带有 PHP 的 SQLite3 如何使用在其他列中找到的数据对列进行交叉引用

Question

Here is how my Sample data looks like

hwid: 1502e3c3-b49c-4907-92fd-2ffac2d50ace
ip: 68.197.140.109 
character name: none

hwid: 1502e3c3-b49c-4907-92fd-2ffac2d50ace
ip: 219.100.37.236
character name: none

hwid: 1502e3c3-b49c-4907-92fd-2ffac2d50acd 
ip: 68.197.140.109 
character name: kalgame

Now say I'm logging in with this hwid 1502e3c3-b49c-4907-92fd-2ffac2d50ace and this ip address 219.100.37.236

which has a link to using a previous IP address of 68.197.140.109
and that IP address has a link to a 3rd hwid that has a character name: kalgame .

Thats what I need to do cross referencing with all IP's linked to all hwid's and then check if any of those hwid's have a character_name . If they have a character_name associated to them return them in a SELECT DISTINCT COUNT(totalMatchesOfUniqueCharacterNames)

I also need to check if the current HWID or current IP you are using has 3 days trial period, if it does then also add it to the SELECT DISTINCT COUNT .. then I could easily check in PHP if SELECT DISTINCT COUNT is greater then 1, then abort giving out any more free trials to the specific user as he already logged in before with a character_name , and also maybe had a 3 day trial period before hand on either his PC hwid or his IP address, note I would also like to exclude blank character_name's as those players didn't login with a character yet.. so they don't count. I also have to match the locale which is which game are they playing if they are playing a different game then they also don't count.

Here is my attempt at cross referencing I been working at this for 6 hours and today is my first day using SQLite or any SQL in general so I have no clue what i'm doing

Here is my table schema DDL for both tables.. they are also both in the same database file.

CREATE TABLE ips (
    hwid           VARCHAR,
    ip             VARCHAR,
    character_name VARCHAR,
    locale         VARCHAR,
    botver         VARCHAR,
    reg_time       DATETIME,
    last_time      DATETIME,
    ping_count     INTEGER  DEFAULT 0
);

CREATE TABLE users (
    hwid      VARCHAR UNIQUE,
    timestamp INTEGER,
    daysToUse INTEGER,
    pcLimit   INTEGER,
    comment   VARCHAR
);

Here is my attempt at putting it all together

SELECT users2.hwid,
       ips2.character_name,
       ips2.ip
  FROM users AS users2
       INNER JOIN
       ips AS ips2 ON ips2.hwid = users2.hwid
 WHERE EXISTS (
           SELECT 1
             FROM users
                  INNER JOIN
                  ips ON ips.hwid = users.hwid
            WHERE ips.character_name != '' AND 
                  ips.locale = ips2.locale AND 
                  (ips.hwid = '1502e3c3-b49c-4907-92fd-2ffac2d50ace' OR 
                   ips.ip = '219.100.37.236') AND 
                  EXISTS (
                          SELECT 1
                            FROM users
                                 INNER JOIN
                                 ips ON ips.hwid = users.hwid
                           WHERE ips.character_name != '' AND 
                                 ips2.ip = ips.ip AND 
                                 ips2.locale = ips.locale
                      )
       )
OR 
       (ips2.hwid = '1502e3c3-b49c-4907-92fd-2ffac2d50ace' AND 
        users2.daysToUse = 3) OR 
       (ips2.ip = '219.100.37.236' AND 
        users2.daysToUse = 3);

Here is what I left in the production website.. which isn't that good.. doesn't do cross referencing that well skips certain columns but atleast it kinda works unlike the top code above which just sounds like it should work but doesn't work at all.

SELECT COUNT(DISTINCT users2.hwid) 
  FROM users AS users2
       INNER JOIN
       ips AS ips3 ON ips3.hwid = users2.hwid
 WHERE ips3.character_name = (
                                 SELECT ips.character_name
                                   FROM users,
                                        ips AS ips2
                                        INNER JOIN
                                        ips ON ips.hwid = users.hwid
                                  WHERE ips.character_name != '' AND 
                                        ips.locale = ips2.locale AND 
                                        (ips.hwid = '$HWID' OR 
                                         ips.ip = '$IP') 
                             )
OR 
       (ips3.hwid = '$HWID' AND 
        users2.daysToUse = 3) OR 
       (ips3.ip = '$IP' AND 
        users2.daysToUse = 3);

If anyone can help me out i will be forever grateful:D

Answer 1

Since your subqueries share same FROM and JOIN clauses, simplify to a compact WHERE clause:

SELECT i.ip, i.character_name, u.hwid, u.daysToUse, i.locale 
FROM users u
INNER JOIN ips i
   ON i.hwid = u.hwid 
WHERE (i.hwid = '1502e3c3-b49c-4907-92fd-2ffac2d50ace'
        AND u.daysToUse = 3) 
   OR (i.ip = '219.100.37.236' AND u.daysToUse = 3) 
ORDER BY i.last_time 

Answer 2

Solved it myself.. turns out it is very simple to do this, I just had the query done in backwards order I flipped the order and bam it works as expected.

The locale problem was hard.. I couldn't detect which locale is currently used, so I relayed on using ips.lasttime which shows which IP or HWID was used last, and being used last its obviously the locale I want to check

Final result

SELECT ips.ip,
       ips.character_name,
       users.hwid,
       users.daysToUse,
       ips.locale
  FROM users
       INNER JOIN
       ips ON ips.hwid = users.hwid
 WHERE ips.ip IN (
           SELECT ips.ip
             FROM users
                  INNER JOIN
                  ips ON ips.hwid = users.hwid
            WHERE (ips.hwid = '1502e3c3-b49c-4907-92fd-2ffac2d50ace' AND 
                   users.daysToUse = 3) OR 
                  (ips.ip = '219.100.37.236' AND 
                   users.daysToUse = 3) 
       )
AND 
       ips.locale = (
           SELECT ips.locale
             FROM users
                  INNER JOIN
                  ips ON ips.hwid = users.hwid
            WHERE (ips.hwid = '1502e3c3-b49c-4907-92fd-2ffac2d50ace' AND 
                   users.daysToUse = 3) OR 
                  (ips.ip = '219.100.37.236' AND 
                   users.daysToUse = 3) ORDER BY ips.last_time ASC
       );