找到包含某种模式的n个出现的行

Question

我有一个包含类似

的行的文件

A,B,1,2,3,$long,6,"A","",$long,,,,"ABC",,$long,,,,
E,F,2,3,4,$long,$long,$long,$long,,,"A","STRING";123456,,,1,2

我的目标是找到包含n次出现模式的行＆＃34; $ long＆＃34;。

有人知道这场比赛的grep正则表达式吗？

Answer 1

你不需要正则表达式。使用awk，您可以使用$long作为字段分隔符，并检查每行有多少字段：

awk -v count=3 'BEGIN {FS="\\$long"} NF==(count+1)' file

测试

$ awk -v count=3 'BEGIN {FS="\\$long"} NF==(count+1)' a
A,B,1,2,3,$long,6,"A","",$long,,,,"ABC",,$long,,,,
$ awk -v count=4 'BEGIN {FS="\\$long"} NF==(count+1)' a
E,F,2,3,4,$long,$long,$long,$long,,,"A","STRING";123456,,,1,2
$ awk -v count=5 'BEGIN {FS="\\$long"} NF==(count+1)' a
$

Answer 2

awk solution by Fedorqui应该可以正常工作。您也可以使用grep：

grep -E '(.*\$long){4}' file
E,F,2,3,4,$long,$long,$long,$long,,,"A","STRING";123456,,,1,2

Answer 3

$ awk -v n=3 'gsub(/\$long/,"&")==n' file
A,B,1,2,3,$long,6,"A","",$long,,,,"ABC",,$long,,,,

$ awk -v n=4 'gsub(/\$long/,"&")==n' file
E,F,2,3,4,$long,$long,$long,$long,,,"A","STRING";123456,,,1,2

但如果$long可以出现在除了作为其自己的字段之外的上下文中，例如：

A,B,1,2,3,$long,6,"A","",$long,,,,"ABC$longDEF",,$long,,,,

并且你只想在它自己的领域中计算它然后你需要更多的东西：

$ awk -F, -v n=3 '{c=0; for (i=1;i<=NF;i++) if ($i=="$long") c++} c==n' file

e.g：

$ cat file
A,B,1,2,3,$long,6,"A","",$long,,,,"ABC",,$long,,,,
E,F,2,3,4,$long,$long,$long,$long,,,"A","STRING";123456,,,1,2
A,B,1,2,3,$long,6,"A","",$long,,,,"ABC$longDEF",,$long,,,,

错：

$ awk -v n=3 'gsub(/\$long/,"&")==n' file
A,B,1,2,3,$long,6,"A","",$long,,,,"ABC",,$long,,,,

$ awk -v n=4 'gsub(/\$long/,"&")==n' file
E,F,2,3,4,$long,$long,$long,$long,,,"A","STRING";123456,,,1,2
A,B,1,2,3,$long,6,"A","",$long,,,,"ABC$longDEF",,$long,,,,

右：

$ awk -F, -v n=3 '{c=0; for (i=1;i<=NF;i++) if ($i=="$long") c++} c==n' file
A,B,1,2,3,$long,6,"A","",$long,,,,"ABC",,$long,,,,
A,B,1,2,3,$long,6,"A","",$long,,,,"ABC$longDEF",,$long,,,,

$ awk -F, -v n=4 '{c=0; for (i=1;i<=NF;i++) if ($i=="$long") c++} c==n' file
E,F,2,3,4,$long,$long,$long,$long,,,"A","STRING";123456,,,1,2