PHP SQL變量

Question

試圖加載$位置到SQL查詢，但只是返回無效：

//query to Equipment 

$Location= $_GET["id"]; 

$sql = mysqli_query($con,"SELECT * FROM `Equipment` WHERE `Location` = `".$Location."`"); 

$result = $sql; 
if (!$result) 
{ 
    die('Invalid query: ' . mysqli_error()); 

} 
while ($row = mysqli_fetch_array($result)) 
{ 

的$_GET['id']從另一個頁面調用。我仔細檢查了鏈接說ID

Answer 1

@戴夫·貝克嘗試類似如下：

<?php 
    $Location = isset($_GET["id"]) ? $_GET["id"] : 0; 
    if($Location){ 
     $sql = mysqli_query($con,"SELECT * FROM Equipment WHERE Location = $Location"); //if location type is integer else use below one 
     $sql = mysqli_query($con,"SELECT * FROM Equipment WHERE Location = '$Location'"); //if location type is string 

     $result = $sql; 
     if (!$result) 
     { 
      die('Invalid query: ' . mysqli_error()); 

     } 
     while ($row = mysqli_fetch_array($result)) 
     {} 

    } 
    else{ 
     echo "id is not set"; 
    }