bash 打印字段编号大于 AWK 的行
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Printing lines which have a field number greater than, in AWK
提问by FatalError
I am writing a script in bash which takes a parameter and storing it;
我正在用 bash 编写一个脚本，它接受一个参数并存储它；
threshold =
I then have sample data that looks something like:
然后我有看起来像这样的示例数据：
5 blargh
6 tree
2 dog
1 fox
9 fridge
I wish to print only the lines which have their number greater than the number which is entered as the parameter (threshold).
我希望仅打印其编号大于作为参数（阈值）输入的编号的行。
I am currently using:
我目前正在使用：
awk '{print > $threshold}' ./file
But nothing prints out, help would be appreciated.
但没有打印出来，帮助将不胜感激。
回答by FatalError
You're close, but it needs to be more like this:
你很接近，但它需要更像这样：
$ threshold=3
$ awk -v threshold="$threshold" ' > threshold' file
Creating a variable with -vavoids the ugliness of trying to expand shell variables within an awkscript.
创建一个变量-v避免了在awk脚本中尝试扩展 shell 变量的丑陋。
EDIT:
编辑：
There are a few problems with the current code you've shown. The first is that your awkscript is single quoted (good), which stops $thresholdfrom expanding, and so the value is never inserted in your script. Second, your condition belongs outside the curly braces, which would make it:
您显示的当前代码存在一些问题。第一个是您的awk脚本是单引号（好），它停止$threshold扩展，因此该值永远不会插入到您的脚本中。其次，您的条件属于花括号之外，这将使它：
> threshold { print }
This works, but the `print is not necessary (it's the default action), which is why I shortened it to
这有效，但不需要`print（这是默认操作），这就是为什么我将其缩短为
> threshold
