如何计算unix行中给定字符的出现次数（crontab） [英] How to count occurrence of a given character in a unix line ( crontab)

问题描述

您好，



我的数据看起来像是浮动的exepmles：



a; kkkhkhk; hgvhgvhgvh; 123; t



a; bbbg; yuj; 1237



x; 25; z; t



我想计算每一行中出现;（分号）的次数。并且，当给定字符的数量大于4时，它应该将相应的行存储到文件中。



我尝试过以下代码：



awk -F';''NF == 31'$ nom'_temp2.​​csv'> $ nom'_temp3.csv'



它在命令行上工作但不在crontab中工作。



等待解决方案。

提前谢谢你



我尝试过：



awk -F';''NF == 31'$ nom'_temp2.​​csv'> $ nom'_temp3.csv'

Hello ,

I have data that look like the floowing exepmles :

a ;kkkhkhk ;hgvhgvhgvh ;123 ;t

a ;bbbg ;yuj ;1237

x ;25 ;z ;t

I want to count the occurrences of ";"(semicolon) in every line. and, when the number of the given character is above 4, it should store the corresponding line into a file .

I have tried the code below :

awk -F ';' 'NF==31' $nom'_temp2.csv' > $nom'_temp3.csv'

it did work on command line but not in crontab.

Waiting for a solution.
Thanks you in advance

What I have tried:

awk -F ';' 'NF==31' $nom'_temp2.csv' > $nom'_temp3.csv'

推荐答案

nom'_temp2.​​csv'>

nom'_temp2.csv' >

nom'_temp3.csv'



它在命令行上工作但不在crontab中工作。



等待解决方案。

提前谢谢你



我尝试过：



awk -F';''NF == 31'

nom'_temp3.csv'

it did work on command line but not in crontab.

Waiting for a solution.
Thanks you in advance

What I have tried:

awk -F ';' 'NF==31'

nom'_temp2.​​csv'>

nom'_temp2.csv' >

这篇关于如何计算unix行中给定字符的出现次数（crontab）的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！