JSON輸入node.js意外結束jason.parse（）

Question

我正在嘗試解析通過以下（PHP）發送回的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',
            )
        );
    }

這是創建的JSON； {"Username":"nil","Code":"nil","DiscordID":"nil","IsVerified":"nil","RobloxID":"nil"}

我正在使用node.js來獲取返回的內容；

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，每當我嘗試運行該函數時，都會得到錯誤堆棧；

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)

關於我在這里缺少什么的任何想法？ 我完全迷路了。

Answer 1

好吧，調用JSON.parse時發生錯誤。 這意味着返回的正文不是正確的JSON，根據請求文檔，它是有意義的，因為它是描述HTML頁面內容的字符串。 這意味着您正在HTML上運行JSON.parse，並且您的網址也必須不正確，這也是您發出請求的原因。 這可能是因為您正在向example.net發出請求？ 我覺得基本網址應為其他網址。

在URL中查找故障的其他方法是檢查當您在瀏覽器中訪問php頁面並將其傳遞給參數時，該用戶對象是否在屏幕上可見。 然后確保該URL是正確的。 然后，最后應該檢查您的主體對象是否包含JSON。 如果它沒有打印出某種形式的JSON，則應打印正文並通讀HTML，因為這可能是PHP發送的錯誤，並且由於某種原因未被您的錯誤打印捕獲。