嘗試將 sql 結果作為 json 數據類型回顯時出現語法錯誤

Question

I'm trying to output the sql results of the statement in json datatype, but an error was prompt SyntaxError: JSON.parse: unexpected character at line 1 column 1 of the JSON data , I have checked my sql statement and try to executed in sql 編輯器，它可以工作，但我不確定為什么當我嘗試在瀏覽器中回顯它時它不起作用。

So i have a big switch statement block, and this is a small part of it, so basically when i type http://localhost/w11/local-html/part_1/api/schedule it should output json datatype by taking in the arguments來自 url 之類的api/schedule和api是arg_1和schedule是arg_2 。

case 'api':
        {
            header("Content-Type: application/json");
            switch ($param2) {
                case 'schedule':
                {
                    switch ($param3) {
                        case '':
                        {
                            try {
                                $sqlQuery = "SELECT room, type, title, day, time FROM sessions INNER JOIN slots ON sessions.slotsID = slots.id";
                                $response = new JSONRecordSet();
                                $response = $response->getJSONRecordSet($sqlQuery, "");

                                echo $response;
                            } catch (PDOException $e) {
                                echo "Connection Failed:" . $e->getMessage();
                            }
                            break;
                        }
                        default:
                        {
                            //do something
                            break;
                        }

                    }
                    break;
                }

這是 getJSONRecordSet 的getJSONRecordSet

class JSONRecordSet extends RecordSet {

    function getJSONRecordSet($sql, $params = null) {
        $queryResult = $this->getRecordSet($sql, $params);

        $recordSet = $queryResult->fetchAll(PDO::FETCH_ASSOC);
        $nRecords = count($recordSet);
        if ($nRecords == 0) {
            $status =  200;
            $message = array("text" => "No records found");
            $result = '[]';
        }
        else {
            $status = 200;
            $message = array("text" => "");
            $result = $recordSet;
        }
        return json_encode(
            array(
                'status' => $status,
                'message' => $message,
                'data' => array(
                    "RowCount"=>$nRecords,
                    "Result"=>$result
                )
            ),
            JSON_PRETTY_PRINT
        );
    }
}

和父 class

abstract class RecordSet {
    protected $conn;
    protected $queryResult;

    function __construct() {
        $this->conn = pdoDB::getConnection();
    }

    function getRecordSet($sql, $params = null) {
        if (is_array($params)) {
            $this->queryResult = $this->conn->prepare($sql);
            $this->queryResult->execute($params);
        }
        else {
            $this->queryResult = $this->conn->query($sql);
        }
        return $this->queryResult;
    }
}

那么，我該如何解決這個錯誤呢？

Answer 1

這段代碼沒有任何問題。
由於SyntaxError: JSON.parse是一個 JS 錯誤，您應該檢查傳遞給它的數據。 檢查對您的請求的原始響應並進行一些調試。
這可能是您的數據庫連接有問題，或者您的 $param2 或 $param3 條件不滿足。