使用bash（SED / AWK）提取CSV文件中的行和列？ [英] using bash (sed/awk) to extract rows AND columns in CSV files?

问题描述

时的bash能够从CSV文件处理提取的行和列？希望我没有诉诸蟒蛇..

Is bash capable of handling extracting rows and columns from csv files? Hoping I don't have to resort to python..

我的5列csv文件如下：

My 5-column csv file looks like:

Rank,Name,School,Major,Year
1,John,Harvard,Computer Science,3
2,Bill,Yale,Political Science,4
3,Mark,Stanford,Biology,1
4,Jane,Princeton,Electrical Engineering,3
5,Alex,MIT,Management Economics,2

我只是想提取3日，4日，5日和列内容，忽略了第一排，所以输出如下：

I only want to extract the 3rd, 4th, and 5th column contents, ignoring the first row, so output looks like:

Harvard,Computer Science,3
Yale,Political Science,4
Stanford,Biology,1
Princeton,Electrical Engineering,3
MIT,Management Economics,2

到目前为止，我只能拿awk将打印出任何的每一行，还是我的CSV文件中的每一列，但像这种情况下，不特定COLS /行！可以来砸做到这一点？

So far I can only get awk to print out either each row, or each column of my CSV file, but not specific cols/rows like this case! Can bash do this?

推荐答案

猛砸解决方案;

使用IFS

#!/bin/bash
while IFS=',' read -r rank name school major year; do
    echo -e "Rank\t: $rank\nName\t: $name\nSchool\t: $school\nMajor\t: $major\nYear\t: $year\n"
done < file.csv
IFS=$' \t\n'

使用字符串操作和阵列

#!/bin/bash
declare -a arr
while read -r line; do
    arr=(${line//,/ })
    printf "Rank\t: %s\nName\t: %s\nSchool\t: %s\nMajor\t: %s\nYear\t: %s\n" ${arr[@]}
done < file.csv

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