Question

我遇到了一些奇怪的问题。使用我的类我连接到mysql，连接成功但是当我调用一个函数时它不起作用。我使用自己的错误处理程序，如果我在这行中有一些错误，它使用退出;我在调用函数之前使用了echo并且在页面没有查杀之后（这意味着php工作成功）mysql查询我已经检查了它的工作但是当我使用返回'some text'时;它没有回来。有什么功能不起作用的建议吗？

我的班级：

<?php
include('../config.php');
include('../chat_error_handler.php');

class Chat {
    private $mysqli;
    var $con;

    //Constructor open db conn
    function __construct() {
        $this->mysqli = new mysqli('localhost', 'root', '', 'test');
    }

    //destructor close db conn
    function __destruct() {
        $this->mysqli->close();
    }

    public function postNewMessage($user_name, $message, $color, $room_id) {
        $user_name = $this->mysqli->real_escape_string($user_name);
        $message = $this->mysqli->real_escape_string($message);
        $color = $this->mysqli->real_escape_string($color);
        $room_id = $this->mysqli->real_escape_string($room_id);
        $query = 'INSERT INTO chat (posted_on, account, message, color, room_id)' .
        ' VALUES (NOW(), "'.$user_name.'", "'.$message.'", "'.$color.'", "'. $room_id .'")';
        $result = $this->mysqli->query($query);
        return 'Pranesimas';
        $result->close();
    }

我的php我用来调试：

<?php
include('core/chat.class.php');
include('chat_error_handler.php');
$chat = new Chat();
$message = 'testas';
echo $message.'<br/>';
$chat->postNewMessage('testing', $message, '#000000', 1);
echo '<br/>'.$message;
?>

Answer 1

public function postNewMessage($user_name, $message, $color, $room_id) {

    $user_name = $this->mysqli->real_escape_string($user_name);
    $message = $this->mysqli->real_escape_string($message);
    $color = $this->mysqli->real_escape_string($color);
    $room_id = $this->mysqli->real_escape_string($room_id);

    $query = 
       "INSERT INTO 
            chat ( posted_on, 
                   account, 
                   message, 
                   color, 
                   room_id )
        VALUES ( NOW(), 
                 '". $user_name ."', 
                 '". $message ."', 
                 '". $color ."', 
                 '". $room_id ."' )";

    $result = $this->mysqli->query($query);

    if ( $result ) {
        return 'Pranesimas'; 
    } else {
        return false;
    }
}

Answer 2

关于PHP5及更高版本中的'var'。它用于在PHP4中声明类成员变量，不再需要它。它将在PHP5中运行，但会在PHP中从版本E_STRICT到版本5.0.0引发5.1.2警告，因此已被弃用。

<强>更新

public function postNewMessage($user_name, $message, $color, $room_id) {
        return 'Pranesimas1';
        $user_name = $this->mysqli->real_escape_string($user_name);
        $message = $this->mysqli->real_escape_string($message);
        $color = $this->mysqli->real_escape_string($color);
        return 'Pranesimas2';
        $room_id = $this->mysqli->real_escape_string($room_id);
        $query = 'INSERT INTO chat (posted_on, account, message, color, room_id)' .
        ' VALUES (NOW(), "'.$user_name.'", "'.$message.'", "'.$color.', "'. $room_id .'")';
        $result = $this->mysqli->query($query);
        return 'Pranesimas';
        $result->close();
    }

尝试逐步添加返回以检测问题所在，并让我知道结果。

php类无法正常工作

2 个答案: