PROBLEM WITH DIALOG_STYLE_INPUT

PROBLEM WITH DIALOG_STYLE_INPUT - Printable Version

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+--- Thread: PROBLEM WITH DIALOG_STYLE_INPUT (/showthread.php?tid=663634)

PROBLEM WITH DIALOG_STYLE_INPUT - jullyus - 07.02.2019

Код:

if(dialogid==MENUCARROS2){

				if(response==1){
					new num;
                                        new string[5];
                                        format(string,sizeof(string),"%d",inputtext);
  					num=strval(string);
						if(num  == strval(string)){
   							GetPlayerPos(playerid,x,y,z);
							CreateVehicle(num, x+2,y+2,z, 0, 0, 0, 6000);
}
}
}

I've got the next problem, i want know їWhy dont works this code IN THE TOP? and this code does works ON THE DOWN .

Код:

if(dialogid==MENUCARROS2){

				if(response==1){
					new num;
                                                                  
  					num=strval(inputtext);
						if(num  == strval(inputtext)){
   							GetPlayerPos(playerid,x,y,z);
							CreateVehicle(num, x+2,y+2,z, 0, 0, 0, 6000);
}
}
}

Re: PROBLEM WITH DIALOG_STYLE_INPUT - Kaliber - 07.02.2019

What are you trying to check there?

Actually you check

PHP код:



if x == x

Re: PROBLEM WITH DIALOG_STYLE_INPUT - jullyus - 07.02.2019

Quote:

Originally Posted by Kaliber

What are you trying to check there?

Actually you check

PHP код:



if x == x

Hey, i'm new in the world of the pawn language, just i need that you answer me the problem that i present.

The conclusion that i think is the 'format' just take variables and inputtext not is a variable.

Correct me please, if i'm wrong.

Re: PROBLEM WITH DIALOG_STYLE_INPUT - Kaliber - 07.02.2019

Both are Variables.

Actually Array Variables, that are used in that case for string storing.

Here is a good Tutorial for you: https://sampforum.blast.hk/showthread.php?tid=654471

With strval you convert that string to an integer.

And format with %d tries to interpret the Variable like an Integer.

So if you pass an A it gets to 65, because ASCII Values are taken.

So in that case you should directly use strval(inputtext), to get the integer values (if that is your target).

I dont know why the first one is not working, you can check that easily out.

Just make:

PHP код:



printf("%d == %d",num,strval(string));

Before the if.

Then you know, what the exact values are and why it is not working

Re: PROBLEM WITH DIALOG_STYLE_INPUT - Banditul18 - 07.02.2019

PHP код:



new string[5];
format(string,sizeof(string),"%d",inputtext);

This doesnt work cus inputtext is string yet you try to format it using integer format

PHP код:



new string[5];
 format(string,sizeof(string),"%s",inputtext);

Re: PROBLEM WITH DIALOG_STYLE_INPUT - jullyus - 07.02.2019

thank you mate ♥♥♥♥♥♥!

Re: PROBLEM WITH DIALOG_STYLE_INPUT - TheToretto - 07.02.2019

Quote:

Originally Posted by Kaliber

What are you trying to check there?

Actually you check

PHP код:



if x == x

You forgot a colon.

Oh sorry I thought you were coding in python.

You need brackets in an if statement in pawn lang, don't make him grasp wrong stuff.

Re: PROBLEM WITH DIALOG_STYLE_INPUT - Kaliber - 07.02.2019

Its obviously nothing, but Pseudo-Code.