我想使用我在 firebase 中的链接数据打开一堆音乐应用程序链接。我想打开,amazonPrimeMusic、Ganna、Spotify、Wynk、JioSavaan 等等。
Widget buildResultCard(data) {
List items = [Text(data['Ganna']),
IconButton(icon:Icon(Icons.all_inclusive),
onPressed: ()=> {Text("Ganna")}
),
Text(data['Wynk']),
IconButton(icon:Icon(Icons.all_inclusive),
onPressed: ()=> {Text("Ganna")}
),
Text(data['JioSavaan']),
IconButton(icon:Icon(Icons.all_inclusive),
onPressed: ()=> {Text("Ganna")}
),
Text(data['PrimeMusic']),
IconButton(icon:Icon(Icons.all_inclusive),
onPressed: ()=> {Text("Ganna")}
)
];
return ListView.builder(
padding: EdgeInsets.only(top: 20),
itemCount: items.length,
itemBuilder: (BuildContext context, int index) {
return items[index];
},
);
}
当我点击列表中的按钮时,它应该打开链接所在的特定应用程序,例如 AmazonPrimeMusic 链接,它应该打开亚马逊音乐应用程序。