the way you're doing queries in quite outdated. It's a very PHP4 way. Have a look here: http://php.net/manual/en/book.mysqli.php
I guess you probably know this, but just in case. Have you had a look in your browsers inspector, when testing you html page? especially the network tab within that inspector. There you can see the actual response from the server and you can see if it is served from cache or fetched (you can even disable cache there), maybe this helps.
Kind regard,
Mark

Ajax post to php > get php variable with xmlhttprequest

I want to update my table with javascript every couple seconds.
So far I made an ajax post request to my update.php and trigger if is set. Then do a mysql query and put the resultset in a json variable.
After this i get it with a XMLHttpRequest.
The problem is that every XMLHttpRequest example uses echo json. But
when I put my echo json in my isset post check it wont return anything
anymore.
I think the problem is that it can't echo?
This is my code:
PHP:
if (isset($_POST["updateTable"])) {
$result = mysqli_query($con,"SELECT * FROM users");
$row = mysqli_fetch_row($result);
$rsArray[] = array();
while ($row) {
$rsArray[] = $row;
$row = mysqli_fetch_row($result);
}
echo json_encode($rsArray);
}
AJAX:
var updateTable = "true";
setTimeout(function(){
$.ajax({
type: 'POST',
url: '../EXT_PHP/update.php',
data: { updateTable: updateTable }
});
console.log(updateTable);
}, 3000);
xmlhttpRequest:
setTimeout(function(){
var oReq = new XMLHttpRequest();
oReq.onload = function() {
alert(this.responseText);
};
oReq.open("get", "../EXT_PHP/update.php", true);
oReq.send();
}, 3000);
Have you Tried to set the header of you PHP Script to JSON?
header('Content-Type: application/json');
Note that the header must be set before the first output of your Programm (i.E. the first echo)
<?php
header('Content-Type: application/json');
if (isset($_POST["updateTable"])) {
$result = mysqli_query($con,"SELECT * FROM users");
$row = mysqli_fetch_row($result);
$rsArray[] = array();
while ($row) {
$rsArray[] = $row;
$row = mysqli_fetch_row($result);
}
echo json_encode($rsArray);
}
if that doenst work still you should check if there has been any errors within the json_encode function, there are some problems with encoding by time. Use
echo json_last_error_msg();
for this.

Trouble with php variables and ajax javascript

ok I have edited this to another couple of questions I've asked on a similar issue, but I really am in a rush so thought I'd start a new one, sorry if it bothers anyone.
first I have a php script on test.php on the apache server
<?php
//create connection
$con = mysqli_connect("localhost", "user", "password", "dbname");
//check connection
if (mysqli_connect_errno()){
echo "failed to connect to MySQL: " . mysqli_connect_error();
}
$grab = mysqli_query($con, "SELECT * FROM table");
$row = mysqli_fetch_array($grab);
$name = $row["name"];
$color = $row["color"];
$price = $row["price"];
$n1 = $name[0];
$c1 = $color[0];
$p1 = $price[0];
?>
Then I've got this ajax script set to fire onload of page a webpage written in html. so the load() function is onload of the page in the body tag. This script is in the head.
function load(){
var xmlhttp = new XMLHttpRequest();
xmlhttp.open("GET", "test.php", true);
xmlhttp.send();
xmlhttp.onreadystatecahnge = function(){
if(xmlhttp.readyState == 4 && xmlhttp.status == 200){
document.getElementById("itemNameLink1").innerHTML = "<?php echo $n1;?>;
}
}
}
ok so what I want is the $n1 variable in the php script to be used in the javascript ajax code. Where the script is, but I'm not sure where or how to make use of the variable, I've tried a few things. All that happens right now is the innerHTML of itemNameLink1 just disappears.
I'm quite new so any advise would be appreciated, thanks.
The response (this is what you echo in php) returned from request you can get by responseText attribute of XMLHttpRequest object.
So first your JS code should be:
function load(){
var xmlhttp = new XMLHttpRequest();
xmlhttp.open("GET", "test.php", true);
xmlhttp.send();
xmlhttp.onreadystatecahnge = function(){
if(xmlhttp.readyState == 4 && xmlhttp.status == 200){
document.getElementById("itemNameLink1").innerHTML = xmlhttp.responseText;
}
}
}
now in php echo $n1 variable:
....
$grab = mysqli_query($con, "SELECT * FROM table");
$row = mysqli_fetch_array($grab);
$name = $row["name"];
$color = $row["color"];
$price = $row["price"];
$n1 = $name[0];
$c1 = $color[0];
$p1 = $price[0];
// echo it to be returned to the request
echo $n1;
Update to use JSON for multiple variables
so if we do this:
$name = $row["name"];
$color = $row["color"];
$price = $row["price"];
$response = array
(
'name' => $name,
'color' => $color,
'price' => $price
);
echo json_encode($response);
Then in javascript we can parse it again to have data object containing 3 variables.
var data = JSON.parse(xmlhttp.responseText);
//for debugging you can log it to console to see the result
console.log(data);
document.getElementById("itemNameLink1").innerHTML = data.name; // or xmlhttp.responseText to see the response as text
Fetching all the rows:
$row = mysqli_fetch_array($grab); // this will fetch the data only once
you need to cycle through the result-set got from database: also better for performance to use assoc instead of array
$names = $color = $price = array();
while($row = mysqli_fetch_assoc($grab))
{
$names[] = $row['name'];
$color[] = $row['color'];
$price[] = $row['price'];
}
$response = array
(
'names' => $names,
'color' => $color,
'price' => $price
);
You can dynamically generate a javascript document with php that contains server side variables declared as javascript variables, and then link this in the head of your document, and then include this into your document head whenever server side variables are needed. This will also allow you to dynamically update the variable values upon page generation, so for example if you had a nonce or something that needs to change on each page load, the correct value can be passed upon each page load. to do this, you need to do a few things. First, create a php script and declare the correct headers for it to be interpreted as a script:
jsVars.php:
<?php
//declare javascript doc type
header("Content-type: text/javascript; charset=utf-8");
//tell the request not to cache this file so updated variables will not be incorrect if they change
header('Cache-Control: no-cache, no-store, must-revalidate'); // HTTP 1.1.
header('Pragma: no-cache'); // HTTP 1.0.
header('Expires: 0'); // Proxies.
//create the javascript object
?>
var account = {
email: <?= $n1; ?>,
//if you need other account information, you can also add those into the object here
username: <?= /*some username variable here for example */ ?>
}
You can repeat this for any other information you need to pass to javascript on page load, and then reference your data using the namespaced javascript object (using object namespacing will prevent collisions with other script variables that may not have been anticipated.) wherever it is needed as follows:
<script type="text/javascript>
//put this wherever you need to reference the email in your javascript, or reference it directly with account.email
var email = account.email;
</script>
You can also put a conditional statement into the head of your document so it will only load on pages where it is needed (or if any permission checks or other criteria pass as well). If you load this before your other scripting files, it will be available in all of them, provided you are using it in a higher scope than your request.
<head>
<?php
//set the $require_user_info to true before page render when you require this info in your javascript so it only loads on pages where it is needed.
if($require_user_info == TRUE): ?>
<script type="text/javascript" href="http://example.com/path-to-your-script/jsVars.php" />
<?php endif; ?>
<script type="text/javascript" href="your-other-script-files-that-normally-load" />
</head>
You can also do this for any other scripts that have to load under specific criteria from the server.
You should define the PHP variable. And use that variable in your javascript:
<?php
$n1 = "asd";
?>
<html>
<head></head>
<body>
<div id="itemNameLink1"></div>
<script>
function load()
{
var xmlhttp = new XMLHttpRequest();
xmlhttp.open('GET', '/test.php', true);
xmlhttp.send(null);
//Note you used `onreadystatecahnge` instead of `onreadystatechange`
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById("itemNameLink1").innerHTML = '<?=$n1?>';
}
}
}
load();
</script>
</body>
</html>

How to use javascript in PHP? The script tag is not recognized

I don't know why but the script tag is not working, the SELECT query is working but i am not getting the prompt from the javascript.
it is not redirecting anywhere only a blank screen is seen
$qry1="SELECT area, aadhar FROM user where username='$user'";
$result1 = $connector->query($qry1);
if($result1){
$row1=mysql_fetch_array($result1);
$userarea= $row1['area'];
$useraadhar=$row1['aadhar'];
}?>
<body>
<script type="text/javascript">
var inputarea=<?php echo $coursename; ?>;
var userarea=<?php echo $userarea; ?>;
var useraadhar=<?php echo $useraadhar;?>'
if(inputarea==userarea){
<?php/
//date
$today = date("Y-m-d");
//Create INSERT query
$qry = "INSERT INTO complain (user,category,regno,course,lecturer,room,details,address,datein) VALUES ('$userid','$category','$reg','$coursename','$lectname','$roomno','$details','$address','$today')";
//$result = #mysql_query($qry);
$result = $connector->query($qry);
//Check whether difjslk the query was successful or not
if($result) {
$errmsg_arr[] = 'Complain succesfully added, please wait for your response';
$errflag = true;
if($errflag) {
$_SESSION['ERRMSG_ARR'] = $errmsg_arr;
session_write_close();
header("location: _new_complains.php");
exit();
}
header("location: _new_complains.php");
exit();
}else {
die("Query failed, couldn't add the new record");
header("location: _new_complains.php");
exit();
}
?>
}
You are sending data (for example body tag) before header(), therefore PHP creates an error. You just don't see it. Header needs to come before anything is sent to the browser (even a space).
You have multiple JS syntax errors:
var inputarea=<?php echo $coursename; ?>;
var userarea=<?php echo $userarea; ?>;
var useraadhar=<?php echo $useraadhar;?>'
Never EVER dump out raw text from PHP into a Javascript context. You're generating code that looks like
var inputarea=foo;
var userarea=bar;
var useradhar=baz';
The data will be seen as undefined variables, and you've got a stray ' in there. All of these errors will KILL the entire <script> block.
Always use json_encode() to dump from PHP->JS:
var inputarea = <?php echo json_encode($coursename); ?>;
This will GUARANTEE that you're producing correct Javascript code. The above line would produce
var inputarea = 'foo';
and be perfectly valid and executable code.

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