链接(在php中)到Android应用程序中的MainActivity

I got an error on line 16 which is:

   echo "You have been successfully registered. <a href="'.$link_address.'">Link</a>";

My code register.php is:

require "conn.php";

$name = $_POST["name"];
$surname = $_POST["surname"];
$age = $_POST["age"];
$username = $_POST["username"];
$password = $_POST["password"];
$link_address = "http://...";

$mysql_qry = "insert into employee_data (name, surname, age, username, password) values ('$name','$surname','$age','$username','$password');";


if($conn->query($mysql_qry)=== TRUE) {
echo "You have been successfully registered. <a href="'.$link_address.'">Link</a>";
}
else {
echo "Error: " . $mysql_qry . "<br>" . $conn->error;
}
$conn->close();
?>

Meanwhile:

What link should I use if I am trying to make a link to the MainActivity in android app (android studio)?

Maybe wrong quote ? Try :

echo 'You have been successfully registered. <a href="'.$link_address.'">Link</a>';

You are doing : "a" '.$var.' "a"
PHP will be lost ! Hope it helps a bit.

EDIT :

$link_address = "http://google.fr";
echo $link_address;
echo 'You have been successfully registered. <a href="'.$link_address.'">Link</a>';

This is my result :
enter image description here

Link goes on google.fr

Modify your code like this:

    require "conn.php";

    $name = $_POST["name"];
    $surname = $_POST["surname"];
    $age = $_POST["age"];
    $username = $_POST["username"];
    $password = $_POST["password"];
    $link_address = "http://...";

    $mysql_qry = "insert into employee_data (name, surname, age, username, password) values ('$name','$surname','$age','$username','$password');";


    if($conn->query($mysql_qry)=== TRUE) {
    ?>

    You have been successfully registered. <a href="<?php echo $link_address; ?>">Link</a>
   <?php
    }
    else {
    echo "Error: " . $mysql_qry . "<br>" . $conn->error;
    }
    $conn->close();
    ?>

This will work!!

Just use..

echo "You have been successfully registered. <a href=".$link_address.">Link</a>";

without ' there is a syntax error.

try using htmlentities to scape the $link_address var:

echo 'You have been successfully registered. <a href="'.htmlentities($link_address).'">Link</a>';