无法使用PHP和J Query Mobile将数据插入数据库

I can insert data into the database for the several time before this. I'm using web matrix and XAMPP to develop. I have problem while dealing with ports but then after altering something it become okay, but the database does not receive data anymore. My other system using PHP that was been develop is running good but the problem is with the jQuery mobile.

我可以在此之前的几次将数据插入数据库。我正在使用web矩阵和XAMPP来开发。我在处理端口时遇到问题但是在更改了某些内容后它变得没问题,但数据库不再接收数据了。我开发的另一个使用PHP的系统运行良好,但问题在于jQuery mobile。

This is the index.html file

这是index.html文件

<!DOCTYPE html>
<html>
<head>
    <meta charset="utf-8" />
    <meta name="viewport" content="width=device-width, initial-scale=1" />
    <meta name="apple-mobile-web-app-capable" content="yes" />
    <meta name="apple-mobile-web-app-status-bar-style" content="black" />
    <title>
    </title>
    <link rel="stylesheet" href="http://code.jquery.com/mobile/1.2.0/jquery.mobile-1.2.0.min.css" />
    <link rel="stylesheet" href="my.css" />
    <script src="http://code.jquery.com/jquery-1.7.2.min.js">
    </script>
    <script src="http://code.jquery.com/mobile/1.2.0/jquery.mobile-1.2.0.min.js">
    </script>
    <script src="my.js">
    </script>
    <!-- User-generated css -->
    <style>
    </style>
    <!-- User-generated js -->
    <script>
        try {

  $(function() {

});

} catch (error) {
  console.error("Your javascript has an error: " + error);
 }
     </script>
 </head>
 <body>
    <!-- Home -->
    <div data-role="page" id="page1">
        <div data-theme="a" data-role="header" data-position="fixed">
            <h3>
               Student Uitm
            </h3>
        </div>
        <div data-role="content">
            <form action="http://localhost/student/save2db.php" method="post">
                <label>Nama</label><input type="text" name="Nama">
                <label>No Kad Pengenalan</label><input type="text" maxlength="12" name="icno">
                <label>Jantina</label><select name="jantina">
                <option value="L">Lelaki</option>
                <option value="P">Perempuan</option>
            </select>
                <input name="sbt" type="submit" value="save" data-inline="true">
            </form>
        </div>
        <div data-theme="a" data-role="footer" data-position="fixed">
            <h3>
                Footer
            </h3>
        </div>
    </div>
</body>
</html>

and this is the save2db.php file:

这是save2db.php文件:

 <?php
 $con=mysqli_connect("localhost","root"," ","student");
    // Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$sql="INSERT INTO student1 (Nama, icno, jantina)
VALUES
('$_POST[Nama]','$_POST[icno]','$_POST[jantina]')";

 if (!mysqli_query($con,$sql))
 {
  die('Error: ' . mysqli_error($con));
  }
 echo "1 record added";

mysqli_close($con);  
?>

1 个解决方案

#1

The problems:

you check for the db connection, but you let the script run through even when it's failed

检查数据库连接,但是即使脚本失败也让脚本运行

you are not being cautious enough with the received data

你对收到的数据不够谨慎

you don't log errors

你不记录错误

internal problem symptoms leak out to users

内部问题症状泄露给用户

Here's a possible alternative to your script taking care of these issues:

以下是处理这些问题的脚本的替代方法:

<?php
$con = mysqli_connect("localhost","root"," ","student");
 // Check connection
if (mysqli_connect_errno()) {
    trigger_error("Failed to connect to MySQL: " . mysqli_connect_error(), 
                   E_USER_ERROR);
    die("unable to complete request");
} else if (!(isset($_POST['Nama']) &&
         isset($_POST['icno']) && isset($_POST['jantina'])) {
    trigger_error( "incomplete POST data", E_USER_ERROR);
    die("unable to complete request");
} else {
    $nama = mysqli_real_escape_string($con, $_POST['Nama']);
    $icno = mysqli_real_escape_string($con $_POST['icno']);
    $jantina = mysqli_real_escape_string($con,$_POST['jantina']);
    $sql = "INSERT INTO student1 (Nama, icno, jantina) VALUES ('$nama','$icno','$jantina')";
    if (!mysqli_query($con,$sql)) {
        trigger_error("query failed :" . mysqli_error($con), E_USER_ERROR);
        die("unable to complete request");
    }
    echo "1 record added";
    mysqli_close($con);  
}

With the above modification done, check the php logs after unsuccessful updates to know the possible reasons of the failure.

完成上述修改后,请在更新失败后检查php日志,以了解失败的可能原因。

#1