对包含带有制表符的字符串的变量进行grep [英] grep on a variable containing a string with a tab

问题描述

我有一个for循环，循环几行。每行都有两个字符串和一个分隔它们的标签，例如

I have a for loop that loops over a few lines. each line has two strings with a tab separating them e.g

chr11   105804693

当我将变量传递给grep并用文件搜索时，返回105804693：No such file or directory。我的代码如下所示：

when I pass the variable to grep with a file to search in it returns "105804693: No such file or directory". my code looks like this:

换行$（cat list_of_lines）;
做grep $ line file_to_search_in >> output_file;
完成

for line in $(cat list_of_lines); do grep $line file_to_search_in >> output_file; done

我尝试了 printf 变量并返回chr11但是当我回应变量时，它返回chr11 105804693

I tried to printf the variable and it returned "chr11" but when I echoed the variable it returned "chr11 105804693"

如何在变量上调用grep并让它搜索整行？

How can I call grep on the variable and have it search for the full line?

推荐答案

You probably haven't added quotes around your variable having the search pattern,

search_pattern_variable="chr11 105804693"
grep "$search_pattern_variable" input-file

由于您的变量包含 chr11 105804693 ， shell 使用默认的 IFS （字段分隔符）将单词拆分为多个单词已经使用 chr11 作为模式， 105804693 作为文件名，报告没有看到这样的文件。

Since your variable had the words chr11 105804693, shell split up the words into multiple words with the default IFS (field-separator) and had used chr11 as the pattern and 105804693 as the file-name, which it is reporting not seeing such a file.

看看 Word-Splitting 表示 bash 。

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