角Node.js的MySQL的喜欢不喜欢按钮

我创建一个YouTube的风格（比如，不喜欢）按钮我的Twitter类似的网站。但我不得不与实施样不喜欢按钮，我使用Node.js的角度MYSQL，与NgFor环和ngIf条件的问题，我有两个表的帖子{POST_ID，USER_ID，说明}和{喜欢like_id，USER_ID，POST_ID，like_status }与普通按键user_id说明，POST_ID。

如果我们没有在喜欢表，我们怎么能写状态特定用户的任何值，显示基本按钮

<div class="container" *ngFor="let post of posts; let i = index ">
   <h6> {{post.description}} </h6>
   <div class="row">          
     <div class="col-md-3">
       <div *ngFor="let like of likes; ">

         <div *ngIf="((post.user_id==like.user_id) && (post.post_id==like.post_id) && (like.like_status=='like'))">
                <button type="button" class="btn btn-success" (click)=likeSubmit(post.user_id,post.post_id)><i class="fa fa-thumbs-up"
                    aria-hidden="true"></i></button>
              </div>

         <div *ngIf="((post.user_id==like.user_id) && (post.post_id==like.post_id) && (like.like_status=='unlike'))">
                <button type="button" class="btn btn-warning" (click)=likeSubmit(post.user_id,post.post_id)><i class="fa fa-thumbs-up"
                    aria-hidden="true"></i></button>
              </div>

<!---problem with ngIf condition -->
       <div *ngIf="ngIf condition">
                <button type="button" class="btn btn-basic" (click)=likeSubmit(post.user_id,post.post_id)><i
                    class="fa fa-thumbs-up" aria-hidden="true"></i></button>
              </div>

前两个ngIf条件下工作良好，但我不能把完美ngIf contion的最后一个，如果按钮like_status ==一样，与like_status ==不像工作好，如果我们没有像表，我们怎么能写任何值ngIf条件o显示按钮

你正在做的一点点错误，你不需要在你的模板进行内部循环，可以执行只在您的组件的逻辑连接两个表，事后只呈现所需的数据。尝试添加方法到组件联接两个表，然后你可以很容易地做你的模板想要的东西。例：

TS文件

...
    public getPostsWithLikes() {
        this.posts.map(post => {
            const like = this.likes && this.likes.find(plike => post.user_id === plike.user_id && post.post_id === plike.post_id);
            return {post, like};
        });
    }
...

HTML文件

<div class="container" *ngFor="let item of getPostsWithLikes(); let i = index ">
   <h6> {{item.post.description}} </h6>
   <div class="row">          
     <div class="col-md-3">
         <div *ngIf="item.like && item.like.like_status=='like'">
                <button type="button" class="btn btn-success" (click)=likeSubmit(item.post.user_id,item.post.post_id)><i class="fa fa-thumbs-up"
                    aria-hidden="true"></i></button>
              </div>

         <div *ngIf="item.like && item.like.like_status=='unlike'">
                <button type="button" class="btn btn-warning" (click)=likeSubmit(item.post.user_id,item.post.post_id)><i class="fa fa-thumbs-up"
                    aria-hidden="true"></i></button>
              </div>

<!---if you don't have related record in likes table, like would be null (false), so you just need to invert it-->
       <div *ngIf="!item.like">
                <button type="button" class="btn btn-basic" (click)=likeSubmit(item.post.user_id,item.post.post_id)><i
                    class="fa fa-thumbs-up" aria-hidden="true"></i></button>
              </div>

希望帮助。

角Node.js的MySQL的喜欢不喜欢按钮

我创建一个YouTube的风格（比如，不喜欢）按钮我的Twitter类似的网站。但我不得不与实施样不喜欢按钮，我使用Node.js的角度MYSQL，与NgFor环和ngIf条件的问题，我有两个表的帖子{POST_ID，USER_ID，说明}和{喜欢like_id，USER_ID，POST_ID，like_status }与普通按键user_id说明，POST_ID。

如果我们没有在喜欢表，我们怎么能写状态特定用户的任何值，显示基本按钮

<div class="container" *ngFor="let post of posts; let i = index ">
   <h6> {{post.description}} </h6>
   <div class="row">          
     <div class="col-md-3">
       <div *ngFor="let like of likes; ">

         <div *ngIf="((post.user_id==like.user_id) && (post.post_id==like.post_id) && (like.like_status=='like'))">
                <button type="button" class="btn btn-success" (click)=likeSubmit(post.user_id,post.post_id)><i class="fa fa-thumbs-up"
                    aria-hidden="true"></i></button>
              </div>

         <div *ngIf="((post.user_id==like.user_id) && (post.post_id==like.post_id) && (like.like_status=='unlike'))">
                <button type="button" class="btn btn-warning" (click)=likeSubmit(post.user_id,post.post_id)><i class="fa fa-thumbs-up"
                    aria-hidden="true"></i></button>
              </div>

<!---problem with ngIf condition -->
       <div *ngIf="ngIf condition">
                <button type="button" class="btn btn-basic" (click)=likeSubmit(post.user_id,post.post_id)><i
                    class="fa fa-thumbs-up" aria-hidden="true"></i></button>
              </div>

前两个ngIf条件下工作良好，但我不能把完美ngIf contion的最后一个，如果按钮like_status ==一样，与like_status ==不像工作好，如果我们没有像表，我们怎么能写任何值ngIf条件o显示按钮

你正在做的一点点错误，你不需要在你的模板进行内部循环，可以执行只在您的组件的逻辑连接两个表，事后只呈现所需的数据。尝试添加方法到组件联接两个表，然后你可以很容易地做你的模板想要的东西。例：

TS文件

...
    public getPostsWithLikes() {
        this.posts.map(post => {
            const like = this.likes && this.likes.find(plike => post.user_id === plike.user_id && post.post_id === plike.post_id);
            return {post, like};
        });
    }
...

HTML文件

<div class="container" *ngFor="let item of getPostsWithLikes(); let i = index ">
   <h6> {{item.post.description}} </h6>
   <div class="row">          
     <div class="col-md-3">
         <div *ngIf="item.like && item.like.like_status=='like'">
                <button type="button" class="btn btn-success" (click)=likeSubmit(item.post.user_id,item.post.post_id)><i class="fa fa-thumbs-up"
                    aria-hidden="true"></i></button>
              </div>

         <div *ngIf="item.like && item.like.like_status=='unlike'">
                <button type="button" class="btn btn-warning" (click)=likeSubmit(item.post.user_id,item.post.post_id)><i class="fa fa-thumbs-up"
                    aria-hidden="true"></i></button>
              </div>

<!---if you don't have related record in likes table, like would be null (false), so you just need to invert it-->
       <div *ngIf="!item.like">
                <button type="button" class="btn btn-basic" (click)=likeSubmit(item.post.user_id,item.post.post_id)><i
                    class="fa fa-thumbs-up" aria-hidden="true"></i></button>
              </div>

希望帮助。