首页 \ 问答 \ 排序Java集合(Sort Java Collection)

排序Java集合(Sort Java Collection)

 我有一个Java集合： 
 

Collection<CustomObject> list = new ArrayList<CustomObject>();

 

 CustomObject在显示列表之前有一个id字段，我想通过该id对这个集合进行排序。 
 

 有没有办法我这样做？ 

I have a Java collection:
 

Collection<CustomObject> list = new ArrayList<CustomObject>();

 

CustomObject has an id field now before display list I want to sort this collection by that id.
 

Is there any way I could that do that? 

原文：https://stackoverflow.com/questions/6957631
更新时间：2023-05-15 07:05

最满意答案

 一种可能的方案： 
 

dl.xpath('dt').each_with_index do |dt, i|
  dds = dt.xpath("following-sibling::dd[not(../dt[#{i + 2}]) or " +
                     "following-sibling::dt[1]=../dt[#{i + 2}]]")
  puts "#{dt.text}: #{dds.map(&:text).join(', ')}"
end

 

 这依赖于dt元素的值比较，并且当存在重复时将失败。 以下（更复杂的）表达式不依赖于唯一的dt值： 
 

following-sibling::dd[not(../dt[$n]) or 
    (following-sibling::dt[1] and count(following-sibling::dt[1]|../dt[$n])=1)]

 

 注意：您对self使用失败，因为您没有正确使用它作为轴（ self:: 。 另外， self总是只包含上下文节点，因此它将引用由表达式检查的每个dd ，而不是返回到原始dt 

One possible solution:
 

dl.xpath('dt').each_with_index do |dt, i|
  dds = dt.xpath("following-sibling::dd[not(../dt[#{i + 2}]) or " +
                     "following-sibling::dt[1]=../dt[#{i + 2}]]")
  puts "#{dt.text}: #{dds.map(&:text).join(', ')}"
end

 

This relies on a value comparison of dt elements and will fail when there are duplicates. The following (much more complicated) expression does not depend on unique dt values:
 

following-sibling::dd[not(../dt[$n]) or 
    (following-sibling::dt[1] and count(following-sibling::dt[1]|../dt[$n])=1)]

 

Note: Your use of self fails because you're not properly using it as an axis (self::). Also, self always contains just the context node, so it would refer to each dd inspected by the expression, not back to the original dt

相关问答

更多