Question

我有匹配表（其中包含＆＃34; ID，日期，team1_id，team2_id ＆＃34;）

和表团队（其中包含＆＃34; ID，名称＆＃34;）

我需要显示团队名称的PHP或SQL代码（来自表团队）WHERE "team1_id"和team2_id" = team.id

......实际上我有类似的东西：

$deleteMatch = Database::query('SELECT match.id, match.team1_id, match.date, 
       team.id, team.name FROM `match`, `team` WHERE team.id = match.team1_id');
foreach($deleteMatch as $matchinfo)
{
    ?><option value="<?php echo $matchinfo["id"];?>">
    <?php echo $matchinfo["date"]; echo $matchinfo["name"];?></option>
    <?php
}

但我需要向foreach添加team2.id(name)。（日期| TEAM1 NAME | TEAM2 NAME）

希望你了解我。谢谢你的帮助。

Answer 1

使用此查询：

SELECT match.id, match.team1_id, match.date, 
  t1.id as team1ID, t1.name as team1Name, 
   match.team2_id,t2.id as team2ID, t2.name as team2Name 
 FROM `match` JOIN `team` t1 ON t1.id = match.team1_id 
  JOIN `team` t2 ON t2.id = match.team2_id

SQLFiddle

Answer 2

$deleteMatch = Database::query('SELECT `match`.id, `match`.team1_id, `match`.date, `team`.id, `team`.name FROM `match` JOIN `team` ON `team`.id = `match`.team1_id' LEFT JOIN `team` ON `team`.id = `match`.team2_id);
foreach($deleteMatch as $matchinfo)
{
?><option value="<?php echo $matchinfo["id"];?>"><?php echo $matchinfo["date"]; echo $matchinfo["name"];?></option>
<?php
 }

Answer 3

我会在团队桌上使用2个内部联接来执行此操作：

$query = "SELECT match.*, team.* FROM match
          INNER JOIN team ON match.team1_id = team.id
          INNER JOIN team ON match.team2_id = team.id;"

表格看起来像这样：

match.id | match.date | team1_team.name | team2_team.name

Answer 4

您可以尝试此SQL查询：

SELECT match.id, date, t1.name AS team1, t2.name AS team2
FROM match, team AS t1, team AS t2
WHERE team1_id=t1.id AND team2_id=t2.id

诀窍是你需要将每个团队ID分别与其名称相匹配，因此你必须两次包括团队表。为了区分两个表，您可以在AS关键字的帮助下使用别名（在示例中，我将别名表分别为t1和t2）

结果表将包含4列：id（匹配ID），日期，team1（团队1的名称）和team2（团队2的名称），因此您可以使用$matchinfo["team1"]和{{ 1：}在你的PHP代码中。

PHP（SQL）选择名称，其中id =更多值

4 个答案: