【发布时间】:2022-01-31 17:31:00
【问题描述】:
user.component.html
<form #ff="ngForm">
<input type="text" name="fname" placeholder="User Name" [(ngModel)] = "selectedUser.fname"/><br/><br/>
<input type="number" name="age" placeholder="age" [(ngModel)] = "selectedUser.age"/><br/><br/>
<input type="radio" name="sex" value="male" [(ngModel)] = "selectedUser.sex"/> Male<br/><br/>
<input type="radio" name="sex" value="female" [(ngModel)] = "selectedUser.sex"/> Female<br/><br/>
<input type="button" name="submit" value="submit" (click)="createUserData(ff.value)">
</form>
user.component.ts
import { Component, OnInit } from '@angular/core';
import { UserService } from '../../service/user.service';
@Component({
selector: 'app-user',
templateUrl: './user.component.html',
styleUrls: ['./user.component.css']
})
export class UserComponent implements OnInit {
constructor(private us:UserService) { }
user: any;
selectedUser: any = {fname:null, age: null, sex: null};
ngOnInit(): void {
}
createUserData(ff){
this.us.createform(ff).subscribe((user:any)=>{
console.log("Success register");
});
}
}
我是 Angular 的新手,我正在尝试使用 Web API 将表单数据插入数据库中的 mysqli,这工作正常。现在,我想在我的user.component.html 页面中显示成功或失败消息。我希望当我单击提交按钮时页面重新加载并显示success or failure 消息。那么,我该怎么做,请帮助我。
谢谢
【问题讨论】:
-
当您将表单数据提交给 api 时,api 会将结果返回为 success 或 failure 。并基于此,您可以分配模板