测试字符串是否为有效整数 [英] Test whether string is a valid integer

问题描述

我正在尝试做一些足够常见的事情:在 shell 脚本中解析用户输入.如果用户提供了一个有效的整数，脚本会做一件事，如果无效，它会做其他事情.问题是，我还没有找到一种简单(且相当优雅)的方法来做到这一点 - 我不想逐个字符地将它分开.

I'm trying to do something common enough: Parse user input in a shell script. If the user provided a valid integer, the script does one thing, and if not valid, it does something else. Trouble is, I haven't found an easy (and reasonably elegant) way of doing this - I don't want to have to pick it apart char by char.

我知道这一定很容易，但我不知道该怎么做.我可以用十几种语言来做，但不能用 BASH！

I know this must be easy but I don't know how. I could do it in a dozen languages, but not BASH!

在我的研究中，我发现了这一点:

In my research I found this:

用于测试字符串是否包含以 10 为底的有效实数的正则表达式

其中有一个关于正则表达式的答案，但据我所知，这是 C 中可用的函数(以及其他函数).尽管如此，它的答案看起来不错，所以我用 grep 尝试了它，但 grep 不知道如何处理它.我试过 -P 在我的盒子上意味着把它当作一个 PERL 正则表达式 - nada.Dash E (-E) 也不起作用.-F 也没有.

And there's an answer therein that talks about regex, but so far as I know, that's a function available in C (among others). Still, it had what looked like a great answer so I tried it with grep, but grep didn't know what to do with it. I tried -P which on my box means to treat it as a PERL regexp - nada. Dash E (-E) didn't work either. And neither did -F.

明确地说，我正在尝试这样的事情，寻找任何输出 - 从那里，我将修改脚本以利用我得到的任何东西.(IOW，我原以为在重复有效行时，不符合要求的输入不会返回任何内容.)

Just to be clear, I'm trying something like this, looking for any output - from there, I'll hack up the script to take advantage of whatever I get. (IOW, I was expecting that a non-conforming input returns nothing while a valid line gets repeated.)

snafu=$(echo "$2" | grep -E "/^[-+]?(?:.[0-9]+|(?:0|[1-9][0-9]*)(?:.[0-9]*)?)$/")
if [ -z "$snafu" ] ;
then
   echo "Not an integer - nothing back from the grep"
else
   echo "Integer."
fi

有人能说明这是最容易做到的吗?

Would someone please illustrate how this is most easily done?

坦率地说，在我看来，这是 TEST 的一个缺点.它应该有一个这样的标志

Frankly, this is a short-coming of TEST, in my opinion. It should have a flag like this

if [ -I "string" ] ;
then
   echo "String is a valid integer."
else
   echo "String is not a valid integer."
fi

推荐答案

[[ $var =~ ^-?[0-9]+$ ]]

• ^ 表示输入模式的开始
• - 是文字-"
• ? 表示前面 (-) 的 0 或 1"
• + 表示1 个或多个前面的 ([0-9])"
• $ 表示输入模式的结束
• The ^ indicates the beginning of the input pattern
• The - is a literal "-"
• The ? means "0 or 1 of the preceding (-)"
• The + means "1 or more of the preceding ([0-9])"
• The $ indicates the end of the input pattern

所以正则表达式匹配一个可选的-(对于负数的情况)，后跟一个或多个十进制数字.

So the regex matches an optional - (for the case of negative numbers), followed by one or more decimal digits.

参考资料:

• http://www.tldp.org/LDP/abs/html/bashver3.html#REGEXMATCHREF

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