重命名目录中特定类型的所有文件

rename all files of a specific type in a directory

我正在尝试使用 bash 重命名目录中匹配特定模式的所有 .txt 文件。我在下面的两次尝试都从目录中删除了文件并引发了错误。谢谢:)

输入

16-0000_File-A_variant_strandbias_readcount.vcf.hg19_multianno_dbremoved_removed_final_index_inheritence_import.txt
16-0002_File-B_variant_strandbias_readcount.vcf.hg19_multianno_dbremoved_removed_final_index_inheritence_import.txt

期望输出

16-0000_File-A_multianno.txt
16-0002_File-B_multianno.txt

Bash 尝试 1 this removes the files from the directory

for f in /home/cmccabe/Desktop/test/vcf/overall/annovar/*_classify.txt ; do
 # Grab file prefix.
 p=${f%%_*_}
 bname=`basename $f`
 pref=${bname%%.txt}
 mv "$f" ${p}_multianno.txt
done

Bash 尝试 2 Substitution replacement not terminated at (eval 1) line 1.

for f in /home/cmccabe/Desktop/test/vcf/overall/annovar/*_classify.txt ; do
 # Grab file prefix.
 p=${f%%_*_}
 bname=`basename $f`
 pref=${bname%%.txt}
 rename -n 's/^$f/' *${p}_multianno.txt
done

你不需要循环。 rename一个人就能做到:

rename -n 's/(.*?_[^_]+).*/_multianno.txt/g' /home/cmccabe/Desktop/test/vcf/overall/annovar/*_classify.txt

正则表达式的意思大致是, 捕获从开始到第二个 _ 的所有内容, 匹配其余的, 并替换为捕获的前缀并附加 _multianno.txt

使用 -n 标志,此命令将打印其将执行的操作,而无需实际执行。 当输出看起来不错时,删除 -n 并重新运行。