How to delete qualified data in the table using table.remove in tolua
- introduce
- Problem (wrong way to delete data)
- Correct deletion plan
-
- Delete from back to front
- Delete recursively
- Insert new table method
- Expand it
- Summarize
Introduction
Deleting qualified data in a table in Lua is actually very simple, but there is a sequence problem, because data deletion in Lua tables needs to be deleted through table.remove. When you delete the previous one, the index value changes.
Problem (wrong way to delete data)
--Test lua table
local tab1 = {<!-- -->
[1] = {<!-- -->
Id = 101,
value1 = 1,
value2 = 2,
value3 = 3,
},
[2] = {<!-- -->
Id = 105,
value1 = 1,
value2 = 2,
value3 = 3,
},
[3] = {<!-- -->
Id = 101,
value1 = 1,
value2 = 2,
value3 = 3,
},
[4] = {<!-- -->
Id = 108,
value1 = 1,
value2 = 2,
value3 = 3,
},
[5] = {<!-- -->
Id = 101,
value1 = 1,
value2 = 2,
value3 = 3,
},
}
\t
--Wrong way one
for k, v in pairs(tab1) do
if(v.Id == 101) then
table.remove(table,k)
end
end
--Same as the deletion method above (with a different writing method)
for i = 1, #tab1 do
if(tab1[i].Id == 101) then
table.remove(table,i)
end
end
--Wrong way two
local index = 1
for i = 1, #tab1 do
if(tab1[i].Id == 101) then
table.remove(table,index)
index = index - 1
end
index = index + 1
end
The above two methods are wrong, and the final printing is not the actual printing as imagined
Correct deletion plan
Delete from back to front
--Lua table of test data
local tab1 = {<!-- -->
[1] = {<!-- -->
Id = 101,
value1 = 1,
value2 = 2,
value3 = 3,
},
[2] = {<!-- -->
Id = 105,
value1 = 1,
value2 = 2,
value3 = 3,
},
[3] = {<!-- -->
Id = 101,
value1 = 1,
value2 = 2,
value3 = 3,
},
[4] = {<!-- -->
Id = 108,
value1 = 1,
value2 = 2,
value3 = 3,
},
[5] = {<!-- -->
Id = 101,
value1 = 1,
value2 = 2,
value3 = 3,
},
}
this.RemoveTabValue(tab1,101)
for k, v in pairs(tab1) do
logError("k ========>"..tostring(k))
logError("v.Id ========>"..v.Id)
end
function this.RemoveTabValue(tab,Id)
for i = #tab, 1 ,-1 do
if tab[i].Id == Id then
table.remove(tab,i)
end
end
end
Print as follows
Recursive deletion
--Lua table of test data
local tab1 = {<!-- -->
[1] = {<!-- -->
Id = 101,
value1 = 1,
value2 = 2,
value3 = 3,
},
[2] = {<!-- -->
Id = 105,
value1 = 1,
value2 = 2,
value3 = 3,
},
[3] = {<!-- -->
Id = 101,
value1 = 1,
value2 = 2,
value3 = 3,
},
[4] = {<!-- -->
Id = 108,
value1 = 1,
value2 = 2,
value3 = 3,
},
[5] = {<!-- -->
Id = 101,
value1 = 1,
value2 = 2,
value3 = 3,
},
}
this.RemoveTabValue(tab1,101)
for k, v in pairs(tab1) do
logError("k ========>"..tostring(k))
logError("v.Id ========>"..v.Id)
end
--recursive method
function this.RemoveTabValue(tab,Id)
for k, v in pairs(tab) do
if v.Id == Id then
table.remove(tab,k)
this.RemoveTabValue(tab,Id)
break
end
end
end
Print as follows
Insert new table method
--Lua table of test data
local tab1 = {<!-- -->
[1] = {<!-- -->
Id = 101,
value1 = 1,
value2 = 2,
value3 = 3,
},
[2] = {<!-- -->
Id = 105,
value1 = 1,
value2 = 2,
value3 = 3,
},
[3] = {<!-- -->
Id = 101,
value1 = 1,
value2 = 2,
value3 = 3,
},
[4] = {<!-- -->
Id = 108,
value1 = 1,
value2 = 2,
value3 = 3,
},
[5] = {<!-- -->
Id = 101,
value1 = 1,
value2 = 2,
value3 = 3,
},
}
local newtab = {<!-- -->}
\t
for k, v in pairs(tab1) do
if v.Id == 101 then
table.insert(newtab, v)
end
end
--I have not written a method to delete the tab1 table here, which means it still occupies memory, so it is equivalent to opening up a new memory space.
--You can delete the data of the original tab1 table by yourself, or use the above two methods
--This method takes up additional memory space
for k, v in pairs(newtab) do
logError("k ========>"..tostring(k))
logError("v.Id ========>"..v.Id)
end
Print as follows
Expand it
To briefly talk about the knowledge here, if you encounter the following dictionary type Lua table
- The length result of #tab1 is 3, not 5, and [true] and [“a”] are excluded (key-value pairs that are not numbers k are not recognized)
- All key-value pairs can only be read using pairs. If ipairs is used, only [1][2][3] key-value pairs with number k can be read.
local tab1 = {<!-- -->
[1] = {<!-- -->
Id = 101,
value1 = 1,
value2 = 2,
value3 = 3,
},
[2] = {<!-- -->
Id = 105,
value1 = 1,
value2 = 2,
value3 = 3,
},
[3] = {<!-- -->
Id = 101,
value1 = 1,
value2 = 2,
value3 = 3,
},
[true] = {<!-- -->
Id = 108,
value1 = 1,
value2 = 2,
value3 = 3,
},
["a"] = {<!-- -->
Id = 101,
value1 = 1,
value2 = 2,
value3 = 3,
},
}
Summary
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