X：确定一个窗口是否对用户可见，即未被其他人覆盖

Question

xdotool如果窗口对用户不可见，我只想执行某个操作，这包括最小化的窗口，以及其他窗口 100% 覆盖的窗口（如果至少没有使用透明度）。忽略透明度问题是否有一种简单的方法可以做到这一点？

xdotool有一个--onlyvisible选项，但只包括最小化的窗口，不包括覆盖的窗口。当然，可以选择循环浏览所有可见的窗口，获取它们的窗口几何形状并计算它们覆盖了多少感兴趣的窗口，但我确实希望有比在 bash 中执行此操作更简单、更快的解决方案。

这是问题的一个很好的说明，但它只是关于列出窗口，它也适用于 Max OS X。这个问题只有一个提示的答案，但没有显示如何通过列出所有可见窗口及其各自的z-order 并手动计算可见区域。

Answer 1

为了完整起见，这里是天真/蛮力解决方案，我希望它已经在其他一些实用程序中实现。该fullyobscured通知事件在评论声音@Gilles环节非常有前途的，但我不知道如何得到它的工作，这个解决方案也很有趣实现。

该脚本只是计算所有重叠窗口的覆盖区域减去重复计算的区域并检查它是否与窗口区域一样大。因为它正确地包含了框架边框，所以代码看起来比它本来的要复杂一些。如果完全覆盖，则返回退出代码 0，否则返回 1。它需要一个窗口 ID 作为参数。例如调用它if xcovered 0x1a00003; then echo 'covered!'; fi

不算注释、调试注释和错误检查，它可能只有 40 行长，甚至更少。我实际上想使用 bc 而不是 python，但我找不到一种简单的方法将 bash 数组传输到 bc 数组。

#!/bin/bash
# Name: xcovered.sh
# Find out if C is completely or only partially covered by A or B
#  +-----------+
#  |   +===+   |
#  |   | +-------+
#  | C | |   B   |
#  |   | +-------+
#  +---| A |---+
#      +---+
# @return 0 if window ist not visible, 1 if visible
# Note: Only tested with three windows like in sketch above, but
#       it should also work for an arbitrary amount of overlapping windwows
wid=$1
if ! xwininfo -id $wid -stats | 'grep' -q 'IsViewable'; then return 0; fi

# get all stacked window ids after and including the given wid
wids=($(xprop -root | 'sed' -nE "/_NET_CLIENT_LIST_STACKING\(WINDOW\)/{ s|.*($wid)|\1|; s|,||g; p }"))
if [ ${#wids} -eq 0 ]; then
    echo -e "\e[31mCouldn't find specified window id $wid in _NET_CLIENT_LIST_STACKING(WINDOW)"'!'"\e[0m"
    return 2
fi
if [ ${#wids} -eq 1 ]; then return 0; fi

# Gather geometry of all windows in higher zorder / possibly lying on top
coords=(); frames=()
for owid in ${wids[@]}; do
    #xwininfo -id $owid | grep xwininfo
    if xwininfo -id $owid -stats | 'grep' -q 'IsViewable'; then
        # _NET_WM_ICON_GEOMETRY doesn't exist for xfce4-panel, thereby making this more difficult
        #coords=$(xprop -id $owid _NET_WM_ICON_GEOMETRY)
        #frames=$(xprop -id $owid _NET_FRAME_EXTENTS)
        x=($(xwininfo -id $owid -stats -wm | sed -nE '
            s|^[ \t]*Absolute upper-left X:[ \t]*([0-9]+).*|\1|Ip;
            s|^[ \t]*Absolute upper-left Y:[ \t]*([0-9]+).*|\1|Ip;
            s|^[ \t]*Width:[ \t]*([0-9]+).*|\1|Ip;
            s|^[ \t]*Height:[ \t]*([0-9]+).*|\1|Ip;
            /Frame extents:/I{ s|^[ \t}Frame Extents:[ \t]*||I; s|,||g; p; };
        ' | sed ':a; N; $!b a; s/\n/ /g '))
        if [ ! ${#x[@]} -eq 8 ]; then
            echo -e "\e[31mSomething went wrong when parsing the output of 'xwininfo -id $owid -stats -wm':\e[0m"
            xwininfo -id $owid -stats -wm
            exit 1
        fi
        # apply the frame width to the coordinates and window width
        # 0:x 1:y 2:w 3:h, border widths 4:left 5:right 6:top 7:bottom
        coords+=( "${x[0]}-${x[4]}, ${x[1]}-${x[6]}, ${x[2]}+${x[4]}+${x[5]}, ${x[3]}+${x[6]}+${x[7]}" )
    fi
done

IFS=','; python - <<EOF #| python
# Calculates the area of the union of all overlapping areas. If that area
# is equal to the window of interest area / size, then the window is covered.
# Note that the calcualted area can't be larger than that!
#   1
# D---C      => overlap given by H and B
# | H-|---G    x-overlap: max(0, xleft2-xright1)
# A---B   |         -> '1' and '2' is not known, that's why for left and right
#   |  2  |            use min, each
#   E-----F         -> max(0, min(xright1,xright2) - max(xleft1,xleft2) )
#                      Note that because of xleft<xright this can only
#                      result in xright1-xleft2 or xright2-xleft1
# All cases: 1 |     +--+ |   +--+ | +--+   | +--+      |
#            2 | +--+     | +--+   |   +--+ |      +--+ |
#      overlap |    0     |    2   |    2   |     0     |
def overlap( x1,y1,w1,h1, x2,y2,w2,h2, x3=0,y3=0,w3=65535,h3=65535 ):
    return max( 0, min(x1+w1,x2+w2,x3+w3) - max(x1,x2,x3) ) * \
           max( 0, min(y1+h1,y2+h2,y3+h3) - max(y1,y2,y3) )
x=[ ${coords[*]} ]
area=0
# Calculate overlap with window in question
# 0:x 1:y 2:w 3:h, border widths 0:left 1:right 2:top 3:bottom
for i in range( 4,len(x),4 ):
    area += overlap( *( x[0:4]+x[i:i+4] ) )

# subtract double counted areas i.e. areas overlapping to the window
# of interest and two other windows on top ... This is n**2
for i in range( 4,len(x),4 ):
    for j in range( i+4,len(x),4 ):
        area -= overlap( *( x[0:4]+x[i:i+4]+x[j:j+4] ) )

print "area =",area
print "woi  =",x[2]*x[3]
# exit code 0: if not fully covered, 1: if fully covered
exit( area < x[2]*x[3] )
EOF
exit $?