在bash（多行）中获取两个单词之间的文本

Question

我正在解决这个问题。我有这段文字：

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

我想在“依赖项：”和“缓存”之间获取所有这些词。 您可以将此文本视为变量，例如：

回显 $text | grep /dostuff/

要清楚，我想要得到的输出是：

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。

Answer 1

使用 GNU grep：

echo "$text" | grep -Poz 'Dependencies:  \K(.*\n)*(?=Cached:)' | grep -Po '^ +\K.*'

输出：

基础 ==4.*，海牛核心 >=0.1.1，dbus 客户端 ==0.3.*， stm >=2.1.2.0，容器 >=0.3.0.0，gtk >=0.12.0， text >=0.7.1.0, mtl >=1.1.0.2, old-time -any, 旧语言环境 -any，glib >=0.12.0，gio >=0.12.0， 文件路径 >=1.1.0.3，utf8 字符串 >=0.3.4，字节串 -any， 网络 -any，curl >=1.3.7，目录 -any， 模板-haskell -any，派生 -any，二进制 -any， 正则表达式-tdfa -any，dbus-core -any

---

见：The Stack Overflow Regular Expressions FAQ

Answer 2

awk 来救援！

$ 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

您可以像往常一样分配给变量

$ text=$(awk ...)

Answer 3

您可以使用sed 获取两种模式之间的文本：

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 -any

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