Question

我想将两个股票（用户输入）和id传递给一个网址我试试这个未定义索引：股票
是我传递输入值错误的方式？

<?php



session_start();
$sql1 = "SELECT * FROM tbl_customers WHERE customers_id='1'";

$sql = "SELECT * FROM tbl_products";

$get = mysqli_query($conn, $sql1) or die(mysqli_error($conn));
$row = mysqli_fetch_array($get);

$customerName = $row['customer_email'];
echo "Welcome $customerName";



$data= mysqli_query($conn, $sql) or die(mysqli_error($conn));


while ($row = mysqli_fetch_assoc($data)) {
    $pid = $row['pid'];
    echo "<form action='' method='GET'>

 <input type=text name=stock value=1><br>
<a href='stock.php?id=" . $pid . "?stock=" . $_GET['stock'] . "'> Add</a>

                      </form>";




}

?>

谢谢大家的帮助，这就是我所做的修复错误我的获取值总是空的如果我不按回车

echo "<form action='' method='GET'>

                     <input type=text name=stock value=1><br>

                             </form>";
    $_GET['stock'] = 1;
    if (isset($_GET["stock"])) {
        $x = $_GET['stock'];
        echo "<a href='stock.php?id=" . $pid . "&stock=" . $_GET['stock'] . "'> Add</a>";
    }

Answer 1

只是改变？与＆amp;在股票之前。

  echo "<form action='' method='GET'><input type=text name=stock value=1><br>
<a href='stock.php?id=" . $pid . "&stock=" . $_GET['stock'] . "'> Add</a> </form>";

Answer 2

<a href="/php/event-detail.php?event_id=$event_id">

这是您如何使用将值传递到另一个页面。你可以参考这个http://w3schools.invisionzone.com/index.php?showtopic=48611

PHP：如何在url

2 个答案: