Question

您好，我有以下架构：

SELECT new GroupByType(tb2.typeName, tb1.aName, coalesce(COUNT(tb2.typeId)),0)
FROM Tbl_1 tb1
INNER JOIN Tbl_3 tb3 ON t1.aId=t3.aId
LEFT JOIN Tbl_2 tb2 on tb2.typeId=tb1.typeId
WHERE tb3.pId=1
GROUP BY b2.typeName, GroupByType.class).getResultList();

所有收入组均相等，但我必须重复此过程。有没有更有效的方式做到这一点？

Answer 1

function likes() {
$postid = $_GET["videoid"]; 


    $output = '<i  if (userLiked($postid)): class="fa fa-thumbs-up like-btn"  else:  class="fa fa-thumbs-o-up like-btn"  endif  data-id=" echo $postid "></i>
    <span class="likes"> echo getLikes($postid); </span>

    &nbsp;&nbsp;&nbsp;&nbsp;

    <i  if (userDisliked($postid)):  class="fa fa-thumbs-down dislike-btn"  else:  class="fa fa-thumbs-o-down dislike-btn"  endif  data-id=" echo $postid "></i>
    <span class="dislikes"> echo getDislikes($postid); </span>';
    return $output;

}

Answer 2

project.js

var ProjectSchema = new Schema({
    companyName : {type: String , required : true}, 
    projectName : String, 
    revenueGroups: [{type: Schema.Types.ObjectId, ref: 'Revenue'}],
});

module.exports = mongoose.model('Project', ProjectSchema);

revenue.js

var RevenueSchema = new Schema({
    campMan : Number,
    startersPakket : Number, 
    marketingCon : Number, 
});

module.exports = mongoose.model('Revenue', RevenueSchema);

猫鼬模式效率不高

2 个答案: