如何在所有的.py .txt .cpp文件 中grep “abc”和“123”两个字符串？

[方紫涵]

如何在所有的.py .txt .cpp文件 中grep “abc”和“123”两个字符串？

[ForestDB]

grep -Ern '(abc|123)' *.cpp *.py *.txt
grep -Ern '(abc|123)' `find . -regextype posix-egrep -regex '.*(cpp|txt|py)'`

[方紫涵]

[Quote=引用 3 楼 的回复:]

find . -name "*.py" -o -name "*.txt" -o "*.cpp" | xargs grep -E "abc|123"
[/Quote]
Mode$ find . -name "*.py" -o -name "*.txt" -o "*.cpp" | xargs grep -E "abc|123"
find: 路径必须在表达式之前: *.cpp
用法: find [-H] [-L] [-P] [-Olevel] [-D help|tree|search|stat|rates|opt|exec] [path...] [expression]


？

[方紫涵]

[Quote=引用 4 楼 的回复:]

grep -R '(abc|123)' *.cpp *.py *.txt
[/Quote]
Mode$ grep -R '(abc|123)' *.cpp *.py *.txt
grep: *.cpp: 没有那个文件或目录
grep: *.py: 没有那个文件或目录
grep: *.txt: 没有那个文件或目录

？

[mymtom]

[Quote=引用 5 楼 的回复:]

引用 3 楼 的回复:

find . -name "*.py" -o -name "*.txt" -o "*.cpp" | xargs grep -E "abc|123"

Mode$ find . -name "*.py" -o -name "*.txt" -o "*.cpp" | xargs grep -E "abc|123"
find: 路径必须在表达式之前: *.cpp
……
[/Quote]
不好意思! 手误了，试试下面这个:
find . -name "*.py" -o -name "*.txt" -o -name "*.cpp" | xargs grep -E "abc|123"

[方紫涵]

[Quote=引用 1 楼 的回复:]

grep -Ern '(abc|123)' *.cpp *.py *.txt
grep -Ern '(abc|123)' `find . -regextype posix-egrep -regex '.*(cpp|txt|py)'`
[/Quote]

ks$ grep -Ern '(String|main)' *.cpp *.java ./
grep: *.cpp: 没有那个文件或目录
grep: *.java: 没有那个文件或目录
./opt/emoji/EmojiFactory.h:90: // The ownership of the instance remains to this class, so users must not
./opt/emoji/EmojiFactory.h:97: // The ownership of the instance remains to this class, so users must not


好像不对阿。。？

[mymtom]

find . -name "*.py" -o -name "*.txt" -o "*.cpp" | xargs grep -E "abc|123"

[freetstar]

grep -R '(abc|123)' *.cpp *.py *.txt