AJAX call to php not returning value when using post method

Question

PHP Code

<?php
    $servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "ngram";

    $conn = new mysqli($servername, $username, $password, $dbname);
    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    }
    $uid =$_REQUEST["username"];
    $sql = 'select * from user_db where username like '.'"'.$uid.'"';
    $result = $conn->query($sql);
    //echo $sql;
    if ($result->num_rows > 0) 
    {
        $conn->close();
        echo (0);
    }
    else 
    {   
        $conn->query("insert into user_db values('".$_REQUEST["name"]."','".$_REQUEST["username"]."','".$_REQUEST["email"]."','".$_REQUEST["pass"]."')");
        $conn->close();
        echo (1);
    }

?>

function connectDb(formElement)
{
    var xmlhttp = new XMLHttpRequest();
    xmlhttp.open("POST","signup.php",true);
    xmlhttp.onreadystatechange=function() 
    {
        if (xmlhttp.readyState == 4 && xmlhttp.status == 200)
        {
            myFunction(xmlhttp.responseText);
        }
    }
    xmlhttp.send(new FormData (formElement));
}
function myFunction(response)
{   
    if(response != 1)
    {
        window.open("error.html","_self");
    }
    else
    {
        window.open("login.html","_self");
    }
}

Here javascript is able to send request but php is not returning any value. The values gets added to the database but the page does not changes. Only returned value gets displayed in screen.

You are using both "xmlhttp" and "xhttp" in your javascript. I believe you need to use "xmlhttp" everywhere. — Matthew Levy
– Matthew Levy, Commented Oct 25, 2016 at 12:45

Abasaheb Gaware · Accepted Answer · 2016-10-25 12:27:27Z

1

you write return $r but not write a function for access it.write a function or either write echo $r;.

answered Oct 25, 2016 at 12:27

Abasaheb Gaware

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1 Comment

Varun Over a year ago

i have tried with echo $r but it just prints it in a new page.

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