如何解决 PHP 文件中的“Parse error syntax error unexpected end of file”？ [复制]

Question

我查看了我的代码，但我不知道为什么这个错误不断出现。

我得到的确切错误信息 '解析错误：语法错误，第 92 行 C:\wamp\www\includes\functions.inc.php 中的文件意外结束'

这是我的代码：

<?php
#since this file contains solely PHP there is no need for the closing tag
#by convention it is better to check if something fails first then succeeds
function emptyInputSignup($username, $password, $passwordRepeat){
    $result = Null;
    if (empty($username) || empty($password) || empty($passwordRepeat)){
        $result = true;
    }
    else{
        $result = false;
    }
    return result;
}

function invalidUsername($username){
    $result = Null;
    #if there is a mistake
    if (!preg_match("/^[a-zA-Z0-9]*$/", $username)){
        $result = true;
    }
    else{
        $result = false;
    }
    return result;
}

function passwordMatch($password, $passwordRepeat){
    $result = Null;
    #if there has been a mistake
    if ($password !== $passwordRepeat){
        $result = true;
    }
    else{
        $result = false;
    }
    return result;
}

function usernameExists($conn, $username){
    $sql = "SELECT * FROM users WHERE username = ?";
    #we have created a prepared statement to prevent SQL injections
    $stmt = mysqli_stmt_init($conn);
    $result = Null;
    if (!mysqli_stmt_prepare($stmt, $sql)){
        header("location: ../signup.php?error=stmtfailed");
        exit();
    }
    
    mysqli_stmt_bind_param($stmt, "s", $username);
    mysqli_stmt_execute($stmt);
    
    $resultData = mysqli_stmt_get_result();
    
    if($row = mysqli_fetch_assoc($resultData)){ 
        return $row;        
        
    }
    else{
        $result = false;
        return $result;
}

function passwordValid($password){
    $result = Null;
    #if there is a mistake
    if (strlen($password) <= 5 && !preg_match("~[0-9]+~", $password)){
        $result = true;
    }
    else{
        $result = false;
    }
    return result;
}

这是我认为错误所在的函数。

function createUser($conn, $username, $password){
    $sql = "INSERT INTO users (username, password) VALUES (?, ?)";
    #we have created a prepared statement to prevent SQL injections
    $stmt = mysqli_stmt_init($conn);
    #this checks if the statement is feasible
    if (!mysqli_stmt_prepare($stmt, $sql)){
        header("location: ../signup.php?error=stmtfailed");
        exit();
    }
    #returns hash of a password
    #this inbuilt function is constantly updated
    $hashedPassword = password_hash($password, PASSWORD_DEFAULT);   
    mysqli_stmt_bind_param($stmt, "ss", $username, $hashedPassword);
    mysqli_stmt_execute($stmt);
    mysqli_stmt_close($stmt);
    header("location: ../signup.php?error=none");
    exit;   
} #this is line 92 but I do not see where the issue is here

我已经对缺少括号和/或 php.ini 的状态如何导致此错误进行了一些研究，但是 short_open_tag 已设置为打开。 是什么导致了错误？

Answer 1

在function usernameExists 关闭其他

来自：

function usernameExists($conn, $username){
    $sql = "SELECT * FROM users WHERE username = ?";
    #we have created a prepared statement to prevent SQL injections
    $stmt = mysqli_stmt_init($conn);
    $result = Null;
    if (!mysqli_stmt_prepare($stmt, $sql)){
        header("location: ../signup.php?error=stmtfailed");
        exit();
    }
    
    mysqli_stmt_bind_param($stmt, "s", $username);
    mysqli_stmt_execute($stmt);
    
    $resultData = mysqli_stmt_get_result();
    
    if($row = mysqli_fetch_assoc($resultData)){ 
        return $row;        
        
    }
    else{
        $result = false;
        return $result;
}

改成

function usernameExists($conn, $username){
    $sql = "SELECT * FROM users WHERE username = ?";
    #we have created a prepared statement to prevent SQL injections
    $stmt = mysqli_stmt_init($conn);
    $result = Null;
    if (!mysqli_stmt_prepare($stmt, $sql)){
        header("location: ../signup.php?error=stmtfailed");
        exit();
    }
    
    mysqli_stmt_bind_param($stmt, "s", $username);
    mysqli_stmt_execute($stmt);
    
    $resultData = mysqli_stmt_get_result();
    
    if($row = mysqli_fetch_assoc($resultData)){ 
        return $row;        
        
    }
    else{
        $result = false;
        return $result;
      }//<---this line
}