将单独的HTML和PHP MYSQL表单合并为一个文件

Question

我需要的原因很复杂，但我有一个当前正在工作的HTML表单，它将一行插入到我的MYSQL数据库中。但是...我需要将这两个单独的进程合并到一个HTML文件中，但经过许多尝试安排代码后，无法找到可行的解决方案。希望有人能指出我正确的方向。

下面是HTML文件：

<form method="post" action="process.php"> 
<input type="text" name="id" placeholder="Enter ID" /><br /> 
<input type="hidden" name="user_text" id="hiddenField" value="x" /><br /> 
<input type="submit" value="Submit" /> 
</form> 

下面是相关process.php文件：

<?php 
if ($_SERVER["REQUEST_METHOD"] == "POST") { 

    //mysql credentials 
    $mysql_host = "localhost"; 
    $mysql_username = "*"; 
    $mysql_password = "*"; 
    $mysql_database = "*"; 

    $u_name = $_POST["id"]; 
    $u_text = $_POST["user_text"]; 

    if (empty($u_name)){ 
     die("Please enter your id"); 
    } 

    //Open a new connection to the MySQL server 
    $mysqli = new mysqli($mysql_host, $mysql_username, $mysql_password, $mysql_database); 

    //Output any connection error 
    if ($mysqli->connect_error) { 
     die('Error : ('. $mysqli->connect_errno .') '. $mysqli->connect_error); 
    } 

    $statement = $mysqli->prepare("INSERT INTO users_data (contractor_id, status) VALUES(?, ?)"); //prepare sql insert query 
    $statement->bind_param('ss', $u_name, $u_text); //bind values and execute insert query 


} 
?> 

我知道这是不是做的事情以正确的方式，但它不是对于公共场所而言，更多是个人记录项目。先进的谢谢！

Answer 1

你可以简单地有一个这样的文件：

<?php 

if ($_SERVER["REQUEST_METHOD"] == "POST") { 
    process(); 
} 
else { 
    show_form(); 
} 


function show_form() { 
?> 
<form method="post"> 
    <input type="text" name="id" placeholder="Enter ID" /><br /> 
    <input type="hidden" name="user_text" id="hiddenField" value="x" /><br /> 
    <input type="submit" value="Submit" /> 
</form> 
<?php 
} 

function process() { 

    //mysql credentials 
    $mysql_host = "localhost"; 
    $mysql_username = "*"; 
    $mysql_password = "*"; 
    $mysql_database = "*"; 

    $u_name = $_POST["id"]; 
    $u_text = $_POST["user_text"]; 

    if (empty($u_name)){ 
     die("Please enter your id"); 
    } 

    //Open a new connection to the MySQL server 
    $mysqli = new mysqli($mysql_host, $mysql_username, $mysql_password, $mysql_database); 

    //Output any connection error 
    if ($mysqli->connect_error) { 
     die('Error : ('. $mysqli->connect_errno .') '. $mysqli->connect_error); 
    } 

    $statement = $mysqli->prepare("INSERT INTO users_data (contractor_id, status) VALUES(?, ?)"); //prepare sql insert query 
    $statement->bind_param('ss', $u_name, $u_text); //bind values and execute insert query 


} 

如果不设置action的形式，在默认情况下，它提交给自己