Question

Let say I submit data to a form with the following code

var xhr = new XMLHttpRequest(), formData = new FormData();  
formData.append("img", img);
formData.append("user", localStorage["username"]);
formData.append("pass", localStorage["password"]);

xhr.onreadystatechange = function (event) {
    if (xhr.readyState === 4 && xhr.status === 200) {
       var value = xhr.responseText; // value should equal "1234"
        alert( "value = " + value );
    }else{
    alert("none");
}
};
xhr.open("POST", "http://joubin.me/uploads3/upload_file.php", true);
xhr.send(formData);

After upload.php is done, it redirects to another page called giveid.php and the only thing it displays is a text string with an id

say 1234

How can I with javascript capture this exact id.

Keep in mind, a different upload.php redirect will have a different id number on giveid.php?

I looked into the xmlhtml responses and could not figure it out.

Here is what form goes

$password = $_REQUEST['pass'];
$username =  $_REQUEST['user'];
$image = $_REQUEST['img'];
echo $password;
echo "<br/>";
echo $username;
echo "<br/>";
echo $image;
$con = mysql_connect("localhost","ya","right");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
        echo "could not connect";
  }
$asql = "SELECT * FROM `ashkan`.`users` where user='$username' and pass='$password';";

$result = mysql_query($asql);
echo $result;
// Mysql_num_row is counting table row
$count=mysql_num_rows($result);

echo $count;

echo 11;
if($count == 1){
$sql = "INSERT INTO `ashkan`.`goog` (`user`, `pass`, `img`) VALUES ('$username', '$passwo$
}
mysql_query($sql);
mysql_close($con);
header( 'Location: giveid.php' ) ;

and here is the content of giveid.php

1234

Any help would be great.

Thanks

Answer 1

You need to use xhr.onreadystatechange to retrieve the response from the server.

Something like this might work.

var value;
var formData = new FormData();
formData.append("img", img);
formData.append("user", localStorage.username);
formData.append("pass", localStorage.password);

var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function (event) {
    if (xhr.readyState === 4 && xhr.status === 200) {
        value = xhr.responseText; // value should equal "1234"
        alert( "value = " + value );
    }
};
xhr.open("POST", "upload.php", true);
xhr.send(formData);

Info Here: http://www.tizag.com/ajaxTutorial/ajaxxmlhttprequest.php

Remember that header() must be called before any actual output is sent. So get rid of all the echos you have in the php file. Once you echo then that starts the output buffer for the response to the client.

Info Here: http://php.net/manual/pt_BR/function.header.php

I think this should be your only echo on the php page.

echo include( 'giveid.php');

Try using Google Chrome Dev Tool Network tab to view the response from your php webpage.

• Launch Google Chrome,
• Hit f12,
• Click the network tab,
• reload your page,
• click on the ajax response page,
• click preview to view the response.

Info Here: https://developers.google.com/chrome-developer-tools/docs/network