英文:
How to close (pop) all the similar routes in flutter application?
问题
在我的应用程序中，我有7个以上的屏幕。我正在使用命名路由。
screen1是默认的第一页。
当从screen4点击返回按钮时，我需要关闭screen4并转到screen3，此时路由树中有多个screen4（连续如1 2 3 4 4 4 4 4）。我希望一次关闭所有的screen4。
我尝试了下面的代码，但是上下文始终返回'screen4'，所以它只关闭到screen2。
while (ModalRoute.of(context)!.settings.name == 'screen4' && Navigator.of(context).canPop()) {
if (mounted) {
Navigator.of(context).pop(true);
}
}
我需要将true传递给screen3，所以我不能在这里使用popuntil。
英文:
In my app I have 7+ screens. I am using named routes.
screen1 is default first page..
I have to close screen4 when tapping back button from screen4 and need to go screen3, in this case there are multiple screen4 there in route tree(back to back like 1 2 3 4 4 4 4 4 ). I want to close all the screen4 in one tap.
I tried below code but the context is always giving 'screen4' so it closing till screen2
while (ModalRoute.of(context)!.settings.name ==
'screen4' && Navigator.of(context).canPop()) {
if (mounted) {
Navigator.of(context).pop(true);
}
}
I need to pass true to screen3 so I can't use popuntil here
答案1
得分: 0
popuntil仍然可以适用于您提到的情况：
Navigator.of(context)
.popUntil((Route route) => route.settings.name != 'screen4');
这将从
1 -> 2 -> 3 -> 4 -> 4 -> 4
变为
1 -> 2 -> 3
如果您有进一步的条件，请在popuntil内应用它们。
英文:
popuntil can still work for the scenario you mention:
Navigator.of(context)
.popUntil((Route route) => route.settings.name != 'screen4');
This will take you from
1 -> 2 -> 3 -> 4 -> 4 -> 4
to
1 -> 2 -> 3
if you have further conditions, apply them within popuntil
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