通過選擇選項標籤從數據庫提取數據，這樣當選擇'所有'數據庫應該選擇'*'

Question

我試圖使用選擇選項標籤，使得如果我選擇'全部'在選擇選項標籤，然後每個該列的數據應該被提取。

以下給出的是我的php的代碼。由於

if(isset($_POST['submit'])) { 
    $area = $_POST['city']; 
    if(isset($_POST['work']!== 'all')) 
     {$work = $_POST['work']; } 
    else {$work = '*';} ; 

    $sel2 = "SELECT * FROM `userdata` WHERE `area`='".$area."' AND `work`='".$work."'" ; 
    $res2 = mysqli_query($con,$sel2); 

Answer 1

既然你是想獲取任何與工作相關的所有數據時work等於all，你沒有明確提到，在SQL查詢。僅當您想要獲取特定的工作相關數據時，提及即。所以，你的代碼應該是這樣的：

if(isset($_POST['submit'])) { 
    $area = $_POST['city']; 
    $sel2 = "SELECT * FROM `userdata` WHERE `area`='".$area."'"; 
    if(isset($_POST['work']) && $_POST['work'] !== 'all') 
     $sel2 .= " AND `work`='".$_POST['work']."'"; 
    $res2 = mysqli_query($con,$sel2); 

---

旁註：瞭解prepared statement因爲現在你的查詢是容易受到SQL注入攻擊。另見how you can prevent SQL injection in PHP。

Answer 2

設置這樣

if(isset($_POST['submit'])) { 
    $area = $_POST['city']; 
    if(isset($_POST['work']!== 'all')) 
$data=addslashes($_POST['work']); 
     {$work = "AND `work`='".$data."'"; 
} 
    else {$work = ""; 
} 
///your query should be like this 
$sel2 = "SELECT * FROM `userdata` WHERE `area`='".$area."' ".$work.""; 
//////so if work is not specified then you don't need to select column WORK in the query. So it works as select * from userdata where area='".$area."';"; 

    $res2 = mysqli_query($con,$sel2);