suggestion check
#1

Hello, well i have a table in database name `suggestion`. Anyway, i wanna check who suggested but anytime when i try the command, it insets me a new row with my name and in suggestion nothing.

PHP код:
CMD:checksuggestionplayeridparams[] )
{
    new 
Query[700], name[24], suggestion[200], string[1000], aux[1000];
    
format(Querysizeof(Query), "SELECT * FROM `suggestions`");
    
mysql_query(Query);
    
mysql_store_result();
    if(
mysql_num_rows() != 0)
    {
        while(
mysql_fetch_row(Query))
        {
            
mysql_fetch_field_row(name"user");
            
mysql_fetch_field_row(suggestion"suggestion");
            
format(auxsizeof(aux), "{00FF00}User: {FFFFFF}%s\t\t{00FF00}Suggestion: {FFFFFF}%s\n"namesuggestion);
            
strcat(stringaux);
            
ShowPlayerDialog(playerid20DIALOG_STYLE_MSGBOX"{FFFFFF}Suggestion List"string"Ok","");
        }
    }
    
mysql_free_result();
    return 
1;

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Messages In This Thread
suggestion check - by Bogdan1992 - 17.02.2012, 12:54
Re: suggestion check - by Bogdan1992 - 17.02.2012, 21:52
Re: suggestion check - by dice7 - 17.02.2012, 22:23
Re: suggestion check - by Bogdan1992 - 18.02.2012, 06:39
Re: suggestion check - by dice7 - 18.02.2012, 10:53
Re: suggestion check - by Bogdan1992 - 18.02.2012, 11:54
Re: suggestion check - by Bogdan1992 - 18.02.2012, 12:19

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