Grep精确匹配字符串，其中空格为变量 [英] Grep exact match string with spaces as variable

问题描述

我有:

file.csv

其中包含

2,1,"string with spaces",3,4,5
2,1,"some other string",3,4,5
2,1,"string with spaces more than this",3,4,5
2,1,"yet another",3,4,5
2,1,"string with spaces too",3,4,5

当我这样做时:

grep '"string with spaces",' file.csv

它产生想要的结果:

2,1,"string with spaces",3,4,5

现在我需要在while循环中这样做:

Now I need to do this in a while loop:

while read p; do

    grep '"$p",' file.csv

done < list.txt

位置:

list.txt

包含:

string with spaces
yet another

我想要的输出是:

2,1,"string with spaces",3,4,5
2,1,"yet another",3,4,5

问题是我的while循环返回为空或部分匹配.如何遍历list.txt&得到我想要的输出?

The problem is that my while loop comes back empty, or matches partially. How do I loop through list.txt & get my desired output?

推荐答案

如果您对awk没问题，那么这应该很简单.

If you are ok with awk this should be an easy one for it.

awk 'FNR==NR{a[$0];next} ($4 in a)' list.txt FS="[,\"]" file.csv

OR(根据Ed ir先生的评论，将字段分隔符设置为逗号并保持更清晰，可以尝试跟随)

OR(as per Ed sir's comment to make field separator as comma and keep it clearer, one could try following)

awk -F, 'FNR==NR{a["\""$0"\""];next} $3 in a' list.txt file.csv

输出如下.

2,1,"string with spaces",3,4,5
2,1,"yet another",3,4,5

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