Alert command within php not working? - javascript

I want to pass a variable in header .if the variable is exist than give an alert.
but its not working
login.php
<?php
include "db.php";
$user=$_POST['t1'];
$pass=$_POST['t2'];
$result=mysql_query("select * from registor where username='$user'")or die(mysql_error());
$row=mysql_fetch_row($result);
if($user=='' || $pass==''){
header("location:account.php?wrong");
}
?>
account.php
<?php
if(isset($_GET['wrong']))
{
?>
<script>alert(Please enter detials !!);</script>
<?php
}
if(isset($_GET['user']))
{
?>
<script>alert(Now, Login Please !!);</script><?php
}
?>

You need quotes around strings :
<script>alert("Now, Login Please !!");</script><?php
But you should have seen the error in the console. Whenever you have something not working client side, always look at the console first.

You're actually passing nothing in your header. Make it like this and it'll work:
header("location:account.php?wrong=wrong");
and use single or double qoutes in your alert message:
<?php
if(isset($_GET['wrong']))
{
?>
<script>alert('Please enter detials !!');</script>
<?php
}
if(isset($_GET['user']))
{
?>
<script>alert('Now, Login Please !!');</script><?php
}
?>

Related

How to call javascript function from form in a php echo?

what the bellow code does is making sure the user isn't allowed to submit a comment unless he's signed in by using $_SESSION['login_user'] supervariable. But it's giving me an error. I think the problem is because I'm calling a javascript function in onsumbit="return(checkUser())". There's something wrong there but I don't know why.
I have the following code:
<script type="text/javascript">
// notice the quotes around the ?php tag
function checkUser() {
<?php
if(isset($_SESSION['login_user'])){
$isExist = true;
}
?>
else{
$isExist= false;
alert( "Please register first!" );
}
var htmlString="<?php echo $isExist; ?>";
return isExist;
}
</script>
...
...
<?php
echo "<form method='POST' onsubmit="return(checkUser());" action='".setComments($connection, $res['post_id'])."'>
//echo "<form method='POST' action='".setComments($connection, $res['post_id'])."'>
<input type='hidden' name='uid' value='".$_SESSION['login_user']."'>
<input type='hidden' name='date' value='".date('Y-m-d H:i:s')."'>
<textarea name='message'> </textarea><br>
<button type='submit' name='commentSubmit'>Comment</button>
</form>";
getComments($connection, $res['post_id']);
?>
....
If this is not the right method to stop the user from submitting a comment, then what could be another method?
In addition to what #RonaldT said, you need to understand that the PHP code is executed on the server before being sent to the browser. So checking for $_SESSION['login_user'] inside a Javascript function is kind of silly, since it will always be the same until the user refreshes the page (only then will PHP re-check the value).
So your function can be simplified like this:
<script type="text/javascript">
// on page load, define a global variable using PHP
var isLoggedIn = <?php echo isset($_SESSION['login_user']) ? "true" : "false"; ?>;
function checkUser() {
// check that global variable every time checkUser() is called in Javascript
if (!isLoggedIn) {
alert( "Please register first!" );
}
return isLoggedIn;
}
</script>
Keep in mind that this kind of "security" is extremely easy to fool (any user can just open their browser console, type isLoggedIn = true; and voila), so be sure to check on the server as well when the form is submitted.
Or better yet: if a user is not allowed to do something, don't give them the opportunity. Why display the form at all if the user will not be allowed to submit it anyway?
<?php
if (!isset($_SESSION['login_user'])) {
echo "Please register to add a comment";
} else {
echo "<form [...everything else...] /form>";
}
getComments($connection, $res['post_id']);
?>

Accessing Through PHP a Posted Javascript Variable

I realize that there are several similar questions that have been asked, but none of those have been able to get me over the top. Maybe what I wnat to do is just not possible?
I have a page on which there is an order form. The admin can create an order for any user in the database by selecting them in the dropdown menu and then fill out the form. But each user may have a PriceLevel that will give them a discount. So I need to be able to make a database call based on the username selected in the dropdown and display their price level and be able to use the username and pricelevel variables in my PHP.
I have the an add_order.php page on which the form resides, and an ajax.php which makes a quick DB call and returns the results in a json format.
The problem I am running into is actually getting the information from jQuery into the PHP. I have tried using the isset method, but it always comes back as false.
Here's what I have:
add_order.php
<?php
// $username = $_POST['orderUser']['Username'];
$username = isset($_POST['orderUser']) ? $_POST['orderUser']['Username'] : 'not here';
echo 'hello, ' . $username;
?>
...
$('#frm_Username').change(function() {
orderUser = $(this).val();
$.post('/admin/orders/ajax.php', {
action: 'fetchUser',
orderUser: orderUser
}
).success(function(data) {
if(data == 'error') {
alert('error');
} else {
console.log(data);
}
})
})
ajax.php
<?php
$action = $_POST['action'];
if($action == "fetchUser"):
$un = $_POST['orderUser'];
/*if($un):
echo $un;
exit;
endif;*/
// SET THE REST UP WITH MYSQL
if($un):
$qid = $DB->query("SELECT u.Username, u.PriceLevel FROM users as u WHERE u.Username = '" . $un . "'");
$row = $DB->fetchObject($qid);
// $row = jason_decode($row);
echo json_encode($row);
exit;
endif;
echo "error";
endif;
?>
I am logging to the console right now and getting this:
{"Username":"dev2","PriceLevel":"Tier 2"}
Any help would be appreciated. Thanks.
After calling $.post('/admin/orders/ajax.php', ...), the PHP code which sees your POSTed variable is ajax.php.
You need to check in there (inside ajax.php), whereas currently your isset check is in add_order.php, which does not see the POST request you send.
You do seem to have some logic in ajax.php, but whatever you've got in add_order.php is not going to see the data in question.

JavaScript not working in mPDF

I am using mPDF for PDF generation in PHP. It is working fine with no issue.
What I want if user is not logged in then I would like to show error in alert box and after that redirect to index.php.
But due to some reason that I don't know it is not showing any alert box nor redirect. It seems like JavaScript is not working.
Here is the code:
<?php
session_start();
$uid=$_SESSION['uid'];
if($uid== "" or $uid == NULL)
{
echo '<script type="text/javascript">window.alert("Please login first to access this page."); </script>';
echo '<script type="text/javascript">window.location.href("/index.php");</script>';
}
Above code is top of the file and now below I have these code for mPDF:
include("pdf/mpdf.php");
$mpdf=new mPDF('');
$stylesheet = file_get_contents('pdf/tablecss.css');
$mpdf->WriteHTML($stylesheet,1);
//==============================================================
//$mpdf->WriteHTML($html);
$mpdf->SetDisplayMode('fullpage');
$mpdf->SetWatermarkText(' www.somewebsite.com ');
$mpdf->watermarkTextAlpha = 0.1;
$mpdf->watermark_font = 'DejaVuSansCondensed';
$mpdf->showWatermarkText = true;
$mpdf->WriteHTML($html);
$html = '
<html>
<head>
<style>
....
I fixed that. What i did is i put mPdf code inside else.
Like this and it works.
if($uid== "" or $uid == NULL)
{
echo '<script type="text/javascript">window.alert("Please login first to access this page."); </script>';
echo '<script type="text/javascript">window.location.replace("/index.php");</script>';
}else{
mpdf code goes here

Echo after header

This is a part of my code. The code is moving to the location, but the echo below is not working.
I am sure this header is affecting it somehow
else {
header('location: index.php');
echo '<script>
document.getElementById("nick").value = "invalid email";
document.getElementById("nick").className = "invalidemail";
</script>';
If you read something about header in php functions, you will came to know that nothing gets printed after header as you are giving instructions to redirect on index.php
However to achieve the same, you can use session variables.
So, on the page where you are redirecting to index.php:
$_SESSION['error']=true;
header('location: index.php');
On index.php, check if session variable is true:
if($_SESSION['error']== true)
{
echo '<script type="text/javascript">
document.getElementById("nick").value = "invalid email";
document.getElementById("nick").className = "invalidemail";
</script>';
}

how to make the result appear in popup widow?

I want to make the result comes out in a popup window.
Code:
<?php
$con=mysqli_connect("localhost","root","1234","fyp");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT password FROM admin WHERE email = '$_POST[email]' AND Admin = '$_POST[admin]'");
while($row = mysqli_fetch_array($result))
echo "your password is : " . " $row['password']" ;
mysqli_close($con);
?>
Is it possible to make it echoed in popup window like javascript alert messages ??
I have tried this but still not working
echo "<script> alert ("<?php echo 'your password is: ' . '$row['password']'?>")</script>";
I found a maybe strange solution for this a while back.
echo "<style onload=\"jsfunc($row['password']);\"></style>";
//in your html or javascript add this function
<script type='text/javascript'>
function jsfunc(data){
alert(data);
}
</script>
the style tag is invisible and will run the function onload, so when its gets echoed. Also I use the style tag because its one of the few tags where the onload works when you echo it like this.
Its a strange solution but it works.
There are some serious flaws in your code, including it being open to SQL Injection. I'd recommend changing it to look more like this:
$con = new MySQLi("localhost","root","1234","fyp");
if($con->connect_errorno) {
echo "Failed to connect to MySQL: ".$sql->connect_error;
}
$email = $sql->real_escape_string($_POST['email']);
$admin = $sql->real_escape_string($_POST['admin']);
$query = "SELECT password FROM admin WHERE email = '$email' AND Admin = '$admin'";
$result = $sql->query($query);
if($result) {
$row = mysqli_fetch_assoc($result);
$pass = $row['password'];
echo '<script> alert("Your password is '.$pass.'");</script>';
} else {
//do some error handling
}
//the closing of a mysqli connection is not required but
mysqli_close($con);
Real escape string is not 100% proof against injection but is a good place to start with sanitising your inputs. Additionally I would strongly advise against storing your passwords in plain text. Please take a look at the PHP sha1() function or the SQL equivalent when storing the passwords initially.
Use this code
<?php
$alert='your password is: '.$row['password'];
?>
<script>
alert("<?php echo $alert; ?>");
</script>
It will work for sure.

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