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Bash 如何从“diff”输出中“剪切”小于“符号?_Bash_Grep_Diff - Fatal编程技术网

Bash 如何从“diff”输出中“剪切”小于“符号?

Bash 如何从“diff”输出中“剪切”小于“符号?,bash,grep,diff,Bash,Grep,Diff,我怎样才能切到最后您可以尝试下面的sed命令 $ echo '#20 17 03 45 * root bash /media/data/loop.sh <' | sed 's/ [^ ]*$//' #20 17 03 45 * root bash /media/data/loop.sh 它将删除最后一个空格和以下字符。所以你的命令是 diff filecron filecrontmp | grep "<" | head -n 1 | sed 's/ [^ ]*$//' 使用BAS

我怎样才能切到最后您可以尝试下面的sed命令

$ echo '#20 17 03 45 * root bash /media/data/loop.sh <' | sed 's/ [^ ]*$//'
#20 17 03 45 * root bash /media/data/loop.sh
它将删除最后一个空格和以下字符。所以你的命令是

diff filecron filecrontmp | grep "<" | head -n 1 | sed 's/ [^ ]*$//'

使用BASH字符串操作:

s='#20 17 03 45 * root bash /media/data/loop.sh <'
echo "${s%<*}"
#20 17 03 45 * root bash /media/data/loop.sh
diff filecron filecrontmp | grep "<" | head -n 1 | sed 's/ [^ ]*$//'
s='#20 17 03 45 * root bash /media/data/loop.sh <'
echo "${s%<*}"
#20 17 03 45 * root bash /media/data/loop.sh