Item: i1-i2-i3-i4-i5 Shortcut: k1-k2-k3-k4-k5 ### After swapped (i2,i3): --- Item: i1-i3-i2-i4-i5 Shortcut: k1-k2-k3-k4-k5 (no change) Expected: k1-k3-k2-k4-k5 (if: k2 != default_shortcut_for_item_#2 || k3 != default_shortcut_for_item#3 -> swap the shortcuts) ### After removed i2: --- Item: i1-i3-i4-i5 Shortcut: k1-k2-k3-k4-k5 (no change) Expected: k1-k3-k4-k5 How to correct the shortcuts? --- Item: i1-j2-i3-i4-i5 (j2 is an imagined item at #2) Shortcut: k1-k2-k3-k4-k5 (k2 is the default shortcut for the item #2) -> continuously swap (j2,i3) -> (j2,i4) -> (j2,i5) (in mem) Item: i1-i3-i4-i5-j2 Shortcut: k1-k3-k4-k5-k6 # This could be not the fastest algorithm, but at least there is one way to solved the problem. :D
I thinks this one is better: --------------------------------- n 1 2 3 4 5 6 7 Item0: i1 i2 i3 i4 i5 i6 i7 Key0 : k1 k2 k3 k4 k5 k6 k7 ### Move i2 to n=5 Item1: i1 i3 i4 i5 i2 i6 i7 Key1 : k1 - - - - k6 k7 How to correct the shortcuts? --- for (i in [i3-i4-i5-i2]): if (i->key0 == default_key) then: i->key1 == default_key_at-new-position else: i->key1 == i->key0 ### Remove i2 = Move i2 to the end + delete i2 Item1: i1 i3 i4 i5 i6 i7 i2 Key1 : k1 - - - - - - -> Use the above function to correct the shortcuts -> restore the default shortcut at position n=7
sorry I dont like the idea: 1. it would add complexity for only a small portion of users 2. the "activate entry" 's of Latte is just that, are not assigned based on the applet but rather on the position so that would break the logic of the implementation 3. the users that want to fix this they can from their plasma systemsettings