Question

运行这些脚本时，我的MyPHP数据库中没有显示任何记录。

我正在跑步： APACHE 2.4.7 MYSQL 5.6.15 PHP 5.5.8

首先是HTML代码......

<html>

<center>
  <font face="Helvetica">

<u><b>Matthew Gieger's Guestbook</b></u>

<form action="link.php" method="post"/>
<p>Name: </p>
<input type="text" name="Name" required/>
<p>Email: </p>
<input type="email" name="Email" required />
<p>Message: </p>
<p><textarea rows="4" cols="50" name="Message"> </textarea></p>
<input type="submit" value="Submit" />
<input type="reset" value="Reset" />
</form>
</center>
</html>

和PHP脚本。这就是我认为问题所在的地方......

 <?php

$username='root';
$password='';
$database='guestbook';

$name= $_POST['Name'];
$email= $_POST['Email'];
$message= $_POST['Message'];

new mysqli('localhost',$username,$password,$database) or die("could not connect to localhost");

echo"connected";

mysqli:"insert into contacts (Name,Email,Message,Timestamp) values ($name,$email,$message,date())";

?>

运行代码时没有错误。我只是得到了预期＆＃34;连接＆＃34;

Answer 1

您没有在任何地方定义mysqli个对象。并且你没有在任何地方使用mysqli_query()。

Here are the docs to mysqli.

尝试这样的事情：

$mysqli = new mysqli('localhost', $username, $password, $database);

/* check connection */
if ($mysqli->connect_errno) {
    printf("Connect failed: %s\n", $mysqli->connect_error);
    exit();
}

$mysqli->query("INSERT INTO `contacts`(`Name`, `Email`, `Message`, `Timestamp`) VALUES ('". $name ."', '". $email ."', '". $message ."', '". $timestamp ."')");

Answer 2

你没有正确地实例化mysqli类。您需要将对象实例保存到变量中：

$mysqli = new mysqli('localhost',$username,$password,$database);

/* check connection */
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

您的代码中也存在严重的安全问题，因为您没有转义输入变量。最好的解决方案是研究PDO：http://php.net/manual/en/pdo.construct.php和参数绑定。

如果您真的想使用mysqli执行此操作，则应首先在所有输入变量上使用mysqli_real_escape_string：

$name = mysqli_real_escape_string($mysqli, $_POST['Name']);
$email = mysqli_real_escape_string($mysqli, $_POST['Email']);
$message = mysqli_real_escape_string($mysqli, $_POST['Message']);

然后正确运行您的查询：

$mysqli->query("insert into contacts (Name,Email,Message,Timestamp) values ('$name','$email','$message',".date().")";

如果你没有逃避用户输入，那么用一个简单的SQL injection来破解你的数据库真的很容易。

Answer 3

很可能是格式化问题，请尝试这样：

<?php 
$username='root';
$password='';
$database='guestbook';

$name= $_POST['Name'];
$email= $_POST['Email'];
$message= $_POST['Message'];
$db = new mysqli('localhost',$username,$password,$database) or die("could not connect to localhost");

if($db->connect_errno > 0){
die('Unable to connect to database [' . $db->connect_error . ']');
}
$currentDate=date();
$sql = "insert into contacts (Name,Email,Message,Timestamp) values ($name,$email,$message,$currentDate)";

if(!$result = $db->query($sql)){
die('There was an error running the query [' . $db->error . ']');
}

?>

我无法使用HTML表单和PHP将记录插入数据库

3 个答案: