Android MySQL Database Tutorial 1 – Creating Database And Writing PHP Script




Android PHP/MYSQL Tutorial PHP Android Tutorial Part Best tutorial for android PHP and mysql How to connect Android app to MySQL database? Android …

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25 responses to “Android MySQL Database Tutorial 1 – Creating Database And Writing PHP Script”

  1. i followed the entire tutorial, but when i run the app only the alert dialog with the title appears, I do not get the success message. i tried unblocking the port 80 as well thru windows firewall but i cant understand where the problem is …. can some one pls help???

  2. Hi Sir, nice and very knowledgable Tutorial. I follow everything you mintioned, but i just get message "Login Status", indeed not refered if the login is success or not. Thanks and waiting your help.

  3. I have this error although i followed every step?

    Fatal error: Call to undefined function mysqlli_connect() in C:wamp64wwwconn.php on line 6
    Call Stack
    # Time Memory Function Location
    1 0.0005 235712 {main}( ) …conn.php:0

  4. Please help! I am getting the following error when trying to launch localhost/conn.php in the browser:

    Parse error: syntax error, unexpected '$db_name' (T_VARIABLE) in /Applications/XAMPP/xamppfiles/htdocs/conn.php on line 3

  5. thanku so much brother
    i have a question
    can u please tell me how to access a specific function of web api inside my android program?

    if(service==washing)
    then it should fetch the data for me according to api.

    how can i use it un my program?

  6. Hi Guys i fixed the problem for message "Login Status"
    first thing from wamp server/alias/ change "require local" to " require all granted" for 4 files there.
    php code for conn, login:

    <?php
    $db_name="employee";
    $mysql_username="root";
    $mysql_password="";
    $server_name="localhost";
    $conn = mysqli_connect($server_name,$mysql_username , $mysql_password ,$db_name );
    ?>

    <?php
    require "conn.php";
    $user_name=$_POST["user_name"];
    $user_pass=$_POST["password"];
    $mysql_qry="select * from employee_data where username like '$user_name' and password like '$user_pass';";
    $result=mysqli_query($conn ,$mysql_qry);
    if(mysqli_num_rows($result) > 0){
    echo "login success";

    }
    else{
    echo "login not success";
    }

    ?>

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