Layui

How can I get the parent of a view using uiautomator?

匿名 (未验证) 提交于 2018-03-20 02:02:49

问题:

I'm trying to identify the parent view of an ui element so I can navigate through the UI freely.

For example, in Settings app, I can find the view with the text "Bluetooth":

UiObject btView = new UiObject(new UiSelector().text("Bluetooth"));

Now, the part where I get stuck is this one: I want to navigate two levels up and start a new search for the on/off button that enables and disables bluetooth.

Note: I can get the button if I use the code below.

UiObject btButtonView = new UiObject(new UiSelector().className("android.widget.Switch").instance(1));

This searches for switch buttons and returns the second encounter. I want the search to be more precise and look for the button in the linear layout that contains the "Bluetooth" text.

UPDATE: This is the layout of the Settings app (the Bluetooth part that I need):

LinearLayout
    LinearLayout
        ImageView
    RelativeLayout
        TextView (with text = "Bluetooth")
    Switch ()

回答1:

You need to find the UiObject two levels up first using the text. This can be done using the getChildByText() methods in UiCollection or UiScrollable. Then you can easily find the switch. For 'Settings' this code works on my device:

UiScrollable settingsList = new UiScrollable(new UiSelector().scrollable(true));
UiObject btItem = settingsList.getChildByText(new UiSelector().className(LinearLayout.class.getName()),"Bluetooth", true);

UiObject btSwitch = btItem.getChild(new UiSelector().className(android.widget.Switch.class.getName()));
btSwitch.click();


回答2:

Below code works for me.

//Getting the scrollable view

UiScrollable settingsList = new UiScrollable(new UiSelector().scrollable(true));

for (int i=0; i<=settingsList.getChildCount(new UiSelector ().className(LinearLayout.class.getName())); i++) {
//Looping through each linear layout view
UiObject linearLayout = settingsList.getChild(new UiSelector().className(LinearLayout.class.getName()).instance(i));

//Checking if linear layout have the text. If yes, get the switch, click and break out of the loop.
if (linearLayout.getChild(new UiSelector ().text("Bluetooth")).exists()) {
    UiObject btSwitch = linearLayout.getChild(new UiSelector().className(android.widget.Switch.class.getName()));
    btSwitch.click ();
    break;
    }
}


回答3:

If you want to just search for ON/OFF slider -> You can directly search for bluetooth OFF/ON button and click on it to disable/enable bluetooth -

You can check screenshot of the bluetooth page(using command - uiautomatorviewer) in command prompt and see that OFF button will have text in the OFF/ON slider. Then simply use -

   new UiObject(new UiSelector().text("OFF")).click();


回答4:

recently found that we could use getFromParent (for UiObject) and fromParent (for UiSelector) to select an uncle of object for example. If we have such layout:

`LinearLayout
    relative layout
        text View
    relative layout
        check box`

we could get checkbox from textview with this code:

TextViewTitle().getFromParent(new UiSelector()
            .fromParent(new UiSelector()
                    .resourceId("android:id/checkbox")));

where TextViewTitle is a Uiobject with text view