如何从提到的列表中获取文件名

Question

我是linux和/或脚本编程的新手，请耐心等待。我想要一个可以获取Linux目录文件的脚本。这里我试图获取文件名。

for NAME in $(ls -1 *.wav /some/path | cut -d "/" -f3 | cut -d "-" -f1-5)

如果文件名中包含-IN或搬出那么他们将SOX -m和MV后到另一个目录，但如果一些其他文件那么它将只是mv

用于参考，文件名像

1030-04-06-2015-1433414216.wav
1030-04-06-2015-1433414318.wav
1030-04-06-2015-1433414440。 WAV
1043-21-05-2015-1432207256.wav
1043-21-05-2015-1432207457.wav
1046-20-05-2015-1432137944.wav
1046-20-05-2015- 1432138015.wav
1046-20-05-2015-1432138704.wav
1431709157.93900.0-in.wav
1431709157.93900.0-out.wav
1431709157.93900.1-in.wav
1431709157.93900.1-out.wav
1431710008.94059.0-in.wav
1431710008.94059.0-out.wav
1431710008.94059.1 -in.wav
1431710008.94059.1-out.wav
1431710008.94059.1.wav
1431710008.94059.2.wav
1431713190.94698.2-in.wav
1431713190.94698.2-out.wav
1431713190.94698.2.wav
1431721107.96010.0-in.wav
1431721107.96010.0-out.wav
1431721107.96010.1.wav

Answer 1

下面是做到这一点的方法之一。

>cat test.sh 
#!/bin/bash 

destination="./DEST" 

# Loop over every file 
for file in "${@}" ; do 
    # If it is "-in", then sox 
    if [[ "${file}" =~ "-in" ]] || [[ "${file}" =~ "-out" ]] ; then 
    printf "sox -m ${file}; " 
    fi 
    echo "mv ${file} ${destination}" 
done 

运行时，我得到以下输出。

>../test.sh * 
mv 1030-04-06-2015-1433414216.wav ./DEST 
mv 1030-04-06-2015-1433414318.wav ./DEST 
mv 1030-04-06-2015-1433414440.wav ./DEST 
mv 1043-21-05-2015-1432207256.wav ./DEST 
mv 1043-21-05-2015-1432207457.wav ./DEST 
mv 1046-20-05-2015-1432137944.wav ./DEST 
mv 1046-20-05-2015-1432138015.wav ./DEST 
mv 1046-20-05-2015-1432138704.wav ./DEST 
sox -m 1431709157.93900.0-in.wav; mv 1431709157.93900.0-in.wav ./DEST 
sox -m 1431709157.93900.0-out.wav; mv 1431709157.93900.0-out.wav ./DEST 
sox -m 1431709157.93900.1-in.wav; mv 1431709157.93900.1-in.wav ./DEST 
sox -m 1431709157.93900.1-out.wav; mv 1431709157.93900.1-out.wav ./DEST 
sox -m 1431710008.94059.0-in.wav; mv 1431710008.94059.0-in.wav ./DEST 
sox -m 1431710008.94059.0-out.wav; mv 1431710008.94059.0-out.wav ./DEST 
sox -m 1431710008.94059.1-in.wav; mv 1431710008.94059.1-in.wav ./DEST 
sox -m 1431710008.94059.1-out.wav; mv 1431710008.94059.1-out.wav ./DEST 
mv 1431710008.94059.1.wav ./DEST 
mv 1431710008.94059.2.wav ./DEST 
sox -m 1431713190.94698.2-in.wav; mv 1431713190.94698.2-in.wav ./DEST 
sox -m 1431713190.94698.2-out.wav; mv 1431713190.94698.2-out.wav ./DEST 
mv 1431713190.94698.2.wav ./DEST 
sox -m 1431721107.96010.0-in.wav; mv 1431721107.96010.0-in.wav ./DEST 
sox -m 1431721107.96010.0-out.wav; mv 1431721107.96010.0-out.wav ./DEST 
mv 1431721107.96010.1.wav ./DEST 
mv DEST ./DEST 

如果要执行的命令，简单地剪切和这些行粘贴到一个壳，或修改bash脚本来执行这些命令，而不是打印出来。

Answer 2

这应该工作：

#!/bin/bash 
regex='(.*)(-out|-in)(.txt)' 
for file in *.txt;do 
# $file is the full path to the file 
    if [[ $file =~ $regex ]];then 
    #filename in this block contains -in or -out 
    filename="${file##*\/}" 
    inorout="${BASH_REMATCH[2]}" 
    extension="${BASH_REMATCH[3]}" 
    filenamewithoutextension="${filename%.*}" 
    filenamewithoutioroutorextension="${filenamewithoutextension/%$inorout/}" 
    filenamewithoutiorout="${filenamewithoutioroutorextension}$extension" 
    echo "$filename" "$inorout" "$extension" "$filenamewithoutextension" "$filenamewithoutiorout" "$filenamewithoutioroutorextension" 
    #do something here sox-m mv or whatever 
    else 
    #filename in this block doesn't contain -in or -out 
    echo "do something else" 
    fi 
done 

---

说明：

"${file##*\/}"是由$file即基本名（文件名）切割*/留下的字符串。

"${BASH_REMATCH[2]}"是从图案捕获的第二组匹配由[[ $file =~ $regex ]]即-in或-out

"${BASH_REMATCH[3]}"做是第三捕获组即.wav。

"${filename%.*}"是没有.wav

---

资源通过削减从 $file即文件名 .*留下的字符串，你应该检查：

1. Bash Parameter Expansion
2. Pattern Matching
3. Bash Variables