通过Bash循环读取空分隔的字符串

Question

我想迭代一个文件列表,而不关心文件名可能包含哪些字符,所以我使用一个由空字符分隔的列表.代码将更好地解释事情.

# Set IFS to the null character to hopefully change the for..in
# delimiter from the space character (sadly does not appear to work).
IFS=$'\0'

# Get null delimited list of files
filelist="`find /some/path -type f -print0`"

# Iterate through list of files
for file in $filelist ; do
    # Arbitrary operations on $file here
done

从文件读取时,以下代码有效,但我需要从包含文本的变量中读取.

while read -d $'\0' line ; do
    # Code here
done < /path/to/inputfile

Answer 1

在bash中,您可以使用here-string

while IFS= read -r -d '' line ; do
    # Code here
done <<<"$var"

请注意,您应该内联IFS=并使用,-d ''但请确保'd'和第一个单引号之间有空格.另外,添加-r标志以忽略转义.

此外,这不是您的问题的一部分,但我可能会建议您在使用时更好的方式来执行您的脚本find; 它使用流程替换.

while IFS= read -r -d '' file; do
    # Arbitrary operations on "$file" here
done < <(find /some/path -type f -print0)

Answer 2

将它们通过管道传输到xargs -0：

files="$( find ./ -iname 'file*' -print0 | xargs -0 )"

xargs 手册：

-0, --null
    Input items are terminated by a null character instead of
    by whitespace, and the quotes and backslash are not
    special (every character is taken literally).