[Deny1]

Hello everebody, i want to figure how to make dialogs show after dialog, example if i define dialog_test
and make cmd like this

PHP код:

CMD:test(playerid, params[])
{
    ShowPlayerDialog(playerid, DIALOG_TEST, DIALOG_STYLE_MSGBOX, "TEST", "TESTr", "Close", "");
    return 1;
} 


I put under OnDialogResponse

PHP код:

if(dialogid == 1) 
    {
    ShowPlayerDialog(playerid, DIALOG_TEST, DIALOG_STYLE_MSGBOX, "TEST", "TESTr", "Close", "");
    } 


Does i go in right way?and I mean when command show me this dialog and i click close to open me another dialog it isn't matter witch...

[AndreiWow]

Did you atleast try it? Yes it will open you the next dialog if you put it in the dialog response.

[Deny1]

Look how can i make under OnDialogResponse if dialog id 1 resposne to open dialog id number 2 and when i close dialog id 2 to automaticly open dialog 3

[AndreiWow]

Put this at the top

Rename them as you need

PHP код:

#define DIALOG_NAME1 //dialog1
#define DIALOG_NAME2 //dialog2
#define DIALOG_NAME3 //dialog3 


OnDialogResponse

PHP код:

switch (dialogid)
{
     case DIALOG_NAME1:
      {
         if(response)//if player press on the left button
          ShowPlayerDialog...//dialog2
      }
     case DIALOG_NAME2:
     {
       if(response)//if player press on the left button
       ShowPlayerDialog...//dialog3
      }
} 


[Deny1]

Yes, it show only first 2 dialogs not third(dialog 3)..

[Deny1]

PHP код:

if(dialogid == 1)
    {
        switch (dialogid)
        {
            case DIALOG_TEST:
            {
            if(response)//if player press on the left button
            ShowPlayerDialog(playerid, DIALOG_TEST3, DIALOG_STYLE_MSGBOX, "Notice", "222222", "22", "22");
            }
            case DIALOG_TEST2:
           {
           if(response)//if player press on the left button
           ShowPlayerDialog(playerid, DIALOG_TEST3, DIALOG_STYLE_MSGBOX, "Notice", "33333", "333", "33");
           }
    }
    } 


[AndreiWow]

Quote:

Originally Posted by Deny1 PHP код: if(dialogid == 1)     {         switch (dialogid)         {             case DIALOG_TEST:             {             if(response)//if player press on the left button             ShowPlayerDialog(playerid, DIALOG_TEST3, DIALOG_STYLE_MSGBOX, "Notice", "222222", "22", "22");             }             case DIALOG_TEST2:            {            if(response)//if player press on the left button            ShowPlayerDialog(playerid, DIALOG_TEST3, DIALOG_STYLE_MSGBOX, "Notice", "33333", "333", "33");            }     }     }

DIALOG_TEST:

PHP код:

 ShowPlayerDialog(playerid, DIALOG_TEST2, DIALOG_STYLE_MSGBOX, "Notice", "222222", "22", "22"); 


You used DIALOG_TEST3 at both cases...

In first case use DIALOG_TEST2 like I did above.

[Deny1]

it doesn't show again..

[Deny1]

someone help?

[Deny1]

BUMP