Bash：如何为同一字符串的多个实例 grep 一行？
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Bash: How can I grep a line for multiple instances of the same string?
提问by BubbleMonster
I've got several lines that look like this:
我有几行看起来像这样：
aaaaaaaaxzaaaaaaaaaaaaaa
bbbbbbbbbbbbxzbbbbbbxzbb
ccxzcccccccccccccccccxzc
dddddddxzddddddddddddddd
Inside two of those lines, there are twoinstances of xzcharacters. I want grep to look for xztwice in the same line and output the lines that it matches on. If xzappears once, I don't want to know.
在其中两行中，有两个xz字符实例。我希望 grepxz在同一行中查找两次并输出它匹配的行。如果xz出现一次，我不想知道。
Running the following:
运行以下内容：
cat lines | grep "xz"
Tells me every line with xz on, but I only want to see lines with xz appearing twice.
告诉我 xz 打开的每一行，但我只想看到 xz 出现两次的行。
How can I make the pattern search repeat in the same line?
如何在同一行中重复模式搜索？
回答by Wiktor Stribi?ew
You can use
您可以使用
cat lines | grep 'xz.*xz'
Or just
要不就
grep 'xz.*xz' lines
The .*will match optional characters (any but a newline) between 2 xz.
在.*将2之间匹配可选字符（但任何一个新行）xz。
In case you need to use look-arounds, you will need -Pswitch to enable Perl-like regexps.
如果您需要使用环视，则需要-P切换以启用类似 Perl 的正则表达式。
回答by Kent
awk is one way to go:
awk 是一种方法：
awk -F'xz' 'NF==3' file
or
或者
awk 'gsub(/xz/,"")==2' file
another benefit awk brings you is, it is easier to check a pattern matched less then n times, exact n timesor greater than n times. you just change the ==into <, <=, >, >=
awk 为您带来的另一个好处是，可以更轻松地检查匹配的模式less then n times,exact n times或greater than n times. 你只要把它==改成<, <=, >, >=
回答by karakfa
If you want to output the matching lines in full you don't need the options
如果要完整输出匹配的行，则不需要选项
grep 'xz.*xz' filename
will do
会做
