blocks|key|2073049|text|但接下来就这么做吧|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2073050|ends+=+'20080201..20080229'.split('..').map{%7Cd%7C+Integer(d)}
ends[0]..ends[1]|code-block|syntax|javascript|2073051|无论如何，出于安全原因，我不推荐eval|2073052|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

But then just do 

<pre><code>ends = '20080201..20080229'.split('..').map{|d| Integer(d)}
ends[0]..ends[1]
</code></pre>

anyway I don't recommend eval, for security reasons

blocks|key|2073208|text|Range.new(*self.split("..").map(&:to_i))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2073209|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>Range.new(*self.split("..").map(&amp;:to_i))
</code></pre>

blocks|key|2073157|text|不带参数的Inject适用于两个元素数组：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2073158|rng='20080201..20080229'.split('..').inject+{+%7Cs,e%7C+s.to_i..e.to_i+}|code-block|syntax|javascript|2073159|当然，这可以是泛型的|2073160|class+Range
++def+self.from_ary(a)
++++a.inject{%7Cs,e%7C+s..e+}
++end
end

rng+=+Range.from_ary('20080201..20080229'.split('..').map{%7Cs%7C+s.to_i})
rng.class++#+=>+Range|2073161|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Inject with no args works well for two element arrays:

<pre><code>rng='20080201..20080229'.split('..').inject { |s,e| s.to_i..e.to_i }
</code></pre>

Of course, this can be made generic

<pre><code>class Range
 def self.from_ary(a)
 a.inject{|s,e| s..e }
 end
end

rng = Range.from_ary('20080201..20080229'.split('..').map{|s| s.to_i})
rng.class # =&gt; Range
</code></pre>

blocks|key|2072954|text|假设您希望该范围在月份等时间内正确迭代，请尝试|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2072955|require+'date'

ends+=+'20080201..20080229'.split('..').map{%7Cd%7C+Date.parse(d)}
(ends[0]..ends[1]).each+do+%7Cd%7C
++p+d.day
end|code-block|syntax|javascript|2072956|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

assuming you want the range to iterate properly through months etc, try

<pre><code>require 'date'

ends = '20080201..20080229'.split('..').map{|d| Date.parse(d)}
(ends[0]..ends[1]).each do |d|
 p d.day
end
</code></pre>

blocks|key|1859591|text|Ranger使用正则表达式来验证字符串，而不用担心SQL注入，然后使用eval。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1859592|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/Epigene/ranger^0|Z|4|0|6|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

<a href="https://github.com/Epigene/ranger" rel="nofollow noreferrer">Ranger</a> uses regex to validate strings with no SQL injection fear, and then <code>eval</code>.

blocks|key|1859426|text|将@Purfideas答案与StackOverflow上某处的另一个答案组合在一起，我通过使用输入检查将代码包围起来，从而解决了这个问题，因此唯一使用的是有效的enumerable|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1859427|if+!value[/%5E[0-9]%2B\.\.[0-9]%2B$/].nil?
++++ends+=+value.split('..').map{%7Cd%7C+Integer(d)}
++++value+=+ends[0]..ends[1]
end|code-block|syntax|javascript|1859428|它本质上是将字符串值重写为可枚举值。如果您在yaml配置文件中添加一个enumerable字段，这将非常方便。|1859429|如果您的应用程序需要它，您可以使用可选的第三个文字点来扩展正则表达式，这可能是可选的。|1859430|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Combining @Purfideas answer with another answer somewhere on StackOverflow, I solved this by also surrounding the code with an input check, so the only thing used is a valid enumerable 

<pre><code>if !value[/^[0-9]+\.\.[0-9]+$/].nil?
 ends = value.split('..').map{|d| Integer(d)}
 value = ends[0]..ends[1]
end
</code></pre>

It essentially rewrites your string value to a enumerable value. This comes in handy if you add a enumerable field in a yaml config file.

If you need it for your application, you could extend the regex with an optional third literal dot, that could be optional.

blocks|key|1859535|text|如果我们这样做|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1859536|v=+"20140101..20150101"
raise+"Error:+invalid+format:+#{v}"+if+/\d{8}\.\.\d{8}/+!~+v
r=+eval(v)|code-block|syntax|javascript|1859537|而攻击者有一种绕过raise检查的方法(简单地通过操纵运行时来禁用异常)，那么我们就可以得到一个危险的评估，这可能会破坏整个宇宙。|1859538|因此，为了减少攻击向量，我们先检查格式，然后手动进行解析，然后检查结果|1859539|v=+"20140101..20150101"
raise+"Error:+invalid+format:+#{v}"+if+/\d{8}\.\.\d{8}/+!~+v
r=+Range.new(*v.split(/\.\./).map(&:to_i))
raise+"Error:+invalid+range:+#{v}"+if+r.first>+r.last|1859540|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

If we do it like

<pre><code>v= "20140101..20150101"
raise "Error: invalid format: #{v}" if /\d{8}\.\.\d{8}/ !~ v
r= eval(v)
</code></pre>

and the attacker has a way of bypassing the raise check (simply by means of manipulating the runtime to disable exceptions) then we can get a dangerous eval which will potentially destroy the universe.

So for the sake of reducing attack vectors, we check the format, and then do the parsing manually, then check the results

<pre><code>v= "20140101..20150101"
raise "Error: invalid format: #{v}" if /\d{8}\.\.\d{8}/ !~ v
r= Range.new(*v.split(/\.\./).map(&amp;:to_i))
raise "Error: invalid range: #{v}" if r.first&gt; r.last
</code></pre>

blocks|key|1859698|text|在这里，假设您希望将散列存储为系统常量值，并在任何模型中获取它。散列键将是一个范围值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1859699|hash_1+=+{1..5+=>+'a',+6..12+=>+'b',+13..67+=>+'c',+68..9999999+=>+'d'}|code-block|syntax|javascript|1859700|然后创建值为hash_1.to_json的系统常量。.to_json会将你的哈希对象转换成JSON对象。现在在代码中创建一个新的散列hash_2，|1859701|JSON.parse(SystemConstant.get('Constant_name')).each{%7Ckey,val%7C+temp_k=key.split('..').map{%7Cd%7C+Integer(d)};+hash_2[temp_k[0]..temp_k[1]]+=+val}|1859702|新的hash_2将成为所需的hash_1|1859703|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Here suppose you want to store the hash as a system constant value and fetch it in any model. The hash key will be a range value.

<pre><code>hash_1 = {1..5 =&gt; 'a', 6..12 =&gt; 'b', 13..67 =&gt; 'c', 68..9999999 =&gt; 'd'}
</code></pre>

Then create the system constant with value as hash_1.to_json.
.to_json will convert your hash object to JSON object.
Now inside the code create a new hash hash_2,

<pre><code>JSON.parse(SystemConstant.get('Constant_name')).each{|key,val| temp_k=key.split('..').map{|d| Integer(d)}; hash_2[temp_k[0]..temp_k[1]] = val}
</code></pre>

The new hash_2 will be the required hash_1

blocks|key|2073876|text|我有一个类似的要求，尽管在我的例子中，字符串有两种可能的格式，偶尔是单个数字字符串，如"7"，其他时候是范围，如"10-14“。无论采用哪种方式，我都希望将字符串转换为可枚举的数字集合。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2073877|我的方法(受highest+voted+answer的启发)是：|offset|length|2073878|def+numbers(from_string:)
++if+from_string.include?('-')
++++return+Range.new(*from_string.split('-').map(&:to_i))
++else
++++return+[from_string.to_i]+#+put+number+in+an+array+so+we+can+enumerate+over+it
++end
end|code-block|syntax|javascript|2073879|如果你认为这更具可读性，也可以将其作为(长)一行代码来使用：|2073880|from_string.include?('-')+?+Range.new(*from_string.split('-').map(&:to_i))+:+[from_string.to_i]|2073881|我正在处理一长串的已知字符串，而不是处理任意的用户输入，所以这不能防止恶意输入。|2073882|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/3243732/145725^0|0|6|K|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@$D|10|E|11|1|12]]|A|$]]|$1|F|3|G|5|H|7|13|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|15|8|@]|9|@]|A|$I|J]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

I had a similar requirement although in my case the strings were in two possible formats, occasionally they were single number strings such as &quot;7&quot;, other times they were ranges such as &quot;10-14&quot;. Either way I wanted to turn the string into an Enumerable collection of numbers.
My approach (inspired by the <a href="https://stackoverflow.com/a/3243732/145725">highest voted answer</a>) was:

<pre class="lang-ruby prettyprint-override"><code>def numbers(from_string:)
 if from_string.include?('-')
 return Range.new(*from_string.split('-').map(&amp;:to_i))
 else
 return [from_string.to_i] # put number in an array so we can enumerate over it
 end
end
</code></pre>
It can also be done as a (long) one-liner if you think that's more readable:
<pre class="lang-ruby prettyprint-override"><code>from_string.include?('-') ? Range.new(*from_string.split('-').map(&amp;:to_i)) : [from_string.to_i]
</code></pre>
I was processing a long list of known strings, not dealing with arbitrary user input so this doesn't guard against malicious input.

I have some Ruby code which takes dates on the command line in the format:

<pre><code>-d 20080101,20080201..20080229,20080301
</code></pre>

I want to run for all dates between <code>20080201</code> and <code>20080229</code> inclusive and the other dates present in the list.

I can get the string <code>20080201..20080229</code>, so is the best way to convert this to a Range instance? Currently, I am using <code>eval</code>, but it feels like there should be a better way.

<hr>

@Purfideas I was kind of looking for a more general answer for converting any string of type <code>int..int</code> to a Range I guess.

What is the best way to convert a Ruby string range to a Range object

我有一些Ruby代码，它在命令行上采用以下格式的日期：-d 20080101,20080201..20080229,20080301我想运行20080201和20080229之间的所有日期，包括列表中的其他日期。我可以获得字符串20080201..20080229，那么将其转换为Range实例的最佳方法是什么？目前，我正在使用eval，但我觉得应该有更好的方法。@Purfideas我在寻找一种更通

问将Ruby字符串范围转换为range对象的最佳方法是什么
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