PHP创建表单将无法识别索引

Question

我正在创建一个使用HTML / PHP和mySQL应用CRUD功能的表单。我可以删除，阅读和更新，但是无法创建新记录。以下是我的数据库结构，索引页面和创建页面。任何指导表示赞赏。我没有包含配置文件，因为我认为没有必要。数据库名称谢谢！

确切错误

注意：未定义的索引：在第30行的C：\ xampp \ htdocs \ Update \ create.php中排名

create.php

// Include config file
require_once "config.php";

// Define variables and initialize with empty values
$nickname = $lastsubmission = $rank = "";
$nickname_err = $lastsubmission_err = $rank_err = "";

// Processing form data when form is submitted
if($_SERVER["REQUEST_METHOD"] == "POST"){
    // Validate name
    $input_nickname = trim($_POST["nickname"]);
    if(empty($input_nickname)){
        $nickname_err = "Please enter a name.";
    } elseif(!filter_var($input_nickname, FILTER_VALIDATE_REGEXP, array("options"=>array("regexp"=>"/^[a-zA-Z\s]+$/")))){
        $nickname_err = "Please enter a valid name.";
    } else{
        $nickname = $input_nickname;
    }

    // Validate address
    $input_lastsubmission = trim($_POST["lastsubmission"]);
    if(empty($input_lastsubmission)){
        $lsstsubmission_err = "Please enter an address.";     
    } else{
        $lastsubmission = $input_lastsubmission;
    }

    // Validate salary
    $input_rank = trim($_POST["rank"]);
    if(empty($input_rank)){
        $rank_err = "Please enter the salary amount.";     
    } elseif(!ctype_digit($input_rank)){
        $rank_err = "Please enter a positive integer value.";
    } else{
        $rank = $input_rank;
    }

Answer 1

HTML表单中的

您的昵称元素需要 name 属性。可能是复制/粘贴错误？

Same适用于所有其他HTML输入元素。

将nickname="nickname"更改为name="nickname"。

<input type="text" name="nickname" class="form-control" value="<?php echo $nickname; ?>">

Answer 2

这是因为在将$ param_nickname与mysqli_stmt_bind_param绑定后分配了$ param_nickname

//First Set parameters
$param_nickname = $nickname;
$param_lastsubmission = $lastsubmission;
$param_rank = $rank;

//Then Bind variables to the prepared statement as parameters
mysqli_stmt_bind_param($stmt, "sss", $param_nickname, 
$param_lastsubmission, $param_rank);

应该可以解决问题