提交到另一个页面后从数据库获取ID的数据

Question

I m trying to get data on another page by id which is on showdetails.php page.but im unable to please help. 我正在尝试通过id在showdetails.php页面上的另一页面上获取数据。但是我无法帮忙。

Showdetails.php Showdetails.php
this page shows details of user with id and button which will allow user to edit details on another page 此页面显示具有ID和按钮的用户详细信息，允许用户在另一页面上编辑详细信息

<!DOCTYPE html>
<html>
<head>
    <title>Details</title>
</head>
<body>

<?php 
    require('database.php');          
?>

<h1>User Lists</h1>
<?php 
        $select = "SELECT id, firstname, lastname FROM signup";
        $selectdata = $conn->query($select);

        if ($selectdata->num_rows > 0){
            while($row = mysqli_fetch_array($selectdata)) {

            $id = $row['id'];
            $first = $row['firstname'];
            $last = $row['lastname']; 

?>
    <form method="get" action="editdetails.php">
        <p><b>ID: <?php echo $id; ?></b></p>
        <p>Name: <?php echo $first; ?>&nbsp;<?php echo $last; ?></p>
        <?php 
            $edit = "SELECT id FROM signup WHERE id= '" .$id. "'";
            $selectedit = $conn->query($edit);
        ?>
        <p><input type="submit" name="display" value="Edit Details"></p>
    </form>
<?php
        }
    }
?>
</body>
</html>

editdetails.php editdetails.php
On this page, user will be able to edit details, and i want details by id 在此页面上，用户将可以编辑详细信息，而我希望通过ID获得详细信息

<!DOCTYPE html>
<html>
<head>
    <title>Edit User Details</title>
</head>
<body>

<?php 
    require('database.php');

    $select = "SELECT firstname, lastname, age, phone_no, age, username, password FROM signup";
    $selectdata = $conn->query($select);

    if ($selectdata->num_rows > 0){
            $row = mysqli_fetch_array($selectdata); 

            $first = $row['firstname'];
            $last = $row['lastname']; 
            $age = $row['age'];
            $phone_no = $row['phone_no'];
            $username = $row['username'];
            $password = $row['password'];
    }
?>

<?php
    if (isset($_POST['update'])) {
        # code...       
        $first = mysqli_real_escape_string($conn, $_POST['first']);
        $last = mysqli_real_escape_string($conn, $_POST['last']);
        $age = mysqli_real_escape_string($conn, $_POST['age']);
        $phone = mysqli_real_escape_string($conn, $_POST['phone_no']);
        $username = mysqli_real_escape_string($conn, $_POST['user']);
        $password = mysqli_real_escape_string($conn, $_POST['pass']);

        $update = "UPDATE signup SET firstname= '$first', lastname= '$last', age= '$age', phone_no = '$phone', username = '$username', password = '$password' WHERE id= '$id'";
        $updatedata = $conn->query($update);

        if ($updatedata) {
            # code...
            echo $status = "Details Updated";
        }
        else {
            # code...
            echo $status = "Not Updated";
        }
    }

    if (isset($_POST['delete'])) {
        # code...

        $delete = "DELETE FROM signup WHERE firstname = $first";
        $deletedata = $conn->query($delete);

        if ($deletedata) {
            # code...
            echo $status = "Details Deleted";
        }
        else {
            # code...
            echo $status = "Not Deleted";
        }
    }
?>  

    <h1>Edit Details</h1>
    <form method="post" action= "<?php htmlspecialchars($_SERVER['PHP_SELF']) ?>">
    <p>FirstName: <input type="text" name="first" value="<?php echo $first; ?>"></p>
    <p>LastName: <input type="text" name="last" value="<?php echo $last; ?>"></p>
    <p>Phone no: <input type="number" name="phone_no" value="<?php echo $phone_no; ?>"></p>
    <p>Age: <input type="number" name="age" value="<?php echo $age; ?>"></p>
    <p>User: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<input type="text" name="user" value="<?php echo $username; ?>"></p>
    <p>Password: <input type="password" name="pass" value="<?php echo $password; ?>"></p>
    <p><input type="submit" name="update" value="Update">
    <input type="submit" name="delete" value="Delete"></p>
    </form>
    <p><?php echo $status; ?></p>


</body>
</html>

Thank You. 谢谢。

Answer 1

Add in your HTML form a hidden input, like so: 在您的HTML表单中添加隐藏的输入，如下所示：

<input type="hidden" name="id" value="<?php echo $id; ?>">

Then in your editdetails.php file you can access it with: $_GET["id"] . 然后，可以在您的editdetails.php文件中使用$_GET["id"] 。

Update: 更新：

Add the hidden input to the form in Showdetails.php . 将隐藏的输入添加到Showdetails.php中的表单。 Then in editdetails.php add at the top of the page $id = (int)$_GET["id"]; 然后在editdetails.php中 ，在页面顶部添加$id = (int)$_GET["id"];

Then add to your SELECT query in editdetails.php a WHERE statement for selecting the correct user: 然后将WHERE语句添加到editdetails.php中的SELECT查询中，以选择正确的用户：

$select = "SELECT ... FROM signup WHERE id = $id";

For the update query you are then good to go since you are already using there WHERE id = $id . 对于更新查询，那么您就可以开始了，因为您已经在使用WHERE id = $id 。 (but before your $id variable was not defined) （但在未定义$id变量之前）

提交到另一个页面后从数据库获取ID的数据

问题描述

1 个解决方案

解决方案1
1 已采纳 2018-07-06 08:48:47

提交到另一个页面后从数据库获取ID的数据

问题描述

1 个解决方案

解决方案1 1 已采纳 2018-07-06 08:48:47

解决方案1
1 已采纳 2018-07-06 08:48:47