庆典计算人的年龄和阅读到CSV文件？ [英] bash Calculate age of person and read to csv file?

问题描述

我有三列的csv文件，上面写着姓名，职务code，年龄DD / MM / YY。我想从计算今天的日期和产出每个条目的时代到CSV文件的第四列？我知道AWK是很方便的，但我不知道读取和各列的写入数据，并创建一个新的！
例如

I have a csv file with three columns that reads name,postcode,age dd/mm/yy. I would like to calculate the age of each entry from todays date and output to a fourth column of the csv file? I know awk is handy but i dont know to read and write the data from the various column and create a new one! e.g

name,postcode,dob,age
Dave,ws245f,09/12/2000,13

我有以下输入

`cat estimateAge.csv|awk -F'/|,' '{b=mktime($5" "$4" "$3" 00 00 00 00");a (systime()-b)/(365*24*60*60);a=a==int(a)?a:int(a)+1;print $0","a}`'

和这是在输出

Joe Bloggs,0121 545465650,01/03/1982,31

输出应

Joe Bloggs,0121 545465650,01/03/1982,30

请注意年龄计算不正确，因为李四还没有31至三月

Note the age calc is incorrect as the Joe Bloggs is not yet 31 until March

推荐答案

我觉得这个问题的核心部分不加入新时代的领域进行到底。但年龄计算。

I think the core part of this problem is not adding the new age field to the end. but the age calculation.

试试这个：

awk -F'/|,' '{b=mktime($5" "$4" "$3" 00 00 00 00");a=(systime()-b)/(365*24*60*60);a=a==int(a)?a:int(a)+1;print $0","a}' file

很好，也许我不应该把它们放进一行：

well maybe I should not put them into one line:

awk -F'/|,' '{b=mktime($5" "$4" "$3" 00 00 00 00");
              a=(systime()-b)/(365*24*60*60);
              a=a==int(a)?a:int(a)+1;print $0","a}' file

测试与您的示例数据：

test with your example data:

kent$  echo "Dave,ws245f,09/12/2000"|awk -F'/|,' '{b=mktime($5" "$4" "$3" 00 00 00 00");a=(systime()-b)/(365*24*60*60);a=a==int(a)?a:int(a)+1;print $0","a}' 
Dave,ws245f,09/12/2000,13

你可以看到，我没有检查标题行，一个 NR大于1 检查可以很容易地跳过称号。你可以DIY。

as you can see, I didn't check the title line, a NR>1 check could easily skip the title. You could DIY.

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