PROBLEM WITH DIALOG_STYLE_INPUT
#4

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
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Messages In This Thread
PROBLEM WITH DIALOG_STYLE_INPUT - by jullyus - 07.02.2019, 15:47
Re: PROBLEM WITH DIALOG_STYLE_INPUT - by Kaliber - 07.02.2019, 15:51
Re: PROBLEM WITH DIALOG_STYLE_INPUT - by jullyus - 07.02.2019, 15:55
Re: PROBLEM WITH DIALOG_STYLE_INPUT - by Kaliber - 07.02.2019, 16:03
Re: PROBLEM WITH DIALOG_STYLE_INPUT - by Banditul18 - 07.02.2019, 16:05
Re: PROBLEM WITH DIALOG_STYLE_INPUT - by jullyus - 07.02.2019, 16:06
Re: PROBLEM WITH DIALOG_STYLE_INPUT - by TheToretto - 07.02.2019, 17:24
Re: PROBLEM WITH DIALOG_STYLE_INPUT - by Kaliber - 07.02.2019, 17:31

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