我需要帮助让我的 HTML 表单将数据提交到我的数据库 (mysql)。数据库连接正常，但似乎无法将数据桥接到数据库中。我使用 Notepad++ 作为我的文本编辑器和测试器。这是为了在类里面提前完成一个项目，我需要了解如何为它做这个。
HTML
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Add Record Form</title>
</head>
<body>
<form action="insert.php" method="post">
<p>
<label for="firstName">TechID:</label>
<input type="text" name="techid" id="TechID">
</p>
<p>
<label for="lastName">First Name:</label>
<input type="text" name="firstname" id="FirstName">
</p>
<p>
<label for="emailAddress">Last Name:</label>
<input type="text" name="lastname" id="LastName">
</p>
<p>
<label for="emailAddress">Phone:</label>
<input type="text" name="phone" id="Phone">
</p>
<p>
<label for="emailAddress">Email:</label>
<input type="text" name="email" id="Email">
</p>
<p>
<label for="emailAddress">State:</label>
<input type="text" name="state" id="State">
</p>
<p>
<label for="emailAddress">Address:</label>
<input type="text" name="address" id="Address">
</p>
<p>
<label for="emailAddress">Zipcode:</label>
<input type="text" name="zipcode" id="Zipcode">
</p>
<p>
<label for="emailAddress">Date:</label>
<input type="text" name="date" id="Date" placeholder="EX: 2017-7-25">
</p>
<p>
<label for="emailAddress">Course:</label>
<input type="text" name="course" id="Course">
</p>
<p>
<label for="emailAddress">Request:</label>
<input type="text" name="request" id="Request">
</p>
<input class="submit" name="submit" type="submit" value="Insert">
</form>
</body>
</html>
PHP
<?php
$link = mysqli_connect("localhost", "root", "", "student_request");
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
if(isset($_POST['submit'])){
$techid = $_POST['techid'];
$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$phone = $_POST['phone'];
$email = $_POST['email'];
$state = $_POST['state'];
$address = $_POST['address'];
$zipcode = $_POST['zipcode'];
$date = $_POST['date'];
$course = $_POST['course'];
$request = $_POST['request'];
$sql = "INSERT INTO student (TECH_ID, FIRST_NAME, LAST_NAME, PHONE_NUM, EMAIL, STATE, ADDRESS, ZIPCODE, DATE, COURSE, REQUEST_TYPE) VALUES ('$techid','$firstname','$lastname','$phone','$email','$state','$address','$zipcode','$date','$course','$request')";
if(mysqli_query($link, $sql)) {
echo "Records inserted successfully.";
} else {
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
}
mysqli_close($link);
?>
student表的表结构:-
CREATE TABLE `student` (
`REQUEST_ID` int(255) NOT NULL,
`TECH_ID` int(11) NOT NULL,
`FIRST_NAME` varchar(255) NOT NULL,
`LAST_NAME` varchar(255) NOT NULL,
`PHONE_NUM` varchar(255) NOT NULL,
`EMAIL` varchar(255) NOT NULL,
`STATE` varchar(255) NOT NULL,
`ADDRESS` varchar(255) NOT NULL,
`ZIPCODE` varchar(255) NOT NULL,
`DATE` date NOT NULL,
`COURSE` varchar(255) NOT NULL,
`REQUEST_TYPE` text NOT NULL
)
最佳答案
@Donald 在这里和你提问的类似例子。这一定会对您有所帮助。
祝你的项目好运
HTML代码:
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Add Record Form</title>
</head>
<body>
<form action="insertrecords.php" method="post">
<p>
<label for="FirstName">First Name:</label>
<input type="text" name="firstname" id="FirstName">
</p>
<p>
<label for="LastName">Last Name:</label>
<input type="text" name="lastname" id="LastName">
</p>
<p>
<label for="Email">Email:</label>
<input type="text" name="email" id="Email">
</p>
</p>
<input class="submit" name="submit" type="submit" value="Insert">
</form>
</body>
</html>
PHP代码:insertrecords.php
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "demo";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
if(isset($_POST['submit'])){
//USE MYSQLI_REAL_ESCAPE_STRING() TO ESCAPE SINGLE QUOTES
// AND AGAINST SQL INJECTION
$firstname = mysqli_real_escape_string($conn, $_POST['firstname']);
$lastname = mysqli_real_escape_string($conn, $_POST['lastname']);
$email = mysqli_real_escape_string($conn, $_POST['email']);
$sql = "INSERT INTO MyGuests (firstname, lastname, email)
VALUES ('$firstname', '$lastname', '$email')";
if (mysqli_query($conn, $sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
mysqli_close($conn);
}
?>
我还建议您开始学习 MYSQLI->PREPARED 声明，以便更安全地对抗 SQL 注入(inject)。下面是与上面相同的示例，但是使用了 MYSQLI->PREPARED 语句和 参数化查询。
<?php
$sql = $conn->stmt_init();
$query = "INSERT INTO MyGuests (firstname, lastname, email)
VALUES (?,?,?)";
if($sql->prepare($query)){
$sql->bind_param('sss',$firstname,$lastname,$email);
$sql->execute();
echo "New record successfully inserted";
}
else
{
echo "Error inserting the record".$conn->error;
}
?>
尝试代码。欢迎提问
关于php - HTML 表单不通过 php 将数据提交到我的数据库，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/44194080/