Question

标题有点令人困惑，希望很明显我想从我的查询中做些什么。我以为我理解了连接，显然不是，这是我的$查询：

     SELECT DATE(T0.timestamp),
            SUM(T0.total_responses), 
            SUM(T0.responses), 
            T0.metric_id, 
            T1.metric_id
       FROM `personal_aggregates` AS T0
  LEFT JOIN `qrs_metrics` AS T1 
         ON T0.metric_id = T1.qrs_metric_id
      WHERE T0.user_id = 1 AND 
            T0.duration = '1' AND 
            T0.category_id IN (1,2,3,4) AND 
            T0.timestamp >= 'period_duration'
   GROUP BY DATE(T0.timestamp)";

我正在尝试在列metric metric和qrs_metric_id上加入表。但是我从上面的查询中得不到任何结果。当我在PHP中遍历结果并检查任何变量时，例如$result['T0.metric_id']，我得到Undefined index: T0.metric_id

如果有人能对此有所了解，我将非常感激。

Answer 1

您的查询应该是

$query = "SELECT 
                DATE(T0.timestamp) AS T0timestamp, 
                SUM(T0.total_responses) AS total_responses, 
                SUM(T0.responses) AS responses,
                T0.metric_id, 
                T1.metric_id AS metric_id_2
        FROM `personal_aggregates` AS T0

并访问

$result['metric_id'];
$result['metric_id_2'];
$result['T0timestamp'];
$result['total_responses'];

从关联数组访问中排除表别名，例如“T0.XXX”错了。您只能在SQL

中执行此操作

Answer 2

试试这个：

$query = "SELECT DATE(T0.timestamp), SUM(T0.total_responses), SUM(T0.responses),
               T0.metric_id AS metric0, T1.metric_id AS metric1
        FROM `personal_aggregates` AS T0
        LEFT JOIN `qrs_metrics` AS T1 ON T0.metric_id = T1.qrs_metric_id
        WHERE T0.user_id = 1 AND T0.duration = '1'
        AND T0.category_id IN (1,2,3,4)
        AND T0.timestamp >= 'period_duration'
        GROUP BY DATE(T0.timestamp)";

你现在可以通过以下方式发挥作用：

echo $row['metric0'];
echo $row['metric1'];

Answer 3

您需要在所选列中添加一些别名

SELECT DATE(T0.timestamp) as tstamp, SUM(T0.total_responses) as responses...

并按别名引用列：

$result['tstamp']

或者您可以通过数字索引引用列：

$result[0]

如何在列具有相同名称时LEFT JOIN

3 个答案: