如何使用php在下一頁中顯示與單擊元素相關的信息

Question

$sql = "SELECT client_id, firm_name FROM savedata";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        echo "<br> Client_id: ". $row["client_id"]. "<br> <a href ='view.php'>Client Name: ". $row["firm_name"]." </a>";
    }
} else {
    echo "0 results";
}

我正在使用此代碼從我的數據庫中顯示我的客戶名稱，我想在view.php中顯示與單擊的客戶相關的所有信息。

謝謝

Answer 1

更改此代碼

echo "<br> Client_id: ". $row["client_id"]. "<br> <a href ='view.php'>Client Name: ". $row["firm_name"]." </a>";

至

echo "<br> Client_id: ". $row["client_id"]. "<br> <a href ='view.php?cid=".$row["client_id"]."'>Client Name: ". $row["firm_name"]." </a>";

然后在view.php

$sql = "SELECT client_id, firm_name FROM savedata where client_id=".$_GET['cid'];
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    $row = $result->fetch_assoc();
    echo "<br> Client_id: ". $row["client_id"]. "<br> Client Name: ". $row["firm_name"];
 }

Answer 2

<a href ='view.php?clientId=".$row["client_id"]."'>Client Name: ". $row["firm_name"]." </a>

view.php：

if(isset($_GET[`clientId`]) {
    $clientId=$_GET[`clientId`];
    $sql = "SELECT * FROM savedata WHERE id=`$clientId`";
    ....
}else{
    ...
}