Hello.
Immediately apologize for the English language.
The question seems clear and understandable. How to make through the old menu, two pages?
Code of players do not need a menu, then the problem lies elsewhere.
When you create a menu of the players in the loop, only one line formatex (possibly with conditions)
Code:
for(new i = iStart; i < iEnd; i++) {
iPlayer = g_iMenuPlayers[i];
get_user_name(iPlayer, szName, charsmax(szName));
if(iPlayer == id) iLen += formatex(szMenu[iLen], charsmax(szMenu) - iLen, "");
else {
iKeysMenu |= (1 << iCount);
iLen += formatex(szMenu[iLen], charsmax(szMenu) - iLen, "");
}
}
Here we see that in the loop body is executed only one formatex (one of the two conditions).
But, in my case in the loop may be more than 9 formatex.
Code:
for(new i = iStart; i < iEnd; i++) {
iLen += formatex(szMenu[iLen], charsmax(szMenu) - iLen, "");
iLen += formatex(szMenu[iLen], charsmax(szMenu) - iLen, "");
iLen += formatex(szMenu[iLen], charsmax(szMenu) - iLen, "");
iLen += formatex(szMenu[iLen], charsmax(szMenu) - iLen, "");
}
Here cat, it's for example. We see that in the loop 4 formatex.
Now the problem itself, because 4 formatex , then the cycle must be performed 4 times !? However, since in the loop formatex 4 and 4 times a loop is executed, then 4x4 = 16 formatex and the menu will not be 4 points and 16.
This is the whole problem of how to make the cycle more than 9 formatex , you need to like the menu with the players was 9 ago, 0 - forward \ output.