position

It looks like a bug in 5 Android(API 21). I need a textview on button, textview should placed above the button. It works correct on Android 4.1(API 16) and incorrect on 5 Android(API 21). There are Screenshots and code: Android 4.1 - It is correct, red textview above the button Android 5 - it is incorrect , red textview under the button! Code: <?xml version="1.0" encoding="utf-8"?> <RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android" android:id="@+id/rlBottom" android:layout_width="fill_parent" android:layout_height="match_parent" > <FrameLayout android:layout_width="wrap

fixed定位的元素，如果父级有transform样式，值不为none，那么fixed定位就会失效。 解决方法：使用transform样式的元素，不要包含fixed定位的子元素。 来源： https://www.cnblogs.com/xiaobaibubai/p/11791487.html

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I'm trying to move a object to the mouse position. But it's giving me large x value like 300 but at that place the pre placed object's x position is -4. rigidBody.velocity = new Vector3(Input.mousePosition.x, EndPointY, 0)*4; So how can I get the current mouse position? Thank you.. 回答1: That is the current mouse position. The issue is that your objects are in world coordinates and the mouse is using screen coordinates. You need to convert the mouse position using Camera.ScreenToWorldPoint() . 回答2: Input.mousePosition will give you the

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I want to lock the position of a node relative to its parent compound node, such that if I grab and drag the parent node, the child node moves with it, but the child is not individually grabbable. If I set the child to be ungrabbable and/or locked, then it doesn't move with its parent, but if I don't, it can be individually dragged, which I don't want. Can this be done? Failing that, is there a way to programmatically grab/ungrab a node so that I can listen for a grab event and then grab the parent instead/as well? 回答1: Compound grabbing

Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. Your algorithm's runtime complexity must be in the order of O (log n ). If the target is not found in the array, return [-1, -1] . Example 1: Input: nums = [5,7,7,8,8,10], target = 8 Output: [3,4] Example 2: Input: nums = [5,7,7,8,8,10], target = 6 Output: [-1,-1] Accepted 371,213 Submissions 1,076,399 class Solution { public: int bs(vector<int> &n,int t) { int l=0,r=n.size()-1,m; while(l<=r) { m=(l+r)/2; if(n[m]<t) l=m+1; else r=m-1; } return l; } vector<int> searchRange

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: In aggregation i am trying to project an object(which is an array) which is present at position 7 inside an array . For example i have fields as below { "$project" : { "result" : { "$cond" : { "if" : { "$eq" : [ "$coupon_type" , 1 ] }, "then" : [ "$_id" , "$coupon_type" , "$nonce" , "$total_coupons" , "$amount" , "$curr_ctr" ], "else" : [ "$_id" , "$coupon_type" , "$nonce" , "$total_coupons" , "$amount" , "$curr_ctr" , "$coupon_codes" , { indexes : conditions }] } } } }, { "$project" : { obj : { $arrayElemAt : [ "$result" , 7 ] } /

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: I have a form comprising a sequence of <label> , <input> pairs (one per line) as follows: <ol style = " list - style : none ; " > <li class = "normal" > <label> Channel Name </label> <input type = "text" > </li> <li class = "reversed" > <label> Subscribe to this channel </label> <input type = "checkbox" > </li> </ol> I'm looking for a pure-CSS way to modify the second line to display the <input> checkbox to the left of the <label> (i.e. exchange their order without modifying the HTML). The following simple rule works perfectly in

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: I'm using GridBagLayout to place my GUI components by the following code, wanting the components lay one by one in a column, without any gaps : import java . awt . GridBagConstraints ; import java . awt . GridBagLayout ; import javax . swing . JButton ; import javax . swing . JFrame ; import javax . swing . JPanel ; public class TestGUI extends JFrame { public TestGUI (){ JPanel bigPanel = new JPanel ( new GridBagLayout ()); JPanel panel_a = new JPanel (); JButton btnA = new JButton ( "button a" ); panel_a . add ( btnA ); JPanel

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I have used RecyclerView and StaggeredGridLayoutManager. When I scroll from top to the bottom, it looked well. But when I scroll from bottom to the top , the top child view appear a blank. Now, I found when I scroll from top to bottom, it looks : left1 right1 left2 right2 left3 right3 left4 right4 left5 right5 left6 right6 left7 right7 When I scroll from bottom to the top, something happened, it looks like this: right1 left1 right2 left2 right3 left3 right4 left4 left5 right5 left6 right6 left7 right7 and the height of left and right is