地带awk程序前导零 [英] strip leading zeros in awk program

问题描述

所以我pretty丢失我错过了一个星期，我打赶上，但我写一个awk程序分辨出来，以天，两个日期之间。

So I'm pretty lost as I missed a week and am playing catch up, but I'm to write an awk program to tell the difference, in days, between two dates.

我或多或少用它做，但是如果用户输入当仁不让的数据验证时，01/01 / YYYY它不喜欢它。我拉了3场出arg的然后将它们传递给一个函数来验证。

I'm more or less done with it, however it doesn't like it if the user inputs 01/01/YYYY when doing my data validation. I pull the 3 fields out of the arg then pass them to a function to validate.

这是给我麻烦的部分是，在我的验证：

the part that's giving me trouble is, in my validation:

if ( day > Days[month] ){
invalid stuff here
}

其中，天[] 是一个数组控股[1] = 31，[2] = 28（或29，如果飞跃）等。

where Days[] is an array holding [1]=31, [2]=28 (or 29 if leap) etc.

由于天[01]没有定义它击中如此。我可以去解决它通过定义[01]因为同为[1]我想，但我宁愿东西更优雅。

It hits true because Days[01] is not defined. I could go around it by defining a [01] as the same as [1] I suppose but I'd rather something more elegant.

我怎样才能剥去前导零，awk程序中，变量一个月？我发现一切都在一个shell脚本使用和awk调用，但这并不能真正帮助我。

How can I strip the leading zeros, within the awk program, of the variable month? Everything I've found is for use in a shell script and invokes awk, but that doesn't really help me.

感谢您的帮助！

推荐答案

您需要这个，假设你的变量称为月：

You need this, assuming your variable is called month:

gsub ("^0*", "", month);

的 ^ 是启动锚和 0 * 表示零个或多个 0 字符。因此，这有效地在变量的开始删除所有 0 字符。

The ^ is the start anchor and 0* means zero or more 0 characters. So this effectively removes all 0 characters at the start of the variable.

通过举例的方式（这也说明了一个办法把它做对的中的数量为好，除了拆分日之前，看到第二个 GSUB 为）：

By way of example (which also shows a way to do it to the middle number as well, before splitting the date apart, see the second gsub for that):

pax> echo '1/1/2013
03/12/2014
02/02/1965' | awk '{gsub ("^0*", "", $0); gsub ("/0*", "/", $0); print}'

1/1/2013
3/12/2014
2/2/1965

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