有这个查询：
SELECT brand,
ROUND(AVG(rating), 0) AS ar,
model
FROM engines
WHERE rating IS NOT NULL
AND rating != ''
AND category='Road'
GROUP BY brand
ORDER BY atr DESC
问题是模型列。
如何使结果显示评分最高的模型？ （目前它只显示与过滤器匹配的第一个模型）。
即。如果数据是：
brand1, model1, 3
brand1, model2, 5
brand1, model3, 4
如何使结果如下：
brand1, 4, model2
（因为model2的评分最高，为5）
谢谢！
答案 0 :(得分：1)
在<script type="text/javascript">
function centerGalleryOverlay(){
var top = ($('.image-container').height() / 2).toFixed() ;
$('.image-container').css('marginTop', '-' + top + 'px');
var left = ($('.image-container').width() / 2).toFixed() ;
$('.image-container').css('marginLeft', '-' + left + 'px');
}
$(document).ready(function(){
var galleryTitle = null;
$('.cm-gallery').click(function(e){
e.preventDefault();
galleryTitle = $(this).data('gallery');
$('.gallery-overlay .image-container img').attr( 'src', $(this).attr('href') );
$('.gallery-overlay').fadeIn(500);
centerGalleryOverlay();
var galleryItems = null;
galleryItems = [];
$('.cm-gallery').each(function(){
if($(this).data('gallery') == galleryTitle){
galleryItems.push($(this).attr('href'))
}
});
var currentIndex = null;
currentIndex = 0;
if(galleryItems.length > 0){
$('.gallery-overlay .image-container .gallery-next').show();
}
$('.gallery-overlay .image-container .gallery-next').click(function(){
currentIndex = currentIndex + 1;
console.log(currentIndex);
if(currentIndex >= (galleryItems.length - 1) ){
$('.gallery-overlay .image-container .gallery-next').hide();
}
$('.gallery-overlay .image-container .gallery-prev').show();
$('.gallery-overlay .image-container img').attr( 'src', galleryItems[currentIndex] );
})
$('.gallery-overlay .image-container .gallery-prev').click(function(){
currentIndex = currentIndex - 1;
console.log(currentIndex);
if(currentIndex <= 0 ){
$('.gallery-overlay .image-container .gallery-prev').hide();
}
$('.gallery-overlay .image-container .gallery-next').show();
$('.gallery-overlay .image-container img').attr( 'src', galleryItems[currentIndex] );
})
})
$('.gallery-overlay .gallery-close').click(function(){
$('.gallery-overlay').hide();
galleryItems = null;
currentIndex = null;
galleryTitle = null;
});
})
</script>
中，您可以执行此操作。
sql server在SELECT brand,
Round(Avg(rating), 0) AS ar,
Model = (SELECT TOP 1 E.model
FROM engines E
ORDER BY rating DESC)
FROM engines
WHERE rating IS NOT NULL
AND rating != ''
AND category = 'Road'
GROUP BY brand
ORDER BY atr DESC
中将子查询更改为
mysql答案 1 :(得分：1)
标准SQL将是：
select e.*
from (select e.*, row_number() over (partition by brand order by rating desc) as seqnum
from engines e
) e
where seqnum = 1;
没有窗口功能的替代方案是：
select e.*
from engines e join
(select brand, max(rating) as maxr
from engines
group by brand
) ee
on e.brand = ee.brand and e.rating = ee.rating;