当用户在输入框中输入唯一代码时，从mysql数据库中检索用户配置文件详细信息

The following code displays a BLANK PAGE when I hit the submit button. I have not a single idea what's wrong. Help check please..

The code below:

//The html code

<form method="post" action="dutydata.php">
   <input type="text" placeholder="provide unique code">
   <input type="submit" name="verify">
</form>

//the php code

<?php
    $conn = mysqli_connect("localhost", "root", "", "army_duty");
    $set = $_POST['verify'];
    if($set) {
        $show = "SELECT * FROM profile where military_number = '$set' ";
        $result = mysqli_query($conn, $show);
        while ($row = mysqli_fetch_array(result)) {
            echo $row['military_number'];
            echo $row['first_name'];
            echo $row['last_name'];
            echo $row['paygrade'];
            echo $row['duty_status'];
            echo $row['photo'];
            echo "<br/>";
        }
    } else {
      echo "Military Number not found";
    }
?>

First, your input for the code needs a name:

<input type="text" name="code" placeholder="provide unique code">

This is what you will use in the query because $_POST['verify'] does not contain any value that you would want to use.

Second, you will want to use the $_POST['code'] number in your query:

$code = $_POST['code'];
if($set) {
    $show = "SELECT * FROM profile where military_number = '$code' ";

Make sure to check for errors:

$result = mysqli_query($conn, $show) or die(mysqli_error($conn));

If you want to know if something was returned from the query you test the result. Since you are only getting one record you can skip the while() loop:

if($result) {
    $row = mysqli_fetch_array($result);
    echo $row['military_number'];
    echo $row['first_name'];
    echo $row['last_name'];
    echo $row['paygrade'];
    echo $row['duty_status'];
    echo $row['photo'];
    echo "<br/>";
} else {
    echo "Military Number not found";
}

Warning!

Little Bobby says your script is at risk for SQL Injection Attacks. Learn about prepared statements for MySQLi. Even escaping the string is not safe!