bash - 从另一个文件中删除包含 ID 的每一行

Question

我有一个 csv 文件 (list.csv)，其中包含类似的内容:

"1","10","1","2"
"2","22","20","2"
"3","33","5","2"
"4","36","225","2"
"5","36","225","2"

我有另一个文件(delete.txt):

"1"
"4"

我想从 delete.txt 文件中删除包含 ID 的每一行。在我的示例中，我想显示:

"2","22","20","2"
"3","33","5","2"
"5","36","225","2"

我该怎么做？

Answer 1

单次使用 grep(如果您想查看使用的查询，可以在末尾添加 echo "$arg"):

#!/bin/bash
arg=""
for i in `cat delete.txt`; do
        if [ -n "$arg" ]; then
                arg="$arg\|^$i"
        else
                arg="^$i";
        fi
done;
grep -v "$arg" input.txt