我有一个字符串
var=refs/heads/testing/branch
我想使用shell脚本摆脱字符串中的refs/heads/
,这样我就只有:
var=testing/branch
我尝试过的命令(每行一个):
echo $(var) | awk -F\\ {'print $2'}
echo $var | sed -e s,refs/heads/,,
echo "refs/heads/testing/branch" | grep -oP '(?<=refs/heads/\)\w+'
echo "refs/heads/testing/branch" | LC_ALL=C sed -e 's/.*\\//'
echo "refs/heads/testing/branch" | cut -d'\' -f2
echo refs/heads/testing/branch | sed -e s,refs/heads/,,
答案 0 :(得分:0)
有很多选择,尝试简单的选择:
echo $var | cut -d "/" -f 3,4
echo $var | awk -F"/" '{print $3"/"$4}'
答案 1 :(得分:0)
Shell参数扩展:从变量内容中删除前缀“ refs / heads /”
$ var=refs/heads/testing/branch
$ echo "${var#refs/heads/}"
testing/branch