如何使用HTML表单POST方法

我有这样的形式进行编辑或更新数据的信息，但是当我点击该按钮用于更新查询$id值为0如何使用HTML表单POST方法

这里通过从数据库ID号是我的代码：

<form action="edit_crew.php?id=$id" method="POST"> 
      <?php 
      while(mysqli_stmt_fetch($stmt1)) { 
      echo "First Name: <input type=\"text\" style=\"border: none; border-color: transparent;\" value=\"$first_name\" name=\"first_name\"><br/>"; 
      echo "Middle Name: <input type=\"text\" style=\"border: none; border-color: transparent;\" value=\"$middle_name\"><br/>"; 
      echo "Last Name: <input type=\"text\" style=\"border: none; border-color: transparent;\" value=\"$last_name\"><br/>"; 
      echo "Age: <input type=\"text\" style=\"border: none; border-color: transparent;\" value=\"$age\"><br/>"; 
      echo "<button type=\"button\" class=\"btn btn-success\" onclick=\"document.location = 'edit_crew.php?id=$id'; \">Continue</button></center>"; 
      } 
?> 
</form>

这里是我的UPDATE查询代码：

<?php 

include '../session.php'; 
require_once 'config.php'; 


    include 'config.php'; 
    $query_update = "UPDATE `crew_info` SET `first_name` = ?, `middle_name` = ?, `last_name` = ?, `age` = ?, `month` = ?, `day` = ?, `year` = ?, `birth_place` = ?, `gender` = ?, `martial_status` = ?, `religion` = ?, `nationality` = ?, `email_address` = ?, `address_1` = ?, `address_2` = ?, `course` = ?, `school_graduated` = ?, `remarks` = ? WHERE `id` = ?"; 
    $stmt2 = mysqli_prepare($conn, $query_update); 
    $first_name = $_POST['first_name']; 
    mysqli_stmt_bind_param($stmt2, 'sssisiisssssssssssi', $first_name, $middle_name, $last_name, $age, $month, $day, $year, $birth_place, $gender, $martial_status, $religion, $nationality, $email_address, $address_1, $address_2, $course, $school_graduated, $remarks, $_GET['id']); 
    mysqli_stmt_execute($stmt2); 

?>

我想查询是正确的，唯一的问题是id是空

下面是我如何获得我使用的mysqli的$id准备

<?php 

include '../session.php'; 
require_once 'config.php'; 
include 'header.php'; 

    $query1 = "SELECT * FROM `crew_info` WHERE `id` = ?"; 
    $stmt1 = mysqli_prepare($conn, $query1); 
    mysqli_stmt_bind_param($stmt1, 's', $_GET['id']); 
    mysqli_stmt_execute($stmt1); 
    mysqli_stmt_bind_result($stmt1, $id, $first_name, $middle_name, $last_name, $age, $month, $day, $year, $birth_place, $gender, $martial_status, $religion, $nationality, $email_address, $address_1, $address_2, $course, $school_graduated, $remarks, $date_added, $crew_status, $image_name, $passport); 


?>

来源

2016-06-14 Adi

Try'

still'id' is empty – Adi

你是如何得到'$ id'的？显示该代码（在原始问题中）。 – Tigger

变化：

<form action="edit_crew.php?id=$id" method="POST">

要：

<form action="edit_crew.php?id=<?=$_GET['id']?>" method="POST">

来源

2016-06-14 10:09:47 Tigger

不应该因为他在'mysqli_stmt_bind_result（$ stmt1，$ id，...'line中设置'$ id'， – Epodax

呃，实际上这个工作正常 – Adi

尝试使用GET方法，因为你获取的数据来更新“方法=” GET“>

来源

2016-06-14 10:09:32

如何使用HTML表单POST方法

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