为什么这个 mysql 查询在 Workbench 中执行但不使用 JDBC 调用时工作正常？答案

【问题标题】：Why this mysql query works fine when is executed in Workbench but not using JDBC call?为什么这个 mysql 查询在 Workbench 中执行但不使用 JDBC 调用时工作正常？
【发布时间】：2018-06-21 02:02:25
【问题描述】：

我有一个类似的查询：

INSERT INTO groups(
    participants, count1, count2, count3) 
SELECT 
0 as participants,
(count1/participants) as count1,
(count2/participants) as count2,
(count3/participants) as count3 WHERE id = 22;

现在假设在表中你有：

组

id    participants         count1       count2       count3
22    5                    10000        10000        10000

该查询将插入具有以下值的新行：

id    participants         count1       count2       count3
23    0                    2000         2000         2000

正如我所说，它在 Workbench 中运行良好，但如果我在 JDBC (Spring-Java) 中运行，我会奇怪地得到这个

id    participants         count1       count2       count3
23    0                    2000         NULL         NULL

这是实际查询或类似查询，这是原始查询，我只是展示上面的例子来帮助理解问题。

INSERT INTO registrations (
  numberOfParticipants,
  eventId,
  eventName,
  STATUS,
  createdOn,
  source,
  sourceDetail,
  paymentType,
  groupId,
  couponId,
  eventModality,
  eventHour,
  groupName,
  couponCode,
  additionalProductQuantities,
  productsPaid,
  discount,
  insurance,
  baseTotal,
  COMMENT,
  processingFee,
  isVolunteer,
  participantType,
  ipNumber,
  refererCodeId
) 
SELECT 
  0 AS numberOfParticipants,
  (SELECT 
    eventId 
  FROM
    groups 
  WHERE id = 27) AS eventId,
  (SELECT 
    lastEventName 
  FROM
    groups 
  WHERE id = 27) AS eventName,
  STATUS,
  createdOn,
  source,
  sourceDetail,
  paymentType,
  27,
  couponId,
  eventModality,
  eventHour,
  (SELECT 
    NAME 
  FROM
    groups 
  WHERE id = 27) AS groupName,
  couponCode,
  additionalProductQuantities,
  (
    productsPaid / numberOfParticipants
  ) AS productsPaid,
  (discount / numberOfParticipants) AS discount,
  (insurance / numberOfParticipants) AS insurance,
  (baseTotal / numberOfParticipants) AS baseTotal,
  NULL,
  (
    processingFee / numberOfParticipants
  ) AS processingFee,
  isVolunteer,
  participantType,
  ipNumber,
  refererCodeId 
FROM
  registrations 
WHERE id = 15787;

【问题讨论】：

我的猜测是您的 JDBC 调用有问题。你能分享那个代码吗？
不要调用你的 SQL 表group，这是一个保留关键字。由于这个和其他几个原因，您的原始 MySQL 代码甚至不会执行。
请向我们展示您实际运行的确切代码。 Java 代码也不错，但您可以从原始 MySQL 代码开始。

标签： mysql jdbc workbench

【解决方案1】：

你确定你的 MySQL 查询的语法是正确的吗？它应该在查询中包含“FROM”语句。 “组”也是 MySQL 中的保留关键字。

INSERT INTO group(
participants, count1, count2, count3) 
SELECT 
0 as participants,
(count1/participants) as count1,
(count2/participants) as count2,
(count3/participants) as count3 
FROM group 
WHERE id = 22;

没有提供足够的信息。这是我能给你的最好的。也许分享java源代码可以帮助我们提供更好的解决方案。

【讨论】：