Question

我做了一个查询，只选择每个用户的最后日期。它可以在phpmyadmin中运行，但是当我想在PHP中的mysqli_query（）中执行它时，它不会返回任何内容，甚至不会出错。

代码是：

select * from table t inner join ( select User_ID, max(Date) as MaxDate from table group by User_ID ) tm on t.User_ID = tm.User_ID and t.Date = tm.MaxDate

如果您有任何想法请告诉我：）

EDIT PHP代码是：

$id = $_SESSION["ID"];  
    $SqlQuery = "SELECT * from 'tablename' t inner join ( select 'User_ID', max('Date') as 'MaxDate' from 'tablename' group by 'User_ID' ) tm on 't.User_ID' = 'tm.User_ID' and 't.Date' = 'tm.MaxDate'";
    $Result = mysqli_query($link, $SqlQuery) or die ("not possible to execute query: $sql on $link");

    if ($Result->num_rows > 0) {
     while($row = $Result->fetch_assoc()) {
         echo "<br> id: ". $row['Sick_ID']. " - UserID: ". $row['User_ID']. "- Reason " . $row['Reason'] . "<br>";
     }
} else {
     echo "0 results";
}

Answer 1

将架构所有者前缀添加到选择查询。它通常发生在PHP上执行mysql查询时。

Select * from data.table_name as t1 inner join data.table_name_2 as t2 .....

如果更好：

Select data.t1.id, data.t2.name from data_table_name as t1 .....

SQL查询在phpmyadmin中工作但不在php中

1 个答案: