无法将html5表单连接到我的sql数据库？

debugcn 发表于 Dev

古萨曼·普拉布（Gowthaman Prabhu）

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的密码为“ mysql”。

提交表单时，出现以下错误：

“致命错误：未被捕获的错误：调用C：\ xampp \ htdocs \ test1 \ SQLConnectionProcess.php：3中未定义的函数mysql_connect（）：堆栈跟踪：＃0 {main}抛出于C：\ xampp \ htdocs \ test1 \ SQLConnectionProcess中。第3行的php”

请帮助我。谢谢....

夏拉

请使用PDO或MySQLi。mysql已贬值，不应在新代码上使用

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

尝试此链接，对我有很大帮助：@ Your-Common-Sense提供的phpdelusions.net/pdo。

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应该如下所示：

    <?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。如果失败，则整个代码将失败。以此为起点。

谢谢。

本文收集自互联网，转载请注明来源。

如有侵权，请联系[email protected] 删除。