sed怎么匹配文件中含有9個(gè)不連續(xù)逗號(hào)的行

4578004

如題，應(yīng)用生成的文件中怎樣匹配到含有9個(gè)不連續(xù)逗號(hào)的行并刪除這行數(shù)據(jù)？謝謝

43206133,2013-06-02 20:00:00,19,19,9_1_5,,2013-06-02,20
43443893,2013-06-02 23:00:00,5,1,9_1_5,1379,2013-06-02,23
43443894,2013-06-02 23:00:00,1,1,m0_1674_109_001,1647,2013-06-02,23
43443895,2013-06-02 23:00:00,5,5,9_1_5,,2013-06-02,23
43443896,2013-06-02 23:00:00,5,1,9_1_5,1341,2013-06-02,23
43443897,2013-06-02 23:00:00,2,1,33,1369,2013-06-02,23
43443898,2013-06-02 23:00:00,4,4,m0_1674_118_001,1647,2013-06-02,23
43206140,2013-06-02 20:00:00,0,1,,1331,2013-06-02,20
43443899,2013-06-02 23:00:00,0,111,1000,1674,2013-06-02,23
43443900,2013-06-02 23:00:00,5,3,9_1_5,1394,2013-06-02,23
43443901,2013-06-02 23:00:00,0,1,33, 33, 33,1368,2013-06-02,23                     ＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝〉匹配這行并刪除
43443902,2013-06-02 23:00:00,5,1,9_1_5,1370,2013-06-02,23
43443903,2013-06-02 23:00:00,2,1,33,1377,2013-06-02,23
43443904,2013-06-02 23:00:00,0,3,ch27,30,2013-06-02,23
43443905,2013-06-02 23:00:00,5,3,9_1_5,1368,2013-06-02,23
43443906,2013-06-02 23:00:00,4,1,,1368,2013-06-02,23
43443907,2013-06-02 23:00:00,0,1,null,1375,2013-06-02,23

yuloveban

sed -ne '/\,[0-9]+\,[0-9]+\,[0-9]+\,[0-9]+\,[0-9]+\,[0-9]+\,[0-9]+\,[0-9]+\,[0-9]+/p' filename

4578004

回復(fù) 2# yuloveban


    數(shù)據(jù)行不是純數(shù)字哦，所以這個(gè)匹配不到－

llh580110

1. egrep  '([^,]+,){9}' file

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dn833

1. awk -F , 'NF==10'

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4578004

回復(fù) 4# llh580110


    嗯 這個(gè)可以匹配到 但是我想在匹配到之后刪除這行數(shù)據(jù)該怎么操作呢

rucypli

awk -F, '{if(NF>9)print $0}' filename

wenhq

5L精簡(jiǎn)。贊一下。

llh580110

回復(fù) 6# 4578004

1. sed -r '/([^,]+,){9}/d' file

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peawseshc

回復(fù) 9# llh580110

1. egrep -v '([^,]+,){9}'

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