我正在努力解决这个问题。我有一条短信:
Executables: manatee-curl
Dependencies: base ==4.*, manatee-core >=0.1.1, dbus-client ==0.3.*,
stm >=2.1.2.0, containers >=0.3.0.0, gtk >=0.12.0,
text >=0.7.1.0, mtl >=1.1.0.2, old-time -any,
old-locale -any, glib >=0.12.0, gio >=0.12.0,
filepath >=1.1.0.3, utf8-string >=0.3.4, bytestring -any,
network -any, curl >=1.3.7, directory -any,
template-haskell -any, derive -any, binary -any,
regex-tdfa -any, dbus-core -any
Cached: No我想把所有这些词放在“属地:”和“缓存”之间。可以将此文本视为变量,例如:
echo $text \ grep /dostuff/
要明确的是,我想得到的输出是:
base ==4.*, manatee-core >=0.1.1, dbus-client ==0.3.*,
stm >=2.1.2.0, containers >=0.3.0.0, gtk >=0.12.0,
text >=0.7.1.0, mtl >=1.1.0.2, old-time -any,
old-locale -any, glib >=0.12.0, gio >=0.12.0,
filepath >=1.1.0.3, utf8-string >=0.3.4, bytestring -any,
network -any, curl >=1.3.7, directory -any,
template-haskell -any, derive -any, binary -any,
regex-tdfa -any, dbus-core -any谢谢。
发布于 2016-07-08 22:44:48
您可以使用sed在两个模式之间获取文本:
text=$(sed -n '/^Dependencies:/,/^Cached:/{$d;s/Dependencies: *//;p;}' file)
echo "$text"
base ==4.*, manatee-core >=0.1.1, dbus-client ==0.3.*,
stm >=2.1.2.0, containers >=0.3.0.0, gtk >=0.12.0,
text >=0.7.1.0, mtl >=1.1.0.2, old-time -any,
old-locale -any, glib >=0.12.0, gio >=0.12.0,
filepath >=1.1.0.3, utf8-string >=0.3.4, bytestring -any,
network -any, curl >=1.3.7, directory -any,
template-haskell -any, derive -any, binary -any,
regex-tdfa -any, dbus-core -anyIDEOne工作演示
发布于 2016-07-09 04:44:24
awk去营救!
$ awk '/^Dependencies:/{$1="";p=1} /^Cached:/{p=0} p{sub(/^ +/,"");print}' file
base ==4.*, manatee-core >=0.1.1, dbus-client ==0.3.*,
stm >=2.1.2.0, containers >=0.3.0.0, gtk >=0.12.0,
text >=0.7.1.0, mtl >=1.1.0.2, old-time -any,
old-locale -any, glib >=0.12.0, gio >=0.12.0,
filepath >=1.1.0.3, utf8-string >=0.3.4, bytestring -any,
network -any, curl >=1.3.7, directory -any,
template-haskell -any, derive -any, binary -any,
regex-tdfa -any, dbus-core -any您可以像往常一样为一个变量赋值。
$ text=$(awk ...)https://stackoverflow.com/questions/38276304
复制相似问题