R sqlite:使用两个表进行更新会产生语法错误“near”。“:

时间:2013-01-11 07:34:27

标签: r sqlite sql-update

我想在sqlite上用两个表做一个UPDATE。

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? 谢谢你的帮助。

3 个答案:

答案 0 :(得分:2)

为什么你使用sqldf进行更新?我认为sqldfselect语句很有用。

我会使用RSQLite

首先我更正你的sql statetemnt。我更喜欢使用sep'\ n',以获得与猫的漂亮请求

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

修改

我在Op澄清(FROM x1,x2变为FROM x2

之后编辑我的答案

答案 1 :(得分: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"))

,并提供:

> 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_having_problems_with_update? 其他要提醒的事情:显然别名不起作用,例如UPDATE x1 a ... 谢谢你的帮助。

答案 2 :(得分:0)

我认为你的内在SELECT没有说出它的选择。单独尝试。我认为它看起来应该更像:

SELECT * FROM x1,x2 WHERE x1.id = x2.id AND x1.t < x2.start