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.get() 请求一直运行良好，直到我将 mysql 添加到组合中，我调试了 SQL 并确保它连接完美，但它仍然无法正常工作。如果有人能告诉我哪里出错了，那就太好了。

JavaScript -

if (command !== '') {
    term.echo('Waiting on response from server...');
    if (command.indexOf(' ') !== -1) {
        var ext_start = command.indexOf(' ');
        var ext_end = command.length;
        var ext = command.substr(ext_start, ext_end);

        $.get('response.php', {command: command, ext: ext}, function(data) {
            term.echo("Response from command: " + command + "->" + data);
        });
    }
    else {
        $.get('response.php', {command: command}, function(data) {
            term.echo("Response from command: " + command + "->" + data);
        });
    }

    try {
        var result = window.eval(command);
        if (result !== undefined) {
            term.echo(new String(result));
        }
    } catch(e) {
        //term.error(new String(e));
    }

} else {
   term.echo('');
}

响应.php -

<?php
if (isset($_GET['command'])) {
    require 'config.php';
    $cmd = $_GET['command'];
    if (isset($_GET['ext'])) {
        //Things to do if a double command was executed (e.g. !screenshot botname)
        $ext = $_GET['ext'];
        echo "Double Command Successful.";
    }
    else {
        //Things to do if just a single command was executed (e.g. !info)
        $query = mysqli_query($connect, "SELECT * FROM commands WHERE command='$cmd'");
        $numrows = mysqli_num_rows($query);
        if ($numrows !== 1) {
            //If the command doesn't exist
            echo "Command non-existant";
        }
        else {
            //If the command does exist*/
            mysqli_query($connect, "UPDATE commands SET status=1 WHERE command='$cmd'");
            echo "Command does exist";
        }
    }
}
else {
    echo "Error";
}
?>

哦，请不要发布/评论“你的代码容易受到 SQL 注入的攻击，bla bla bla。”，我知道，我还没有到添加检查的地步。

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0 回答 0