打印由变量中的数字给出的行

I try to get a line by a giving number in a variable so i try awk command:

我尝试通过变量中的给定数字得到一行，所以我尝试awk命令：

i did :

我做了：

DATE_ISSUE_OPIT_SIP="31456
12390
11232
12222"

COUNTER_NUMBER=1
line=$(awk 'NR==$COUNTER_NUMBER' <<< "$Date_ISSUE_OPIT_SIP")

Output is empty.

输出为空。

I want the output like this for example :

我希望输出像这样：

COUNTER_NUMBER=2
line=$(awk 'NR==$COUNTER_NUMBER' <<< "$Date_ISSUE_OPIT_SIP")
echo $line
11232

Thank's for help

感谢帮助

3 个解决方案

#1

The simplest fix will be replacing single quotes with double quotes

最简单的修复方法是用双引号替换单引号

... awk "NR==$COUNTER_NUMBER" <<< "$DATE_ISSUE_OPIT_SIP"

note also that the variable names you used didn't match, it's case sensitive.

另请注意，您使用的变量名称不匹配，它区分大小写。

#2

Awk has a wonderly easy way of setting an internal variable by passing it on the command line:

通过在命令行上传递内部变量，awk有一种非常简单的方法来设置内部变量：

txt="31456
12390
11232
12222"

n=2
line=$(awk 'NR==counter' counter=$n <<< "$txt")
echo "$line"
12390

It also means you can't easily have a file with an = in the name.

这也意味着您不能轻易拥有名称中带有=的文件。

You can also do:

你也可以这样做：

line=$(awk -v counter=$n 'NR==counter' <<< "$txt")

Note: NR starts at 1. So to get the 3rd line, you would pass 3 in your example.

注意：NR从1开始。因此，要获得第3行，您将在示例中传递3。

#3

It's OK with sed too

sed也没问题

sed "$COUNTER_NUMBER"'!d' <<< "$DATE_ISSUE_OPIT_SIP"

#1