php - 如何获取特定用户的最后 7 个投注单 (BetSlipID)

Question

我当前的代码显示最近 7 天的投注单，但无论日期如何，我都需要显示最近 7 个投注单。

select 
  betsliphistory.MatchID,Team1.Name as HomeTeam ,
  Team2.Name as AwayTeam, BetSlipID , userid , TipID , 
  matches.ResultTipID ,betsliphistory.`Date`
from betsliphistory
inner join matches on betsliphistory.MatchID = matches.MatchID
inner join teams as Team1 on matches.HomeTeamID = Team1.TeamID
inner join teams as Team2 on matches.AwayTeamID = Team2.TeamID
where userid =".$user." 
  and betsliphistory.`Date` between CURDATE()-7 and CURDATE() 
order by BetSlipID , MatchID;

Answer 1

但无论日期如何，我都需要出示最近 7 次投注单。

• WHERE从子句中删除日期条件。

• 将ORDER By BetSlipID DESCwithLIMIT 7用于前 7 名BetSlipID，如下所示：

   ... -- Your current query
   WHERE userid = ".$user." 
   ORDER BetSlipID  DESC
   LIMIT 7;
  

请注意：您的代码以这种方式容易受到SQL Injection的攻击。请改用 PDO 或准备好的语句。有关更多详细信息，请参见：

• PHP：防止 SQL 注入的最佳方法？.

Answer 2

试试这个：：

 select betsliphistory.MatchID,Team1.Name as HomeTeam , 
                      Team2.Name as AwayTeam, BetSlipID , userid , TipID , matches.ResultTipID ,betsliphistory.`Date`
                      from betsliphistory
                      inner join matches on betsliphistory.MatchID = matches.MatchID
                      inner join teams as Team1 on matches.HomeTeamID = Team1.TeamID
                      inner join teams as Team2 on matches.AwayTeamID = Team2.TeamID
                      inner join (Select DISTINCT(BetSlipID) tempBetSlip from betsliphistory where userid =".$user."  order by `Date` limit 7) temp_betHistory on temp_betHistory.tempBetSlip = BetSlipID
                          where userid =".$user." order by `BetSlipID`