无法将 html5 表单连接到我的 sql 数据库？

Question

我有一个 HTML5 文档用于请求用户输入表单（保存为 employee.php）。

我还创建了一个 php 文档 (SQLConnectionProcess.php) 用于将表单连接到本地数据库。我正在使用 XAMPP 和 PHPmyAdmin。

employee.php:

<html>
<body>

<form name="EmployeeDatabase" action="SQLConnectionProcess.php" method="post">

<link rel="stylesheet" href="css.css">

<h1>EMPLOYEE DATABASE</h1>

Employe Card NO: <input type="text" name="cardNO" ><br><br>
Employee NO: <input type="text" name="employeeNO" ><br><br>
Employee Name: <input type="text" name="employeename"><br><br>
Nationality: <input type="text" name="nationality"><br><br>
Profession: <input type="text" name="profession"><br><br>
DOB: <input type="text" name="DOB"><br><br>
DOJ: <input type="text" name="DOJ"><br><br>
DOA(VisitVisa): <input type="text" name="DOA"><br><br>
Company Code: <input type="text" name="companycode"><br><br>
Sponsor Code: <input type="text" name="sponsorcode"><br><br>
Visa Type: <input type="text" name="visatype"><br><br>
Status: <input type="text" name="status"><br><br>

<input type="submit" name="formSubmit" value="Submit">

<?php

?>

</form>

</body>
</html>

SQLConnectionProcess.php:

<?php

$con = mysql_connect('localhost','root','mysql');
mysql_select_db('employee_info',$con);


if(isset($_POST['formSubmit'])){
  $cardNO= isset($_POST['cardNO']) ? $_POST['cardNO'] : 0;
  $employeeNO= isset($_POST['employeeNO']) ? $_POST['employeeNO'] : 0;
  $employeename= isset($_POST['employeename']) ? $_POST['employeename'] : "";
  $nationality= isset($_POST['nationality']) ? $_POST['nationality'] : "";
  $profession= isset($_POST['profession']) ? $_POST['profession'] : "";
  $DOB= isset($_POST['DOB']) ? $_POST['DOB'] : "";
  $DOJ= isset($_POST['DOJ']) ? $_POST['DOJ'] : "";
  $DOA= isset($_POST['DOA']) ? $_POST['DOA'] : "";
  $companycode = isset($_POST['companycode']) ? $_POST['companycode'] : 0;
  $sponsorcode= isset($_POST['sponsorcode']) ? $_POST['sponsorcode'] : 0;
  $visatype= isset($_POST['visatype']) ? $_POST['visatype'] : "";
  $status= isset($_POST['status']) ? $_POST['status'] : "";


  $sql = "INSERT INTO employee_info info(EmployeeCardNO,EmployeeNO,EmployeeName,Nationality,Profession,DOB,DOJ,DOA,CompanyCode,SponsorCode,VisaType,Status) VALUES ($cardNO,$employeeNO,$employeename,$nationality,$profession,$DOB,$DOJ,$DOA,$companycode,$sponsorcode,$visatype,$status)";
  mysql_query($sql);
}
?>

PHPmyAdmin 密码是 "mysql"。

当我提交表格时出现以下错误：

"Fatal error: Uncaught Error: Call to undefined function mysql_connect() in C:\xampp\htdocs\test1\SQLConnectionProcess.php:3 Stack trace: #0 {main} thrown in C:\xampp\htdocs\test1\SQLConnectionProcess.php on line 3"

请帮助我。谢谢....

Answer 1

请使用 PDO 或 MySQLi。 mysql 已贬值，不应用于新代码

http://php.net/manual/en/function.mysql-query.php

试试这个 link，它对我帮助很大：phpdelusions。net/pdo @Your-Common-Sense。

我会这样编码：

1: 数据库连接文件:

    <?php

    $db = new PDO('mysql:host=yourhost;dbname=dbname', 'username', 'password', array(PDO::MYSQL_ATTR_INIT_COMMAND => "SET NAMES 'UTF8'")); 

    $db->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
    $db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

    ?>

您的 SQLConnectionProcess.php 应如下所示：

    <?php

require ('path/to/db/con.php');

if(isset($_POST['formSubmit'])){
        $cardNO         =   isset($_POST['cardNO']) ? $_POST['cardNO'] : 0;
        $employeeNO     =   isset($_POST['employeeNO']) ? $_POST['employeeNO'] : 0;
        $employeename   =   isset($_POST['employeename']) ? $_POST['employeename'] : "";
        $nationality    =   isset($_POST['nationality']) ? $_POST['nationality'] : "";
        $profession     =   isset($_POST['profession']) ? $_POST['profession'] : "";
        $DOB            =   isset($_POST['DOB']) ? $_POST['DOB'] : "";
        $DOJ            =   isset($_POST['DOJ']) ? $_POST['DOJ'] : "";
        $DOA            =   isset($_POST['DOA']) ? $_POST['DOA'] : "";
        $companycode    =   isset($_POST['companycode']) ? $_POST['companycode'] : 0;
        $sponsorcode    =   isset($_POST['sponsorcode']) ? $_POST['sponsorcode'] : 0;
        $visatype       =   isset($_POST['visatype']) ? $_POST['visatype'] : "";
        $status         =   isset($_POST['status']) ? $_POST['status'] : "";


        $stmt       =   $db->prepare("INSERT INTO employee_info

                                        (EmployeeCardNO,
                                        EmployeeNO,
                                        EmployeeName,
                                        Nationality,
                                        Profession,
                                        DOB,
                                        DOJ,
                                        DOA,
                                        CompanyCode,
                                        SponsorCode,
                                        VisaType,
                                        Status)

                                        VALUES 

                                        ($cardNO,
                                        $employeeNO,
                                        $employeename,
                                        $nationality,
                                        $profession,
                                        $DOB,
                                        $DOJ,
                                        $DOA,
                                        $companycode,
                                        $sponsorcode,
                                        $visatype,
                                        $status)"

                                    );

        $stmt->execute();

        }
else{

    //something went wrong

}

?>

你有很多isset()条件。如果一个失败整个代码失败。试试这个作为起点。

谢谢。

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