文件名的 awk/sed 部分

Question

每当我将文件名作为参数放入命令行时，我只想 awk/sed 文件名的前缀。

例如，

我有多个文件：

a.fastq.gz
b.fastq.gz
c.fastq.gz
d.fastq.gz

如果我执行：

sh test.sh --INFILE b.fastq.gz

我想要的输出是：

b

我尝试过但失败的是，

prefix="sed 's/.fastq//' ${INFILE}"

Answer 1

Using shell parameter expansion (assuming you are assigning your filename to INFILE):

INFILE=b.fastq.gz
prefix=${INFILE%%.*}

Or if your suffix is sure to be fixed and you want to be more precise (always recommended when possible):

prefix=${INFILE%.fastq.gz}

${parameter%word}

${parameter%%word}

The word is expanded to produce a pattern and matched according to the rules described below (see Pattern Matching). If the pattern matches If the pattern matches a trailing portion of the expanded value of parameter, then the result of the expansion is the value of parameter with the shortest matching pattern (the ‘%’ case) or the longest matching pattern (the ‘%%’ case) deleted. If parameter is ‘@’ or ‘’, the pattern removal operation is applied to each positional parameter in turn, and the expansion is the resultant list. If parameter is an array variable subscripted with ‘@’ or ‘’, the pattern removal operation is applied to each member of the array in turn, and the expansion is the resultant list.

Answer 2

使用标准basename实用程序删除已知后缀：

$ basename b.fastq.gz .fastq.gz
b

With a variable:

$ pathname="/some/path/name.fastq.gz"
$ basename "$pathname" .fastq.gz
name

Assigning to a variable:

$ prefix=$( basename "$pathname" .fastq.gz )
$ printf 'Prefix is "%s"\n' "$prefix"
Prefix is "name"

In a loop (over all the .fastq.gz files in the current directory):

for filename in ./*.fastq.gz; do
    prefix=$( basename "$filename" .fastq.gz )
    # Do things using "$prefix" here
done