I can't insert data. I don't know what went wrong.

Question

<?php

// Determine duplicate passwords

if(trim($_POST['pwd']) != trim($ _POST['rpwd'])){

exit('The two passwords are inconsistent, please return to the previous page');

}

// Ready to write data

$username = trim($_POST['username']); // Visible data

$password = md5($_POST[' pwd']); // Visible data, md5 is a way to encrypt passwords

$time = time(); // Invisible data, returns unix timestamp , the user's registration time

$ip = $_SERVER['REMOTE_ADDR']; // Invisible data, returns the IP address, the user's registered IP, we can use ip2long to convert it to integer storage

// Connect to the database server, determine errors, select the database and set the character set

$conn = mysqli_connect('localhost', 'root', '123' );

if(mysqli_errno($conn)){

echo mysqli_error($conn);

exit;

}

##mysqli_select_db($conn, 'user');

mysqli_set_charset($conn, 'utf8');

// Combined SQL statement

$sql = "insert into user(username, password, createtime, createip) values('" . $username ."', '" . $password . "', '" . $time . "', '" . $time . "', '" . $ip . "')";

// Send statement and determine status

$result = mysqli_query($conn, $sql);

if($result){

echo 'success' . "<br />";

}else{

echo 'failure' . "<br />";

}

// Use mysqli_insert_id() to print out the auto-incremented primary key ID

echo 'The ID inserted by the current user is:' . mysqli_insert_id ($conn);

// Close the database connection

mysqli_close($conn);

##?>

Answer 1

There is nothing wrong with the PHP code. You can print out whether the data submitted in the form exists, and then test whether the $result value exists to find errors. Also, is your database table really set to have an ID auto-increment? Anyway, I succeeded, haha

Run result:

Answer 2

What is the execution result?