SQL / PHP如果不滿足條件，則選擇所有內容

Question

我的網頁上有兩個搜索參數（例如類型和位置）的輸入。 輸入參數后，PHP將搜索SQL數據庫並輸出滿足輸入參數的結果。

$sql = "
SELECT MainID
     , MainName
     , MainLocation
     , MainType 
  FROM Main 
 WHERE MainType = 'typeIWant' 
   AND MainLocation = 'LocationIWant' 
";

但是，當用戶僅選擇一個參數時，它應顯示僅滿足所選參數的所有結果。

就像是：

$sql = "SELECT MainID, MainName, MainLocation, MainType FROM Main WHERE MainType= 'typeIWant' AND MainLocation = '*' ";

除非這行不通。

它應該顯示所有滿足MainType的結果，而不管它們的MainLocation是什么。

怎么做？

Answer 1

下一行：

$sql = "SELECT MainID, MainName, MainLocation, MainType FROM Main WHERE MainType= 'typeIWant' AND MainLocation = '*' ";

正在搜索名為（或稱為）的MainLocation *

如果只想在MainType上搜索， MainType應該執行以下操作：

$sql = "SELECT MainID, MainName, MainLocation, MainType FROM Main WHERE MainType= 'typeIWant' ";

Answer 2

您可以使用if statement進行檢查，例如：

$sql = "
SELECT MainID
     , MainName
     , MainLocation
     , MainType 
  FROM Main 
 WHERE MainType = 'typeIWant'  
";

if (isset($_GET['MainLocation']) && $_GET['MainLocation'] != '') {
  $sql .= " AND MainLocation = 'LocationIWant'"
}