Player Age Dialog
#1

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#2

if(strlen(inputtext) < 18 || strlen(inputtext) > 100) return ShowDialog(playerid, 7);

change to

if(strval(inputtext) < 18 || strval(inputtext) > 100) return ShowDialog(playerid, 7);


even more, put new string = strval(inputtext); above that compare and use

if(string < 18 || string > 100) return ShowDialog(playerid, 7);
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#3

Your format should be like this (idk, i don't use to use mySQL)

PHP код:

format
(querysizeof(query), "UPDATE `playerdata` SET sex = '%d', age = '%d' WHERE username = '%s'"PlayerData[playerid][pSex], string); 
ShowDialog also doesn't exist. Use ShowPlayerDialog instead (except if you defined it before, in this situation, sorry )
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#4

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