Question

我有3个表let reachabilityManager = NetworkReachabilityManager() reachabilityManager.listener = { status in switch status { case .notReachable: print("The network is not reachable") self.onInternetDisconnection() case .unknown : print("It is unknown whether the network is reachable") self.onInternetDisconnection() // not sure what to do for this case case .reachable(.ethernetOrWiFi): print("The network is reachable over the WiFi connection") self.onInternetConnection() case .reachable(.wwan): print("The network is reachable over the WWAN connection") self.onInternetConnection() } }，scan_1，scan_2。这是SQL模式的结构：

scan_3

我想将scan_1: scan_2: scan_3: P_no work P_no work P_no work 1 YES 1 YES 1 NO 2 NO 2 NO 2 NO 3 YES 3 YES 3 NO计算在P_no。但是，如果在work ='YES'的{{1}}的2个位置发生了，则P_no = 1和scan_1中的“是”，则必须计为1。

我的查询是：

scan_2

Answer 1

据我所知，这可能对你有用。使用union来消除重复。

    select count(*) AS `ab1` FROM
    (
    select `P_no`,`work` from `scan_1` s1 where `work`= 'YES' 
     union
    select `P_no`,`work` from `scan_2` s2 where `work`= 'YES'
     union
     select `P_no`,`work` from `scan_3` s3 where `work`= 'YES' 
    ) as final

Answer 2

尝试：

SELECT COUNT(*) AS total_count 
FROM
    (
        SELECT P_no, work FROM scan_1 WHERE work = 'YES'
        UNION
        SELECT P_no, work FROM scan_2 WHERE work = 'YES'
        UNION
        SELECT P_no, work FROM scan_3 WHERE work = 'YES'
    ) AS total

Answer 3

您可以使用 UNION 获得所需的结果，如下所示：

const functions = require('firebase-functions');
var iapReceiptValidator = require('iap-receipt-validator');
const password = 'xxxx'; // Shared Secret from iTunes connect
const production = false; // use sandbox or production url for validation
var validateReceipt = iapReceiptValidator(password, production);

exports.validate = functions.https.onRequest((request, response) => {
  try {
    const validationData = await validateReceipt(request.body.receiptData);
    response.send(JSON.stringify(validationData));
  } catch(err) {
      console.log(err.valid, err.error, err.message)
  }
 });

SQL HERE

注意使用 UNION合并的三个查询。它合并结果并从结果中排除重复记录。

Answer 4

试试这个，

SELECT COUNT(DISTINCT P_No) AS ab1
FROM (
    SELECT P_no, work FROM scan_1
    UNION
    SELECT P_no, work FROM scan_2
    UNION
    SELECT P_no, work FROM scan_3
) T
WHERE T.Work='YES'

希望这会对你有所帮助。

从不同的sql表中计算同名列

4 个答案: