grep 忽略模式中的字符

Question

模式来自长度为 14 的变量。该字符串是 grep 的模式。现在文本文件的行每行包含 13 个字符。

比如长度为14的pattern是

pattern =   58244804671021

并且文本文件包含

3823480467102
4724470467102

如何让 grep 忽略模式中的最后一个字符？

Answer 1

假设您的模式在 $pattern 中并且您正在使用 bash，您可以

grep ${pattern%?} file

从变量中删除最后一个字符。

您还可以将 cut 与字符 1 到 13 一起使用：

grep $(echo "$pattern" | cut -c 1-13 -) file

甚至更好 bash 和 ksh 作为此处字符串

grep $(cut -c 1-13 <<<$pattern) file

Answer 2

你的意思是这样的：

cat file
5824480467102
4534324435455
8244804671021

所有行=13字符

pattern="58244804671021"

图案=14字符

awk -v p="$pattern" '==substr(p,1,13)' file
5824480467102

这将删除模式的最后一个字符并针对字段 #1

进行测试

Answer 3

您可以使用参数扩展作为

$ grep  ${pattern:0:13} filename

From Bash Manual

If offset evaluates to a number less than zero, the value is used as an offset in characters from the end of the value of parameter. If length evalu‐ ates to a number less than zero, it is interpreted as an offset in characters from the end of the value of parameter rather than a number of charac‐ ters, and the expansion is the characters between offset and that result. Note that a negative offset must be separated from the colon by at least one space to avoid being confused with the :- expansion.

${variable:offset:lenght}
- pattern是变量
- 0偏移量，或开始
- 13长度

测试

$ cat input
3823480467102
4724470467102
5824480467102

$ grep  ${pattern:0:13} input
5824480467102

grep 忽略模式中的字符

grep ignore characters in the pattern

regex

unix

command-line

grep