使用PHP从关系数据库获取数据时出错

Question

我设计了一个简单的关系数据库。当我试图从它抛出一个错误的服务器中的数据：（我跳过了一些代码，使其简单）

这是我使用的SQL语法：

$sql = "SELECT lead.id, lead.name, lead.phone, lead.email, treatment.name, source.name, status.name FROM lead join treatment join source join status on treatment.id = lead.treatment_id and source.id = lead.source_id and status.id = lead.status_id"; 

这是

echo " 
 
<tr> 
 
<td>".$row["lead.id"]."</td> 
 
<td>".$row["lead.name"]."</td> 
 
<td>".$row["lead.email"]."</td> 
 
<td>".$row["treatment.name"]."</td> 
 
<td>".$row["source.name"]."</td> 
 
<td>".$row["status.name"]."</td> 
 
</tr>";

此代码是给了一个错误，当我改变$row["lead.id"] to $row["id"]它的工作原理，但我需要：HTML里面使用提到表名，因为我几乎在所有表中都有相同的列名。

有没有办法使用表名来做到这一点？

Answer 1

您有在错误的地方，并用不正确的，条件 条件，你应该在条件为每个表使用

$sql = "SELECT 
      lead.id 
     , lead.name 
     , lead.phone 
     , lead.email 
     , treatment.name 
     , source.name 
     , status.name 
     FROM lead 
     join treatment on treatment.id = lead.treatment_id 
     join source on source.id = lead.source_id 
     join status on status.id = lead.status_id"; 

和索引尝试使用别名 避免表名和点符号

$sql = "SELECT 
      lead.id as lead_id 
      , lead.name as lead_name 
      , lead.phone as lead_phone 
      , lead.email as lead_email 
      , treatment.name as treatment_name_ 
      , source.name as source_name 
      , status.name as status_name 
     FROM lead 
     join treatment on treatment.id = lead.treatment_id 
     join source on source.id = lead.source_id 
     join status on status.id = lead.status_id";