有一个按钮可以打开侧面菜单。我想根据是否打开的菜单将图标更改为按钮。也就是说,如果窗口处于活动状态,则按钮pi pi-lock
上的图标否则为pi pi-lock-open
。怎么做?
html:
...
<p-button (click)="_toggleOpened()" icon="pi pi-lock"></p-button>
...
ts:
private _opened: boolean = true;
private _toggleOpened(): void {
this._opened = !this._opened;
}
答案 0 :(得分:0)
将icon属性设置为属性绑定,并将.pi pi-lock分配给.ts中的变量,
将您的条件作为您的要求
<p-button (click)="_toggleOpened()" [icon]="pi pi-lock"></p-button>
ts file:
variable="pi pi-lock"
private _opened: boolean = true;
private _toggleOpened(): void {
this._opened = !this._opened;
if(!this.opened)
{
this.variable="pi pi-lock-open";
}
else{
this.variable="pi pi-lock"
}
}
答案 1 :(得分:0)
尝试一下:
HTML:
<p-button (click)="_toggleOpened()" [icon]="icon_val"></p-button>
TS:
icon_val: string="pi pi-lock"
private _opened: boolean = false;
private _toggleOpened(): void {
this._opened=!this.opened
if (this._opened)
this.icon_val="pi pi-lock-open"
else
this.icon_val="pi pi-lock"
}