Question

请温柔地对待我，我只是一个新手。请告诉我为什么这个功能不起作用的错误。没有错误，只有空白页。

<?php 

include("db.php");

$id = (integer)$_GET['id'];
echo $id;

function updateshowdata($stmt, $con){

$stmt = $con->prepare("SELECT `firstname` FROM `MyGuests` where id =?");
$stmt->bind_param ('i',$id);
$stmt->execute ();
$stmt->bind_result ($firstname);


$updateshow = "<form method='post' action='updateprocess.php'>";

while ($stmt->fetch()) {
$updateshow .= "<input type='text' name='firstname'  value='{$firstname}' >";

}

$updateshow .= "</form>";

echo $updateshow;

}
updateshowdata($stmt, $con);

?>

如果我将其更改为下面的删除功能，它将开始工作

<?php 

include("db.php");

$id = (integer)$_GET['id'];
echo $id;

$stmt = $con->prepare("SELECT `firstname` FROM `MyGuests` where id =?");
$stmt->bind_param ('i',$id);
$stmt->execute ();
$stmt->bind_result ($firstname);


$updateshow = "<form method='post' action='updateprocess.php'>";

while ($stmt->fetch()) {
$updateshow .= "<input type='text' name='firstname'  value='{$firstname}' >";

}

$updateshow .= "</form>";

echo $updateshow;


?>

Answer 1

有效吗？

function updateshowdata($stmt, $con){
    $id = (int)$GET['id'];

    $stmt = $con->prepare("SELECT `firstname` FROM `MyGuests` where id =?");
    $stmt->bind_param ('i',$id);
    $stmt->execute ();
    $stmt->bind_result ($firstname);    


    $updateshow = "<form method='post' action='updateprocess.php'>";

    while ($stmt->fetch()) {
        $updateshow .= "<input type='text' name='firstname'      value='{$firstname}' >";

}

函数不使用表单元素

1 个答案: