使用grep提取文本文件中的字符串

Question

我的file.txt每行有一个名字，如下所示：

ABCB8
ABCC12
ABCC3
ABCC4
AHR
ALDH4A1
ALDH5A1
....

我想从input.txt文件中grep其中的每一个。

手动我一次这样做

grep "ABCB8" input.txt > output.txt

有人可以帮助从input.txt自动grep file.txt中的所有字符串并将其写入output.txt。

Answer 1

您可以使用Bash, Linux, Need to remove lines from one file based on matching content from another file

中所述的-f标记

grep -o -f file.txt input.txt > output.txt

标志

• -f FILE，--file=FILE：

  

从FILE获取模式，每行一个。空文件   包含零模式，因此不匹配。 （-f是   由POSIX指定。）

• -o，--only-matching：

  

仅打印匹配行的匹配（非空）部分   每个这样的部分都在一个单独的输出线上。

Answer 2

for line in `cat text.txt`; do grep $line input.txt >> output.txt; done

text.txt的内容：

ABCB8
ABCC12
ABCC3
ABCC4
AHR
ALDH4A1
ALDH5A1

修改：

阅读时更安全的解决方案：

cat text.txt | while read line; do grep "$line" input.txt >> output.txt; done

编辑2 ：

示例text.txt：

ABCB8
ABCB8XY
ABCC12

示例input.txt：

You were hired to do a job; we expect you to do it.
You were hired because ABCB8 you kick ass;
we expect you to kick ass.
ABCB8XY You were hired because you can commit to a rational deadline and meet it;
ABCC12 we'll expect you to do that too.
You're not someone who needs a middle manager tracking your mouse clicks

如果您不关心行的顺序，快速解决方法是通过sort | uniq管道解决方案：

cat text.txt | while read line; do grep "$line" input.txt >> output.txt; done; cat output.txt | sort | uniq > output2.txt

结果是output.txt。

编辑3 ：

 cat text.txt | while read line; do grep "\<${line}\>" input.txt >> output.txt; done

没关系吗？