使用PHP從關係數據庫獲取數據時出錯

Question

我設計了一個簡單的關係數據庫。當我試圖從它拋出一個錯誤的服務器中的數據：（我跳過了一些代碼，使其簡單）

這是我使用的SQL語法：

$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"; 

這是

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>";

此代碼是給了一個錯誤，當我改變$row["lead.id"] to $row["id"]它的工作原理，但我需要：HTML裏面使用提到表名，因爲我幾乎在所有表中都有相同的列名。

有沒有辦法使用表名來做到這一點？

Answer 1

您有在錯誤的地方，並用不正確的，條件 條件，你應該在條件爲每個表使用

$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"; 

和索引嘗試使用別名 避免表名和點符號

$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";