JSON输入node.js意外结束jason.parse（）

Question

我正在尝试解析通过以下（PHP）发送回的JSON；

$UserData = $con->query("SELECT * FROM Discord WHERE RobloxID=".$UserId) or trigger_error($mysqli->error);
    $result = $UserData->fetch_array();
    if (!is_null($result[RobloxID])) {
        echo json_encode(array(
            'Username'  =>  $result[Username],
            'Code'      =>  $result[Code],
            'DiscordID' =>  $result[DiscordID],
            'IsVerified'=>  $result[IsVerified],
            'RobloxID'  =>  $result[RobloxID],
            )
        );
    } else {
        //$code = substr(md5(uniqid(mt_rand(), true)) , 0, 8);
        //$UserData->query("INSERT INTO Discord (Username,Code,DiscordID,IsVerified,RobloxID) VALUES ('$username','$code','$discordID','false','$UserId')");
        echo json_encode(array(
            'Username'  =>  'nil',
            'Code'      =>  'nil',
            'DiscordID' =>  'nil',
            'IsVerified'=>  'nil',
            'RobloxID'  =>  'nil',
            )
        );
    }

这是创建的JSON； {"Username":"nil","Code":"nil","DiscordID":"nil","IsVerified":"nil","RobloxID":"nil"}

我正在使用node.js来获取返回的内容；

request('http://example.net/API/verify.php?token=Nf_327&username=' + Username + '&UserId=' + UserID + '&discordid='+ msg.author.id, function (error, response, body) {
        console.log('error:', error); // Print the error if one occurred 
        console.log('statusCode:', response && response.statusCode); // Print the response status code if a response was received 
        console.log('body:', body); // Print the HTML for the Google homepage.
        var info = JSON.parse(body);
        if (info.IsVerified == 'nil') {
            request('http://example.net/API/verify.php?token=Nf_328&username=' + Username + '&UserId=' + UserID + '&discordid='+ msg.author.id, function (error, response, body) {
                console.log('error:', error); // Print the error if one occurred 
                console.log('statusCode:', response && response.statusCode); // Print the response status code if a response was received 
                console.log('body:', body); // Print the HTML for the Google homepage.
            });
            msg.reply("Looks like its your first time trying to verify, I have DM'd you instructions on how to verify your account");
        }
    });

但是，每当我尝试运行该函数时，都会得到错误堆栈；

Unexpected end of JSON input
at JSON.parse (<anonymous>)
at Request._callback (/root/InfiusBot/commands/group/verify.js:41:22)
at Request.self.callback (/root/node_modules/request/request.js:188:22)
at emitTwo (events.js:125:13)
at Request.emit (events.js:213:7)
at Request.<anonymous> (/root/node_modules/request/request.js:1171:10)
at emitOne (events.js:115:13)
at Request.emit (events.js:210:7)
at IncomingMessage.<anonymous> (/root/node_modules/request/request.js:1091:12)
at Object.onceWrapper (events.js:314:30)
at emitNone (events.js:110:20)
at IncomingMessage.emit (events.js:207:7)
at endReadableNT (_stream_readable.js:1047:12)
at _combinedTickCallback (internal/process/next_tick.js:102:11)
at process._tickCallback (internal/process/next_tick.js:161:9)

关于我在这里缺少什么的任何想法？ 我完全迷路了。

Answer 1

好吧，调用JSON.parse时发生错误。 这意味着返回的正文不是正确的JSON，根据请求文档，它是有意义的，因为它是描述HTML页面内容的字符串。 这意味着您正在HTML上运行JSON.parse，并且您的网址也必须不正确，这也是您发出请求的原因。 这可能是因为您正在向example.net发出请求？ 我觉得基本网址应为其他网址。

在URL中查找故障的其他方法是检查当您在浏览器中访问php页面并将其传递给参数时，该用户对象是否在屏幕上可见。 然后确保该URL是正确的。 然后，最后应该检查您的主体对象是否包含JSON。 如果它没有打印出某种形式的JSON，则应打印正文并通读HTML，因为这可能是PHP发送的错误，并且由于某种原因未被您的错误打印捕获。