Question

有人可以指出我在这里出错的地方，我一直在我的sql语句中收到错误。我不得不使用别名，因为我试图在夹具列表中显示两个足球队的名字。到目前为止我的sql看起来像这样：

actions:
 - id
 - name
 - data

phrases:
 - id
 - action_id
 - phrase

和我的PHP代码回应这个是这样的：

$sql = "SELECT homeTeam, awayTeam, roundID, teamID, teamName, logo, groundName, away.abbreviatedName AS abbreviatedNameAway, home.abbreviatedName AS abbreviatedNameHome, matchDate, matchTime, venue FROM fixtures 
      INNER JOIN team home ON fixtures.homeTeam = home.teamID 
      INNER JOIN team away ON fixtures.awayTeam = away.teamID 
      INNER JOIN ground ON ground.groundID = home.groundID";

我正在使用的表是：

    <td><?php echo $row ["home.abbreviatedName"]?></td>
    <td> V  </td>
    <td><?php echo $row ["away.abbreviatedName"] ?></td>

homeTeam＆amp; awayTeam本质上是teamID＆，场地是groundID 我收到的错误是：

team           | fixtures   | ground
------         | ------     | groundID
teamID         | matchID    | groundName
abbreviatedname| homeTeam   |
groundID       | awayTeam   |
               | venue

Answer 1

away.abbreviatedName AS abbreviatedNameAway
home.abbreviatedName abbreviatedNameHome

应该是

$row ["abbreviatedNameAway"] 
$row ["abbreviatedNameHome"]

在PHP中调用sql别名

1 个答案: