kcd-clan Posted March 22, 2006 Share Posted March 22, 2006 Havent tested yet but im shur it wont work I need it to make $8 to add like so each one $8_1,$8_2,$8_3 How could i do it $8 = InputBox("Step 8-", "", "What will fast tp key be?", "5", -1, -1, 0, 0) For $8 = 1 To $8 $k=$k+1 $8_$k=InputBox("Step "&$k&"-"&$8, "","The "&$k&"of "&$8&"Message", "Tp Up", -1, -1, 0, 0) Next Visit mEMy programs made.Iul - IulG-V Console - G-V Console_RandomLetter - _RandomLetter()Saftey Kill - Saftey Killcolorzone() = colorzone() Link to comment Share on other sites More sharing options...
PsaltyDS Posted March 22, 2006 Share Posted March 22, 2006 Havent tested yet but im shur it wont work I need it to make $8 to add like so each one $8_1,$8_2,$8_3 How could i do it $8 = InputBox("Step 8-", "", "What will fast tp key be?", "5", -1, -1, 0, 0) For $8 = 1 To $8 $k=$k+1 $8_$k=InputBox("Step "&$k&"-"&$8, "","The "&$k&"of "&$8&"Message", "Tp Up", -1, -1, 0, 0) Next I have NO IDEA what you're trying to do, but I tried to set variables with Eval(): $8 = 'A' $k=1 Eval($8 & "_" & $k) = 'Test' MsgBox(64, "Results", "$A_1 = " & $A_1) Doesn't work. I'm not sure you can specify a variable with a variable. How about an array instead? #include <array.au3> $Count = InputBox("Input", "How many values for the table?") If Not StringIsInt($Count) Or $Count < 1 Then MsgBox(16, "Error", "Value must be a positive integer!") Exit EndIf Dim $8[$Count + 1] $8[0] = $Count For $k = 1 To $Count $8[$k] = InputBox("Input", "Input value for table index " & $k & ":") Next _ArrayDisplay($8, "Resulting array:") Hope that helps! Valuater's AutoIt 1-2-3, Class... Is now in Session!For those who want somebody to write the script for them: RentACoder"Any technology distinguishable from magic is insufficiently advanced." -- Geek's corollary to Clarke's law Link to comment Share on other sites More sharing options...
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