php - mysqli 左连接重复行

Question

桌面电脑

id | com       | id_status  
1    testing 1   1  
2    testing 2   1  
3    testing 3   2  
4    testing 4   2  

表格更新

id_status | update  
1           update 1  
2           update 2  
3           update 3  


SELECT `update`.`update`, `com`.`com` as comen
FROM `update` 
    LEFT JOIN `com` ON `update`.`id_status`=`com`.`id_status`  

结果:

update 1  
testing 1  
update 1  
testing 2  
update 2  
testing 3  
update 2  
testing 4  

更新表结果重复

我已经使用了 group by update.id_status，结果是:

update 1  
testing 1  
testing 2  
update 3  

表 com 仅结果 1 行

编辑--
我从中得到了语法 MySQL LEFT JOIN display duplicate rows

$first = true;  
while($row = $query->fetch_object()){  
if($first){  
    echo $row->update;  
    $first = false;  
    echo "<br>";  
}  
echo $row->comen;  
echo "<br>";  
}  

update 1  
testing 1  
testing 2  
testing 3  
testing 4  

它只获取 1 个表行

我希望结果如下所示:

update 1  
testing 1  
testing 2  
update 2  
testing 3  
testing 4  
update 3  
...  

--

正确的语法或查询如何工作？

Answer 1

试试这个，我想这会解决:-

SELECT `update`.`update`, group_concat(`com`.`com`) as comen
FROM `update` 
    LEFT JOIN `com` ON `update`.`id_status`=`com`.`id_status` 
Group by `update`.`update`

并使用您的 php，如下所示:-

while($row = $query->fetch_object()){  
    echo $row->update;  
    $comen = explode(",", $row->comen);
    foreach($comen as $values){
     echo "<br>";  
     echo $values;
    }
    echo "<br>";  
}