问题描述
我正在学习用 Bash 编写脚本.
I am learning to script in Bash.
我有一个 5 列 22 行的 CSV 文件.我有兴趣从第二列中获取数据并将其放入数组中.
I have a CSV file with 5 columns and 22 rows. I am interested in taking the data from the second column and put it into an array.
我想要的是第一个名称在 array[0] 中，第二个在 array[1] 中，依此类推.
What I want is that the first name be in array[0], the second in array[1] and so on.
Bash 脚本:
#!/bin/sh
IFS=","
eCollection=( $(cut -d ',' -f2 MyAssignment.csv ) )
printf "%s\n" "${eCollection[0]}"
CSV 看起来像这样.没有标题行.
The CSV looks like this. There is no line with headers.
带有 Vl18xx 数字的列是我想要拆分为数组的内容.
The column with the Vl18xx numbers is what I want to split into an array.
John,Vl1805,VRFname,10.9.48.64/28,10.9.48.78
John,Vl1806,VRFname,10.9.48.80/28,10.9.48.94
John,Vl1807,VRFname,10.9.48.96/28,10.9.48.110
John,Vl1808,VRFname,10.9.48.112/28,10.9.48.126
John,Vl1809,VRFname,167.107.152.32/28,167.107.152.46
bash 脚本没有将第二列放入数组中，我做错了什么?
The bash script is not placing the 2nd column into the array, what am I doing wrong?
推荐答案
移除 IFS="," 赋值，从而让默认的 IFS 值空格、制表符、换行符应用
Remove the IFS="," assignment thereby letting the default IFS value of space, tab, newline apply
#!/bin/bash
eCollection=( $(cut -d ',' -f2 MyAssignment.csv ) )
printf "%s\n" "${eCollection[0]}"
解释:由于外括号，eCollection 变量是一个数组.它使用来自 $(cut -d ',' -f2 MyAssignment.csv) 子外壳的 IFS 分隔(想想函数或命令行参数)词中的每个元素进行初始化，这是cut 命令与 ',' 分隔符一起使用并打印 MyAssignment.csv 文件中的第二个字段 -f2.
Explained: The eCollection variable is an array due to the outer parenthesis. It is initialized with each element from the IFS-separated (think function or command line arguments) words which come from the $(cut -d ',' -f2 MyAssignment.csv) subshell, which is the cut command used with the ',' delimiter and printing the second field -f2 from the MyAssignment.csv file.
printf 语句只是显示如何按索引打印任何项目，您也可以尝试 echo "${eCollection[@}]}" 来查看所有元素.
The printf statement just shows how to print any item by index, you could also try echo "${eCollection[@}]}" to see all of the elements.
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