Question

我有以下SQL查詢：

$workspace = $_GET['workspace']; 
$location = $_GET['location']; 

$query="SELECT * FROM `offices` WHERE (`workspace` LIKE '%".$workspace."%') AND (`county` = '$location') AND (`active` = 'yes') LIMIT 0, 30 "; 
$result=mysql_query($query); 

的$位置值從一種形式返回先前的網頁，其中有32個縣名單上，編號從1-32。我想要一個選項來選擇「所有縣」，這將返回一個匹配查詢和任何縣的其他標準的結果。縣的列表存儲爲SQL表中列中1-32的數字。

如果最近被問到，但無法找到任何東西，道歉。謝謝！

Answer 1

$workspace = mysql_real_escape_string($_GET['workspace']); 

$location = mysql_real_escape_string($_GET['location']); 

$query="SELECT * FROM `offices`"; 

$conditions = array("`active` = 'yes'"); 

if(!empty($workspace)) 
{ 
$conditions []= "`workspace` LIKE '%".$workspace."%'" 
} 

if(!empty($location)) 
{ 
$conditions []= "(`county` = '$location')"; 
} 

if(!empty($conditions)) 
{ 
$query .= " WHERE " . implode(' AND ', $conditions); 
} 

$query .= "LIMIT 0, 30 "; 

$result=mysql_query($query); 

這是一個有點粗糙和準備，但應該這樣做

在貴國名單剛剛碰到這樣的：

<select name="location"> 
<option id="0">All Countries</option> 
<option id="1">...</option> 
</select> 

Answer 2

我沒聽清楚你們的問題。但下面就給你做什麼簡單的想法：

$locationCondition = "(`county` = '$location') AND "; 
if ($location == "ALL") 
    $locationCondition = ""; 
$query="SELECT * FROM `offices` WHERE (`workspace` LIKE '%".$workspace."%') AND ".$locationCondition."(`active` = 'yes') LIMIT 0, 30 "; 

Answer 3

我希望這將有助於我認爲

$query="SELECT * FROM `offices` WHERE (`workspace` LIKE '%".$workspace."%') AND (`county` in ('$location')) AND (`active` = 'yes') LIMIT 0, 30 ";