only4tibia
03-20-2014, 10:41 PM
Hey All,
Explained how "for i = 1, #Something do" works to ppgab. Thought this may help others as well so decided to post here.
Really apreciate your efforts man ;)
Care to explain what "for i = 1, #RashidLocation do" does? i understood everything else except this
So, heres how that works. RashidLocation is a table of values. Rashid[i] represents these values. If you put Rashid[1] then you will get the first value of the table, Rashid[2] will give you 2nd value of the table. Let me make a smaller table.
Items = { "Wands", "Blue Djiin",
"Rods", "Green Djiin"
}
For this table, Items[1] is "Wands", Items[2] is "Blue Djiin", Items[3] is "Rods", Items[4] is "Green Djiin".
Now, looking at the following:
ItemToSell = "Rods"
for i = 1, #Items do
if ItemToSell == Items[i] then WhereToSell = Items[i+1] end
end
"for i = 1, #Items do" <--- for i equaling numbers 1 to 4 (4 being total number of Items table values), "do" whatever follows.
This means lets put i in to the equation each time and see if its true.
if ItemToSell == Items[i] then WhereToSell = Items[i+1] end <-----Original
if ItemToSell == Items[1] then WhereToSell = Items[1+1] end <-- With i = 1, "if ItemToSell == "Wands" then WhereToSell = "BlueDjiin" (Items[2] value)
if ItemToSell == Items[2] then WhereToSell = Items[2+1] end
if ItemToSell == Items[3] then WhereToSell = Items[3+1] end
if ItemToSell == Items[4] then WhereToSell = Items[4+1] end
ItemToSell is "Rods" and will only be true when i = 3, as Items[3] value is "Rods". Therefore we will end up with WhereToSell = Items[3+1] or WhereToSell = "Green Djiin"
Hope this makes sense and helps some people.
Regards,
O4T
Explained how "for i = 1, #Something do" works to ppgab. Thought this may help others as well so decided to post here.
Really apreciate your efforts man ;)
Care to explain what "for i = 1, #RashidLocation do" does? i understood everything else except this
So, heres how that works. RashidLocation is a table of values. Rashid[i] represents these values. If you put Rashid[1] then you will get the first value of the table, Rashid[2] will give you 2nd value of the table. Let me make a smaller table.
Items = { "Wands", "Blue Djiin",
"Rods", "Green Djiin"
}
For this table, Items[1] is "Wands", Items[2] is "Blue Djiin", Items[3] is "Rods", Items[4] is "Green Djiin".
Now, looking at the following:
ItemToSell = "Rods"
for i = 1, #Items do
if ItemToSell == Items[i] then WhereToSell = Items[i+1] end
end
"for i = 1, #Items do" <--- for i equaling numbers 1 to 4 (4 being total number of Items table values), "do" whatever follows.
This means lets put i in to the equation each time and see if its true.
if ItemToSell == Items[i] then WhereToSell = Items[i+1] end <-----Original
if ItemToSell == Items[1] then WhereToSell = Items[1+1] end <-- With i = 1, "if ItemToSell == "Wands" then WhereToSell = "BlueDjiin" (Items[2] value)
if ItemToSell == Items[2] then WhereToSell = Items[2+1] end
if ItemToSell == Items[3] then WhereToSell = Items[3+1] end
if ItemToSell == Items[4] then WhereToSell = Items[4+1] end
ItemToSell is "Rods" and will only be true when i = 3, as Items[3] value is "Rods". Therefore we will end up with WhereToSell = Items[3+1] or WhereToSell = "Green Djiin"
Hope this makes sense and helps some people.
Regards,
O4T