Question

我收到此错误，为什么它希望参数不是布尔值？我该如何解决？（我知道这是一个常见的错误，所以我想了解为什么会出现这种错误，以便我可以在它再次出现时自行修复它。）

警告：mysqli_stmt_num_rows（）要求参数1为mysqli_stmt，布尔值在第20行的C：\ Users \ James \ Desktop \ Container \ XAMPP \ htdocs \ Triiline1 \ signupuser.php中给出

<!doctype html>
<html>
<head>
<title>User Signup</title>
</head>
<body>
<?php
include 'connect.php';


//if submit is clicked
if (isset($_POST['submit'])) {
    //then check if all fields are filled
    if (empty($_POST['username']) | empty($_POST['password']) | empty($_POST['firstname']) | empty($_POST['MI']) | empty($_POST['lastname']) | empty($_POST['email']) | empty($_POST['phonenumber']) | empty($_POST['country']) ) {
        echo('You did not complete all of the required fields'); }

    $username = $_POST['username'];
    $password = $_POST['password'];
    $usernamesquery = mysql_query("SELECT * FROM logins WHERE username='$username'");
    if(mysqli_stmt_num_rows($usernamesquery) > 0) {
        die('This username is already taken.');
    }


} ?>
<form action="" method="post">

Username: <input type="text" name="username" maxlength="30"><br>
Password: <input type="password" name="password" maxlength="30"><br>
First Name: <input type="text" name="firstname" maxlength="30"><br>
Middle Initial: <input type="password" name="MI" maxlength="30"><br>
Last Name: <input type="text" name="lastname" maxlength="30"><br>
Email: <input type="password" name="email" maxlength="50"><br>
Phone Number: <input type="text" name="phonenumber" maxlength="11"><br>
Country: <input type="password" name="country" maxlength="40"><br>
<input type="submit" name="submit">
</form>


</body>
</html>

Answer 1

您在此声明中使用mysql_*而不是mysqli_*

$usernamesquery = mysql_query("SELECT * FROM logins WHERE username='$username'");

尝试：

$usernamesquery = mysqli_query($connection, "SELECT * FROM logins WHERE username='$username'");

修改

您需要在当前的实施中使用mysqli_num_rows($usernamesquery)而不是mysqli_stmt_num_row(...)：

mysqli_stmt_num_rows()用作：

if ($stmt = mysqli_prepare($connection, $query)) { /* execute query */ mysqli_stmt_execute($stmt); /* store result */ mysqli_stmt_store_result($stmt); printf("Number of rows: %d.\n", mysqli_stmt_num_rows($stmt)); }

Answer 2

尝试使用mysqli或mysql不要混淆..你正在使用两者

$usernamesquery = mysqli_query("SELECT * FROM logins WHERE username='$username'");
if(mysqli_stmt_num_rows($usernamesquery) > 0) {
    die('This username is already taken.');
}

错误：期望参数1为mysqli_stmt，给定布尔值？

2 个答案: