PHP MySQL從今天比較$ row的日期

Question

我想知道是否可以將所有$行中的日期從while循環與今天的日期進行比較。讓我解釋。

我創造了這個圖像中顯示此表：

我已經通過連接幾個表中創建這個表，居然有一個while循環while循環（這是我得到了多個遊戲展示但每次只有一個投注金額

我想要做的就是將每個單獨遊戲的日期與今天的日期進行比較，以列出parlay是否活動（在這張圖片中，他們都不會活躍，但那不是重點）。

I wan t取今天的日期，如果它小於遊戲的所有日期，那麼Active =='Future'，或者如果今天的日期介於一組日期之間，那麼Active =='In Progress'，如果今天日期已經過去了所有的日期，然後Active =='過去'。使這個具有挑戰性的部分是，我不知道有多少個別遊戲會出現，所以我想這樣的事情可能工作：

/* fetch row */ 
$firstrow = $result2->fetch_row(); 

$date1=date_create("$firstrow[0]"); 
$date2=date_create(date("Y-m-d")); 
$diff=date_diff($date1,$date2); 

但我不知道有多少行做出測試我的最後日期，或許還有一個

$row_count = $result->num_rows; 
$datelast=date_create("$lastrow[$row-count-1]"); 

不知道如何再實施和使用這個，如果它是正確的邏輯！

那麼有誰知道我可以如何比較每個日期行到今天的當前日期？

這是我最初的選擇查詢： 我省略了從我的形象的一些信息，要儘量使問題更容易的核心邏輯，這就是爲什麼有一些額外的字段這裏

//Create parlay select query  
$query = "SELECT 
     u.first_name AS 'User First Name', 
     u.last_name AS 'User Last Name', 
     b.betting_site_name AS 'Betting Site', 
     p.id AS 'Parlay ID', 
     p.wager AS 'Wager', 
     p.odds AS 'Odds', 
     p.success AS 'Success', 
     DATE_FORMAT(p.creationdate, '%d-%m-%Y') AS 'Date', 
     pg.parlayid AS 'PG Parlay ID', 
     SUM(p.wager * p.odds) AS Winnings 
    FROM parlays p 
    JOIN parlaygames pg ON pg.parlayid = p.id 
    JOIN bonuses b ON p.bettingsiteid = b.id 
    JOIN users u ON p.userid = u.id 
    WHERE userid=$id 
    GROUP BY p.id 
    ORDER BY p.id DESC 
    LIMIT 5"; 

$result = $mysqli->query($query) or die($mysqli->error.__LINE__); 

這是我的表創建代碼（和第二while循環得到個別遊戲）：

<h2>Betting History</h2> 
     <table class="table table-striped userwinnings"> 
     <tr> 
      <th>Parlay Information</th> 
      <th>Wager</th> 
      <th>Active</th> 
     </tr> 
<?php 

while($row = $result->fetch_assoc()) { 
      $output = '<tr>'; 
        $output .= '<td><table><tr><th>Date</th><th>Game</th><th>Bet Info</th></tr>'; 

        $parlayid = $row['Parlay ID']; 

        $query3 = "SELECT 
       pg.parlayid AS ParlayID, 
       g.date AS GameDate, 
       ht.name AS HomeTeam, 
       away.name AS AwayTeam, 
       pg.betinfo AS BetInfo 
      FROM parlaygames pg 
      JOIN parlays p ON pg.parlayid = p.id 
      JOIN games g ON pg.gameid = g.id 
      JOIN teams ht ON g.home_team = ht.id 
      JOIN teams away ON g.away_team = away.id 
      JOIN users u ON u.id = p.userid 
      WHERE p.id = $parlayid"; 

     $result3 = $mysqli->query($query3) or die($mysqli->error.__LINE__); 
     while($row3 = $result3->fetch_assoc()) { 
      $gamescount = $result3->num_rows; 
      $output .= '<tr>'; 
      $output .= '<td>'.$row3['GameDate'].'</td>'; 
      $output .= '<td>'.$row3['HomeTeam'].' vs '.$row3['AwayTeam'].'</td>'; 
      $output .= '<td>'.$row3['BetInfo'].'</td>'; 
      $output .= '</tr>'; 
     }      

      $output .= '</table>'; 
        $output .= '</td>'; 
        $output .= '<td>'.$row['Wager'].'</td>'; 
        $output .= '<td>'.$row['Active'].'</td>'; 
        $output .= '</tr>'; 

        echo $output; 

} 

?> 
</table> 

Answer 1

你好，請更換這些查詢

$query3 = "SELECT 
       pg.parlayid AS ParlayID, 
       g.date AS GameDate, 
       ht.name AS HomeTeam, 
       away.name AS AwayTeam, 
       pg.betinfo AS BetInfo 
      FROM parlaygames pg 
      JOIN parlays p ON pg.parlayid = p.id 
      JOIN games g ON pg.gameid = g.id 
      JOIN teams ht ON g.home_team = ht.id 
      JOIN teams away ON g.away_team = away.id 
      JOIN users u ON u.id = p.userid 
      WHERE p.id = $parlayid"; 

有了這些一個

$query3 = "SELECT 
       pg.parlayid AS ParlayID, 
       g.date AS GameDate, 
       IF((NOW() between MIN(g.date) AND MAX(g.date)),'In Progress',IF((MAX(g.date) > NOW()),'Future','Past')) as Acctive, 
       ht.name AS HomeTeam, 
       away.name AS AwayTeam, 
       pg.betinfo AS BetInfo 
      FROM parlaygames pg 
      JOIN parlays p ON pg.parlayid = p.id 
      JOIN games g ON pg.gameid = g.id 
      JOIN teams ht ON g.home_team = ht.id 
      JOIN teams away ON g.away_team = away.id 
      JOIN users u ON u.id = p.userid 
      WHERE p.id = $parlayid group by pg.parlayid"; 

Answer 2

這是我結束瞭解決這一問題：

while($row3 = $result3->fetch_assoc()) { 
      $gamescount = $result3->num_rows; 

      /* seek to row no. 1 */ 
       $result4->data_seek(0); 

       /* fetch row */ 
       $firstrow = $result4->fetch_row(); 

       $result4->data_seek($gamescount-1); 
       $lastrow = $result4->fetch_row(); 

      $date1=date_create("$firstrow[1]"); 
      $date3=date_create("$lastrow[1]"); 
      $date2=date_create(date("Y-m-d")); 
      $diff=date_diff($date3,$date2); 

      $firstdate = $date3->format('Y-m-d'); 
      $lastdate = $date1->format('Y-m-d'); 
      $today = $date2->format('Y-m-d'); 

      if ($firstdate < $today && $today < $lastdate) { 
       $active = "In Progress"; 
      } else if ($today > $lastdate) { 
       $active = "Past"; 
      } else if ($today < $firstdate) { 
       $active = "Future"; 
      } 

我希望這個邏輯可以幫助其他人。