Question

我以前的代码通过使用“用户名”和“密码”数据收集而成功。但是我尝试添加新数据以在用户注册期间插入，新数据是“名称”。

因此，我已经在phpmyadmin和PHP脚本中添加了“名称”属性。不幸的是，当我检查phpmyadmin时，数据被错误地插入到列中。

<?php
// Include config file
require_once "include/config.php";

// Define variables and initialize with empty values
$username = $name = $password = $confirm_password = "";
$username_err = $name_err = $password_err = $confirm_password_err = "";

// Processing form data when form is submitted
if($_SERVER["REQUEST_METHOD"] == "POST"){

    // Validate username
    if(empty(trim($_POST["username"]))){
        $username_err = "Please enter Staff ID.";
        $name_err = "Please enter Staff Name.";
    } else{
        // Prepare a select statement
        $sql = "SELECT id FROM users WHERE username = ?";

        if($stmt = mysqli_prepare($link, $sql)){
            // Bind variables to the prepared statement as parameters
            mysqli_stmt_bind_param($stmt, "s", $param_username);

            // Set parameters
            $param_username = trim($_POST["username"]);

            // Attempt to execute the prepared statement
            if(mysqli_stmt_execute($stmt)){
                /* store result */
                mysqli_stmt_store_result($stmt);

                if(mysqli_stmt_num_rows($stmt) == 1){
                    $username_err = "Staff ID is already exist.";
                    $name_err = "Staff name is already exist.";
                } else{
                    $username = trim($_POST["username"]);
                }

            } else{
                echo "Oops! Something went wrong. Please try again later.";
            }
        }

        // Close statement
        mysqli_stmt_close($stmt);
    }

    // Validate password
    if(empty(trim($_POST["password"]))){
        $password_err = "Please enter a password.";     
    } elseif(strlen(trim($_POST["password"])) < 6){
        $password_err = "Password must have atleast 6 characters.";
    } else{
        $password = trim($_POST["password"]);
    }

    // Validate confirm password
    if(empty(trim($_POST["confirm_password"]))){
        $confirm_password_err = "Please confirm password.";     
    } else{
        $confirm_password = trim($_POST["confirm_password"]);
        if(empty($password_err) && ($password != $confirm_password)){
            $confirm_password_err = "Password did not match.";
        }
    }

    // Check input errors before inserting in database
    if(empty($username_err) && empty($name_err) && empty($password_err) && empty($confirm_password_err)){

        // Prepare an insert statement
        $sql = "INSERT INTO users (username, name, password) VALUES (?, ?, ?)";

        if($stmt = mysqli_prepare($link, $sql)){
            // Bind variables to the prepared statement as parameters
            mysqli_stmt_bind_param($stmt, "sss", $param_username,  $param_name, $param_password);

            // Set parameters
            $param_username = $username;
            $param_password = password_hash($password, PASSWORD_DEFAULT); // Creates a password hash
            $param_name = $name;

            // Attempt to execute the prepared statement
            if(mysqli_stmt_execute($stmt)){
                // Redirect to login page
                header("location: index.php");
            } else{
                echo "Something went wrong. Please try again later.";
            }
        }

        // Close statement
        mysqli_stmt_close($stmt);
    }

    // Close connection
    mysqli_close($link);
}
?>


<form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>" method="post">
            <div class="form-group <?php echo (!empty($username_err)) ? 'has-error' : ''; ?>">
                <label>Staff ID</label>
                <input type="text" name="username" class="form-control" value="<?php echo $username; ?>">
                <span style="color:#FF0500" class="help-block"><?php echo $username_err; ?></span>
            </div>
            <div class="form-group <?php echo (!empty($name_err)) ? 'has-error' : ''; ?>">
                <label>Staff Name</label>
                <input type="text" name="name" class="form-control" value="<?php echo $name; ?>">
                <span style="color:#FF0500" class="help-block"><?php echo $name_err; ?></span>
            </div>    
            <div class="form-group <?php echo (!empty($password_err)) ? 'has-error' : ''; ?>">
                <label>Password</label>
                <input type="password" name="password" class="form-control" value="<?php echo $password; ?>">
                <span style="color:#FF0500" class="help-block"><?php echo $password_err; ?></span>
            </div>
            <div class="form-group <?php echo (!empty($confirm_password_err)) ? 'has-error' : ''; ?>">
                <label>Confirm Password</label>
                <input type="password" name="confirm_password" class="form-control" value="<?php echo $confirm_password; ?>">
                <span style="color:#FF0500" class="help-block"><?php echo $confirm_password_err; ?></span>
            </div>
            <div class="form-group">
                <input type="submit" class="btn btn-primary" value="Submit">
                <input type="reset" class="btn btn-default" value="Reset">
            </div>
            <p>Already have an account? <a href="index.php">Login here</a>.</p>
        </form>

Answer 1

更改此

// Bind variables to the prepared statement as parameters
mysqli_stmt_bind_param($stmt, "sss", $param_username,  $param_name, $param_password);
// Set parameters
$param_username = $username;
$param_password = password_hash($password, PASSWORD_DEFAULT); // Creates a password hash
$param_name = $name;

对此

// Set parameters
$param_username = $username;
$param_password = password_hash($password, PASSWORD_DEFAULT); // Creates a password hash
$param_name = $name;
// Bind variables to the prepared statement as parameters
mysqli_stmt_bind_param($stmt, "sss", $param_username,  $param_name, $param_password);

问题在于，在您的示例中，$param_*变量中的任何一个都未初始化，然后再将其传递给mysqli_stmt_bind_param()函数。

请问您可以在更改后共享结果吗？

编辑：分配$name变量。在准备SQL语句之前添加此代码。

if(empty(trim($_POST["name"]))){
   /* ERROR_HANDLING */ 
} else{ $name = $_POST["name"]; }

在PHPMyAdmin中错误地插入数据

1 个答案: