1

我想对 sqlite 上的两个表进行更新。

x1 <- data.frame(id = rep(1,3),
                  t = as.Date(c("2000-01-01","2000-01-15","2000-01-31"))
)
x1.h <- 0
x2 <- data.frame(id = 1, start = as.Date("2000-01-14"))

更新是：

sqldf(paste("UPDATE x1"
        ," SET x1.h = 1"
        ," WHERE EXISTS (SELECT *"
        ,"               FROM x2"
        ,"               WHERE x1.id = x2.id"
        ,"                     AND x1.t < x2.start"
        ,"               )"
        )
  )

我收到以下错误：

Error in sqliteExecStatement(con, statement, bind.data) : 
   RS-DBI driver: (error in statement: near ".": syntax error)

有人知道出了什么问题吗？感谢您的帮助。

4

3 回答 3

2

你为什么sqldf要更新？我认为sqldf仅用于select声明。

我会习惯RSQLite这个。

首先，我更正了您的 sql 语句。我更喜欢使用 sep '\n' 来获得与 cat 的漂亮请求

str.update <- paste(" UPDATE x1"
            ," SET h = 1 "              ## here the error
            ," WHERE EXISTS (SELECT * "
             ,"              FROM x2 "             ## here second error 
             ,"              WHERE x1.id = x2.id "
            ,"               AND x1.t < x2.start "
            ,"       )"
      ,sep ='\n'
)


cat(str.update)
 UPDATE x1
 SET h = 1 
 WHERE EXISTS (SELECT * 
              FROM x1,x2    ##
              WHERE x1.id = x2.id 
               AND x1.t < x2.start 
       )

然后你可以这样做：

library(RSQLite)
con <- dbConnect(SQLite(), ":memory:")
dbWriteTable(con, "x1", x1)            # I create my table x1
dbWriteTable(con, "x2", x2)            # I create my table x2
res <- dbSendQuery(con, str.update)   
dbReadTable(con,name='x1')            ## to see the result

编辑

我在操作说明后编辑我的答案（FROM x1,x2 变为FROM x2）

于 2013-01-11T08:34:56.340 回答
1

我找到了这个解决方案：

x1$h <- 0
x1 <- sqldf(c("UPDATE x1
        SET h = 1
        WHERE EXISTS (SELECT x1.id
                        FROM x2
                        WHERE x1.id = x2.id
                          AND x1.t < x2.start
                     )",
        "SELECT * FROM main.x1"))

给予：

> x1

  id          t h
1  1 2000-01-01 1
2  1 2000-01-15 0
3  1 2000-01-31 0

来源： https ://code.google.com/p/sqldf/#8._Why_am_I_have_problems_with_update ？其他要提醒的事情：显然别名不起作用，例如UPDATE x1 a ...感谢您的帮助。

于 2013-01-11T10:14:01.330 回答
0

我认为你内心的 SELECT 并没有说它从哪里选择。分开试试。我认为它应该看起来更像：

SELECT * FROM x1,x2 WHERE x1.id = x2.id AND x1.t < x2.start
于 2013-01-11T08:34:06.560 回答