如何找出并替换多余的双引号

chg1226

有个问题困扰我两天了，请各位大大帮忙看看
我有一个数据文件如下
"name","age","address"
"ZhangSan","31","no.1 fl.2nd, 108 rd"
"LiSi","36","2-4-6, build 3, DW Dist"
"Wangwu","34","room 307, fl. 15, Build 14, wd" square"
问题出来第三行里，address那一列里有一个多余的双引号-wd''，程序处理到这一行的时候就把这个双引号后面的数据给截断了。
我想把那个多余的双引号替换成空格，结果就像这样:
"Wangwu","34","room 307, fl. 15, Build 14, wd  square
请问这个要怎么做呀？谢谢各位了！！

seeLnd

替换成DOUBLE_QUOTE  ？  什么意思？

yinyuemi

回复 1# chg1226

1. awk '{t=gensub(/(.*",").*"/,"\\1",1);s=gensub(/.*","(.*)"/,"\\1",1);gsub(/"/,"DOUBLE_QUOTE",s);print t s "\""}'
   

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qq312873977

1. sed 's/"\([^"]*"$\)/DOUBLE_QUOTE\1/' conf

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yestreenstars

把你想要的结果贴出来，这样大家才知道要替换哪些双引号。

Shell_HAT

回复 2# seeLnd


"Wangwu","34","room 307, fl. 15, Build 14, wd" square"
就是把这个红色位置的"替换成DOUBLE_QUOTE

Shell_HAT

回复 4# qq312873977


    这样会把每行里面那个位置的都给替换掉，以楼主顶楼的数据为例的话，只要求替换最后一行的那个

q1208c

简单点, 分两次,

sed -e 's/^"/#/' -e 's/"\s*,\s*"/#,#/g' -e 's/"$/#/' yourfile | sed -e 's/"/DOUBLE_QUOTE/g' | sed -e 's/#/"/g'

好象有点笨.

dn833

1. root@01 tmp]# sed -r 's/([^",])"([^,"])/\1DOUBLE_QUOTE\2/g' file
   
2. "Wangwu","34","room 307, fl. 15, Build 14, wdDOUBLE_QUOTE square"
   
3. [root@01 tmp]# cat file
   
4. "Wangwu","34","room 307, fl. 15, Build 14, wd" square"

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chg1226

抱歉，更新了贴子内容，希望这回说清楚了。