Question

我想使用PHP在MySQL中更新我的学生表，但它不起作用，我不知道我犯了什么错误：/

这是我的代码。

BTW我在id周围加上引号，因为varchar不是int。

它只是说数据没有更新。

<?php

// PHP code to update data from MySQL database table

if(isset($_POST['update']))
{

   $hostname = "localhost";
   $username = "root";
   $password = "";
   $databaseName = "students";

   $connect = mysqli_connect($hostname, $username, $password, $databaseName);

   // get values form input text and number

   $id = $_POST['id'];
   $fname = $_POST['fname'];
   $lname = $_POST['lname'];
   $age = $_POST['age'];
   $country=$_POST['country'];
    $phone=$_POST['phone'];
    $email=$_POST['email'];

   // mysql query to Update data
   $query = "UPDATE `students` SET `fname`='".$fname."',`lname`='".$lname."',`Nationality`='".$country."',`PhoneNumber`=$phone,`Email`='".$email."',`age`= $age WHERE `id` = '$id'";

   $result = mysqli_query($connect, $query);

   if($result)
   {
       echo 'Data Updated';
   }else{
       echo 'Data Not Updated';
   }
   mysqli_close($connect);
}

?>

<!DOCTYPE html>

<html>

    <head>

        <title>PHP INSERT DATA USING PDO</title>

        <meta charset="UTF-8">

        <meta name="viewport" content="width=device-width, initial-scale=1.0">

    </head>
    <body>
        <form action="updating.php" method="post">
    <input type="text" name="id"  placeholder="Enter new ID"><br><br>
    <input type="text" name="fname"  placeholder="Enter new First Name"><br><br>
    <input type="text" name="lname"  placeholder="Enter new Last Name"><br><br>
    <input type="number" name="age"  placeholder="Enter new age" min="13" max="90"><br><br>
    <input type="text" name="country"  placeholder="Enter new Nationality"><br><br>
    <input type="number" name="phone"  placeholder="Enter new Phone Number"><br><br>
    <input type="text" name="email"  placeholder="Enter new Email"><br><br>
    <input type="submit" name="update" value="update">
    </form>

    </body>

</html>

Answer 1

我认为您的$query变量很糟糕。查看以下查询。这应该可以解决您的问题。

$query = "UPDATE students SET fname= '$fname', lname = '$lname', Nationality = '$country', PhoneNumber = '$phone', Email= '$email', age = '$age' WHERE id = '$id'";

我认为你的错误是你没有用单引号包裹$age。无论是整数还是字符串，只要需要更新它就需要用单引号来包装它。

Answer 2

您的查询写错了：

应该是：

$query = "UPDATE `students` SET `fname`='" . $fname . "',`lname`='" . $lname . "',`Nationality`='" . $country . "',`PhoneNumber`=" . $phone . ",`Email`='".$email."',`age`= " . $age . " WHERE `id` = '" . $id . "'";

<强> N.B。：假设您的年龄和电话是INT类型数据。

此查询执行$ query后将是：

UPDATE `students` SET `fname`='john',`lname`='adams',`Nationality`='usa',`PhoneNumber`=456123789,`Email`='aa@bb.com',`age`= 23 WHERE `id` = '1'

如果年龄和电话是CHAR，则查询将如下所示：

$query = "UPDATE `students` SET `fname`='" . $fname . "',`lname`='" . $lname . "',`Nationality`='" . $country . "',`PhoneNumber`='" . $phone . "',`Email`='".$email."',`age`= '" . $age . "' WHERE `id` = '" . $id . "'";

此查询执行$ query后将是：

UPDATE `students` SET `fname`='john',`lname`='adams',`Nationality`='usa',`PhoneNumber`='456123789',`Email`='aa@bb.com',`age`= '23' WHERE `id` = '1'

希望它能解决问题。

谢谢和问候

PHP / HTML：MySQL表格没有更新

2 个答案: