Groovy的数字字符串比较

Question

，该怎么办错了：

assert 'foo' == 'foo' //PASS 
assert '500' == '500' //PASS 
assert '500' < '1000' //FAIL <-- Supposed to pass 
assert '500' <= '1000' //FAIL <-- Supposed to pass 
assert '1000' > '500' //FAIL <-- Supposed to pass 
assert '1000' >= '500' //FAIL <-- Supposed to pass 

它是一个可定制的 “条件” 的对象：

class Condition { 
    static def compareClosure = [ 
      '==' : { a, b -> a == b}, 
      '!=' : { a, b -> a != b}, 
      '<' : { a, b -> a < b}, 
      '<=' : { a, b -> a <= b}, 
      '>' : { a, b -> a > b}, 
      '>=' : { a, b -> a >= b} 
    ] 

    String comparator 
    def value 

    Condition(String comparator, String value) { 
     this.value = value 
     this.comparator = comparator 
    } 

    boolean isSatisfiedBy(def value) { 
     compareClosure[comparator](value, this.value) 
    } 
} 

所以

assert new Condition('<=', '1000').isSatisfiedBy('500') //FAIL 

有没有一种办法在不将值转换为数值类型的情况下执行此操作

Answer 1

你的问题的简短答案是否定的。

但是，我有一种感觉，这是为了排序的目的。如果是这样的话。这是我为此使用的排序功能。

例子在行动：Groovy Web Console

Closure customSort = { String a, String b -> 
     def c = a.isBigDecimal() ? new BigDecimal(a) : a 
     def d = b.isBigDecimal() ? new BigDecimal(b) : b 


     if (c.class == d.class) { 
     return c <=> d 
     } else if (c instanceof BigDecimal) { 
     return -1 
     } else { 
     return 1 
     } 

    } 

['foo','500','1000', '999', 'abc'].sort(customSort)