Question

这是用于计算DateTimeSolved和DateTimeCreated之间的时差的SQL查询，我需要结果以小时为单位。

$query = "SELECT problem.id, problem.DateTimeCreated, problem.DateTimeSolved TIMEDIFF(HOUR, problem.DateTimeSolved, problem.DateTimeCreated)
From problem";

我已将where子句添加到我的查询中：

$ query =“SELECT problem.status，problem.id，problem.DateTimeSolved，problem.DateTimeCreated，TIMESTAMPDIFF（DAY，problem.DateTimeSolved，problem.DateTimeCreated）从问题 WHERE problem.status ='5' GROUP BY MONTH（problem.DateTimeCreated）= MONTH（CURDATE（））“;这是它工作的查询的当前格式，但我得到的输出是否定的。

当我尝试运行语句时，我仍然收到此错误：

您的SQL语法有错误;

检查手册   对应于您的MySQL服务器版本以获得正确的语法

Answer 1

你错过了一个逗号而且timediff函数只需要两个值，要格式化它（只显示小时）使用以下查询：

SELECT problem.id, problem.DateTimeCreated, problem.DateTimeSolved, TIME_FORMAT(TIMEDIFF(problem.DateTimeSolved, problem.DateTimeCreated), '%k %i')
From problem

Answer 2

错误消息是由于在TIMEDIFF函数之前缺少逗号：

SELECT problem.id, problem.DateTimeCreated, problem.DateTimeSolved, TIMEDIFF(HOUR, problem.DateTimeSolved, problem.DateTimeCreated)

从问题

然后您会发现TIMEDIFF的参数不正确。这应该有效：

SELECT problem.id, problem.DateTimeCreated, problem.DateTimeSolved, TIMEDIFF( problem.DateTimeSolved, problem.DateTimeCreated)
From problem

如果您只需要几个小时：

SELECT problem.id, problem.DateTimeCreated, problem.DateTimeSolved, HOUR(TIMEDIFF( problem.DateTimeSolved, problem.DateTimeCreated)) From problem

Answer 3

试试这个

$query = "SELECT problem.id, problem.DateTimeCreated, problem.DateTimeSolved TIMESTAMPDIFF(HOUR, problem.DateTimeSolved, problem.DateTimeCreated)
From problem";

MYSQL-无法计算到字段之间的timediff并且无法使用where子句

3 个答案: