Question

嗨，这是我的数据结构：

$(document).ready(function() {
  $('#noscript-description').css({
    'display' : 'none'
    
    // Just change the none to block to see the results
  });
});

我想输出：所有类别和每个类别都应显示：

图片数量
最多观看次数
最新图片的日期
最新文件的ID

期望的输出：

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<html>
  <body>
    <p id="noscript-description">This can be seen if you <b>DO NOT</b> have JavaScript enabled.</p>
    <!-- I have added in jQuery, if you never realised -->
  </body>
</html>

到目前为止我的查询：

categories
id   name
----------------
1    dogs    
2    cats
3    birds

images
id   name   cat  datetime     views
---------------------------------------
1    dog1   1    2015-10-01   123
2    dog2   1    2015-10-02   45
3    cat1   2    2015-10-03   678
4    cat2   2    2015-11-01   901
5    cat3   2    2015-11-02   234
6    bird1  3    2015-12-03   456

问题是：我如何获得最后一个字段cat name images views latest id_latest ---------------------------------------------------- 1 dogs 2 123 2015-10-02 5 2 cats 3 901 2015-11-02 4 3 birds 1 456 2015-12-03 6？

感谢

Answer 1

Select t.*, i.id as imageID 
  from 
    (Select 
        categories.*,
        Count(images.id) As images,
        Sum(images.views) As views,
        Max(images.datetime) As latest
      From categories 
      Left Join 
        images 
        On images.catid =  categories.id 
      Group By categories.id
    ) t
  left join 
    images i 
    on i.`datetime`= latest

example on sqlfiddle

Answer 2

你的桌子必须是这样的：

categories
catid   name
----------------
1    dogs    
2    cats
3    birds

images
imgid   name   catid  datetime     views
---------------------------------------
1    dog1   1    2015-10-01   123
2    dog2   1    2015-10-02   45
3    cat1   2    2015-10-03   678
4    cat2   2    2015-11-01   901
5    cat3   2    2015-11-02   234
6    bird1  3    2015-12-03   456

代码：

 Select
  categories.catid,
  categories.name,
  Count(images.imgid) As images,
  Sum(images.views) As views,
  Max(images.datetime) As latest

From
  categories 
  Left Join images On images.catid =  categories.catid 
Group By
  categories.catid, categories.name

mysql查询：查询嵌套（？）

2 个答案:

你的桌子必须是这样的：