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Javascript 角2中停止旋转数的解_Javascript_Angular_Spinner - Fatal编程技术网

Javascript 角2中停止旋转数的解

Javascript 角2中停止旋转数的解,javascript,angular,spinner,Javascript,Angular,Spinner,这里我添加+1 evrytime,它不会停止,当我达到某个数据计数时,它应该停止计数。如果您有硬编码的值,可以使用take()命令 例如: happyCustomers = 0; followers = 0; awardsWinning = 0; photosTaken = 0; arrayOfImages = []; constructor() { } ngOnInit() { Observable.interval(400).subscribe(x => { t

这里我添加+1 evrytime,它不会停止,当我达到某个数据计数时,它应该停止计数。

如果您有硬编码的值,可以使用take()命令

例如:

 happyCustomers = 0;
 followers = 0;
 awardsWinning = 0;
 photosTaken = 0;
 arrayOfImages = [];
 constructor() { }

 ngOnInit() {
   Observable.interval(400).subscribe(x => {
 this.happyCustomers = this.happyCustomers + 1;
  });
   Observable.interval(200).subscribe(x => {
  this.followers = this.followers + 1;
  });
   Observable.interval(700).subscribe(x => {
  this.awardsWinning = this.awardsWinning + 1;
  });
   Observable.interval(300).subscribe(x => {
   this.photosTaken = this.photosTaken + 1;
 });
}
发出500个事件后,上述代码将停止

请参见使用方法

take
方法从源中获取第一个
count
值,然后完成

订阅前先打电话

const certainNumber=10;
可观察。间隔(400)
.take(确定号码)
.subscribe(=>this.happyCustomers++);

take()的任何导入;你应该在订阅之前调用
take
方法。我这边有点打字错误。修正了答案。谢谢你指出这一点!
Observable.interval(400)
          .take(500)
          .subscribe(x => {
               this.happyCustomers = this.happyCustomers + 1;
          });