entityMap|0|type|LINK|mutability|MUTABLE|data|url|http://ionicframework.com/docs/v2/api/navigation/NavController/|blocks|key|c44i9|text|您可以通过使用insertPage方法修改导航控制器堆栈来完成此操作。|unstyled|depth|inlineStyleRanges|entityRanges|floc2|+this.nav.insertPages(0,+[{page:+your_home_page},+{page:+SearchPage}]);
+this.nav.pop();|code-block|syntax|javascript|digqa|这里0指的是您想要插入页面的位置。在上面的代码中，0将是您的主页，1将是您的搜索页。在DetailPage中使用此命令可弹出导航到SearchPage。|cm6cl|有关更多信息，请查看http://ionicframework.com/docs/v2/api/navigation/NavController/|offset|length|5f2e8^0|0|0|0|A|1R|0|0^^$0|$1|$2|3|4|5|6|$7|8]]]|9|@$A|B|C|D|2|E|F|U|G|@]|H|@]|6|$]]|$A|I|C|J|2|K|F|V|G|@]|H|@]|6|$L|M]]|$A|N|C|O|2|E|F|W|G|@]|H|@]|6|$]]|$A|P|C|Q|2|E|F|X|G|@]|H|@$R|Y|S|Z|A|10]]|6|$]]|$A|T|C|-4|2|E|F|11|G|@]|H|@]|6|$]]]]

You can do this by modifying the navigation controller stack using insertPage method.

<pre><code> this.nav.insertPages(0, [{page: your_home_page}, {page: SearchPage}]);
 this.nav.pop();
</code></pre>

Here 0 refers to the position where you want to insert your page. In above code, 0 will be your home page and 1 will be your search page. Use this command in your DetailPage to navigation pop to SearchPage.

For more information check <a href="http://ionicframework.com/docs/v2/api/navigation/NavController/" rel="nofollow noreferrer">http://ionicframework.com/docs/v2/api/navigation/NavController/</a>

entityMap|blocks|key|9jh9m|text|好吧，我找到了一个解决方案。看起来很管用..。至少目前是这样|type|unstyled|depth|inlineStyleRanges|entityRanges|data|cg9i3|this.navCtrl
++++++++.push(SearchPage)
++++++++.then(()+=>+{

++++++++++++const+index+=+this.viewCtrl.index;

++++++++++++for(let+i+=+index;+i+>+0;+i--){
++++++++++++++++this.navCtrl.remove(i);
++++++++++++}

++++++++});|code-block|syntax|javascript|6pma8^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|J|9|@]|A|@]|B|$F|G]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

ok, found a solution. It looks like it works ... at least for the moment

<pre><code>this.navCtrl
 .push(SearchPage)
 .then(() =&gt; {

 const index = this.viewCtrl.index;

 for(let i = index; i &gt; 0; i--){
 this.navCtrl.remove(i);
 }

 });
</code></pre>

entityMap|blocks|key|8oo82|text|试试这个！进入DetailPage后，执行以下操作：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|8n652|this.navCtrl.remove(2,1);+//+This+will+remove+the+'ResultPage'+from+stack.
this.navCtrl.pop();++//+This+will+pop+the+'DetailPage',+and+take+you+to+'SearchPage'|code-block|syntax|javascript|fnpdg^0|7|A|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|M|9|@$A|N|B|O|C|D]]|E|@]|F|$]]|$2|G|4|H|6|I|8|P|9|@]|E|@]|F|$J|K]]|$2|L|4|-4|6|7|8|Q|9|@]|E|@]|F|$]]]]

Try this! Once you are in the <code>DetailPage</code>, do: 

<pre><code>this.navCtrl.remove(2,1); // This will remove the 'ResultPage' from stack.
this.navCtrl.pop(); // This will pop the 'DetailPage', and take you to 'SearchPage'
</code></pre>

entityMap|blocks|key|bnj5|text|+this.nav.push(AnotherPage)
+.then(()+=>+{
++++const+index+=+this.viewCtrl.index;
++++this.nav.remove(index);
+});|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|cidgs|unstyled^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|G|9|@]|A|@]|B|$C|D]]|$2|E|4|-4|6|F|8|H|9|@]|A|@]|B|$]]]]

<pre><code> this.nav.push(AnotherPage)
 .then(() =&gt; {
 const index = this.viewCtrl.index;
 this.nav.remove(index);
 });
</code></pre>

entityMap|blocks|key|av62k|text|this.navController.push(SearchPage).then(()+=>+{
++const+index+=+this.viewCtrl.index;
++this.navController.remove(index,+1);+//this+will+remove+page3+and+page2
});|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|b4ekf|将此代码包含在您的组件中。|unstyled|5fe72^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@]|A|@]|B|$C|D]]|$2|E|4|F|6|G|8|J|9|@]|A|@]|B|$]]|$2|H|4|-4|6|G|8|K|9|@]|A|@]|B|$]]]]

<pre><code>this.navController.push(SearchPage).then(() =&gt; {
 const index = this.viewCtrl.index;
 this.navController.remove(index, 1); //this will remove page3 and page2
});
</code></pre>

Include this code in your component.

entityMap|blocks|key|644dm|text|我也遇到了同样的问题，并找到了解决方案，它类似于公认的答案，但没有循环和更多的细节，所以我希望这能帮助别人理解它。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|gg00|所以流程是这样的：|2rq6v|Home+->+navCtrl.push(SearchPage)+->+navCtrl.push(ResultPage)+->+navCtrl.push(DetailPage)|code-block|syntax|javascript|43bap|下面是各自的索引：|dtius|Home:+0
SearchPage:+1
ResultPage:+2+
DetailPage:+3|53mke|如果您想要从DetailPage返回到SearchPage，您基本上希望在推送ResultPage时将DetailPage从堆栈中删除。|7n9mn|因此，当您推送DetailPage时，请在ResultPage上尝试以下代码：|7kiir|//+First+get+the+index+of+ResultPage,+it+should+return+2.
let+resultIndex+=+this.navCtrl.last().index;+

//+Push+DetailPage.
this.navCtrl.push(DetailPage).then(()+=>+{
++//+Once+it's+pushed,+this+block+is+called+and+now+we+can+remove+the+ResultPage+from+the+stack.
++this.navCtrl.remove(resultIndex,+1);+
++//+second+parameter+is+optional+and+defaults+to+1,+if+you+want+to+remove+more+pages+from+stack+start+with+the+bottom+most+index+you+want+to+remove+and+pass+number+of+pages+you+want+to+remove+starting+from+the+given+index.+

});|7gb3n|希望这对某个地方的人有所帮助。|28ba9^0|0|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|W|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|X|9|@]|A|@]|B|$]]|$2|E|4|F|6|G|8|Y|9|@]|A|@]|B|$H|I]]|$2|J|4|K|6|7|8|Z|9|@]|A|@]|B|$]]|$2|L|4|M|6|G|8|10|9|@]|A|@]|B|$H|I]]|$2|N|4|O|6|7|8|11|9|@]|A|@]|B|$]]|$2|P|4|Q|6|7|8|12|9|@]|A|@]|B|$]]|$2|R|4|S|6|G|8|13|9|@]|A|@]|B|$H|I]]|$2|T|4|U|6|7|8|14|9|@]|A|@]|B|$]]|$2|V|4|-4|6|7|8|15|9|@]|A|@]|B|$]]]]

I also came across to same problem and found the solution, it is similar to the accepted answer but without loop and bit more details so I hope this helps someone to understand it.

So here's the flow:

<pre><code>Home -&gt; navCtrl.push(SearchPage) -&gt; navCtrl.push(ResultPage) -&gt; navCtrl.push(DetailPage)
</code></pre>

And here are the respective index:

<pre><code>Home: 0
SearchPage: 1
ResultPage: 2 
DetailPage: 3
</code></pre>

If you want to go back to SearchPage from the DetailPage, you basically want to remove ResultPage from the stack when you push DetailPage. 

So try this code on the ResultPage when you push DetailPage:

<pre><code>// First get the index of ResultPage, it should return 2.
let resultIndex = this.navCtrl.last().index; 

// Push DetailPage.
this.navCtrl.push(DetailPage).then(() =&gt; {
 // Once it's pushed, this block is called and now we can remove the ResultPage from the stack.
 this.navCtrl.remove(resultIndex, 1); 
 // second parameter is optional and defaults to 1, if you want to remove more pages from stack start with the bottom most index you want to remove and pass number of pages you want to remove starting from the given index. 

});
</code></pre>

Hope this helps someone somewhere.

entityMap|blocks|key|68ik2|text|我在这里回答了一个类似的问题，似乎也适用于你的问题。因为我延迟加载模块，所以popTo()似乎只有在我从NavController.getViews()数组传递对象时才起作用。这样说来，您可以尝试如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|99kl8|let+targetView+=+this._navCtrl.getViews().filter(view=>+view.id+==+'MyAwesomePage')
targetView.length+?+this._navCtrl.popTo(targetView[0])+:+this._navCtrl.pop()|code-block|syntax|javascript|7s35l|或者像这样更冗长的内容：|7ite0|let+index:+number;
let+views:+any[]+=+this._navCtrl.getViews()
let+found:+boolean+=+views.some((view,+i)+=>{
++index+=+i
++return+(view.id+==+'MyAwesomePage')
})
found+?+this._navCtrl.popTo(views[index])+:+this._navCtrl.pop()|2vt77|如果要更改顺序，可以使用getViews()、.reverse()、.filter()或views.reverse().some()。这是离子3和ES5的Array.some()|crkk^0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|O|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|P|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|Q|9|@]|A|@]|B|$]]|$2|J|4|K|6|E|8|R|9|@]|A|@]|B|$F|G]]|$2|L|4|M|6|7|8|S|9|@]|A|@]|B|$]]|$2|N|4|-4|6|7|8|T|9|@]|A|@]|B|$]]]]

I answered a similar question on here and it seems to apply for yours too. Since I'm lazy loading the modules, popTo() only seems to work when I pass an object from the NavController.getViews() array.
With that said, you can try something like this:

<pre><code>let targetView = this._navCtrl.getViews().filter(view=&gt; view.id == 'MyAwesomePage')
targetView.length ? this._navCtrl.popTo(targetView[0]) : this._navCtrl.pop()
</code></pre>

Or something more verbose like this:

<pre><code>let index: number;
let views: any[] = this._navCtrl.getViews()
let found: boolean = views.some((view, i) =&gt;{
 index = i
 return (view.id == 'MyAwesomePage')
})
found ? this._navCtrl.popTo(views[index]) : this._navCtrl.pop()
</code></pre>

If you want to change the order, you may getViews().reverse().filter() or views.reverse().some(). This is with Ionic 3 and ES5's Array.some()

entityMap|blocks|key|amnb2|text|在我的例子中，它是这样工作的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|8bl7j|import+{+NavController,+NavParams+}+from+'ionic-angular';

export+class+MyComponent+{
++++...
++++constructor(private+navParams:NavParams,+private+navCtrl:NavController){
++++++++//If+I+get+the+paramater,+I+know+I+have+to+"skip"+last+view
++++++++let+condition+=+this.navParams.get("condition");
++++++++if+(condition){
++++++++++++this.navCtrl.removeView(this.navCtrl.last());
++++++++}
++++}
++++...
}|code-block|syntax|javascript|eh973|因此，无论您是使用软/硬按钮返回，还是手动弹出()，上一个视图都将被跳过。|fugmf^0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|K|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|L|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|M|9|@]|A|@]|B|$]]|$2|J|4|-4|6|7|8|N|9|@]|A|@]|B|$]]]]

In my case it worked liked this

<pre><code>import { NavController, NavParams } from 'ionic-angular';

export class MyComponent {
 ...
 constructor(private navParams:NavParams, private navCtrl:NavController){
 //If I get the paramater, I know I have to "skip" last view
 let condition = this.navParams.get("condition");
 if (condition){
 this.navCtrl.removeView(this.navCtrl.last());
 }
 }
 ...
}
</code></pre>

So, whether you go back with the soft/hard button or just pop() manually, the previous view will be skipped.

entityMap|blocks|key|angpf|text|最好的方法(因为避免页面之间的转换，直接转到倒数第二个页面)是以下解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5iavv|this.navCtrl.remove(nav.getActive().index+-+1,+2)|code-block|syntax|javascript|1p916|希望这对你有用|aucuc^0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|K|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|L|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|M|9|@]|A|@]|B|$]]|$2|J|4|-4|6|7|8|N|9|@]|A|@]|B|$]]]]

The best way (because avoid transitions between pages and go directly to the second last page) is the follow solution:

<pre><code>this.navCtrl.remove(nav.getActive().index - 1, 2)
</code></pre>

Hope that work for you

I have the following navigation-case:

<pre><code>Home -&gt; navCtrl.push(SearchPage) -&gt; navCtrl.push(ResultPage)
</code></pre>

or 

<pre><code>Home -&gt; navCtrl.push(SearchPage) -&gt; navCtrl.push(ResultPage) -&gt; navCtrl.push(DetailPage)
</code></pre>

I want to navigate back to SearchPage. In first case, there is no problem, I can use 

<pre><code>this.navCtrl.pop()
</code></pre>

But, in second case, I try to use 

<pre><code>this.navCtrl.popTo(SearchPage)
</code></pre>

and this does not work as expected. Ionic navigates only one page back in stack. I know there is an issue with popTo() (<a href="https://github.com/driftyco/ionic/issues/6513" rel="nofollow noreferrer">https://github.com/driftyco/ionic/issues/6513</a>)

How can I solve this problem?

Ionic 2 Angular NavController, pop back to second last page

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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 我有以下导航案例： Home -> navCtrl.push(SearchPage) -> navCtrl.push(ResultPage) 或者 Home -> navCtrl.push(SearchPage) -> navCtrl.push(ResultPage) -> navCtrl.push(DetailPa...

问Ionic 2 Angular NavController，弹回倒数第二页
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