我无法使用HTML表单和PHP将记录插入数据库
运行这些脚本时，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代码我无法使用HTML表单和PHP将记录插入数据库,php,html,mysql,apache,easyphp,Php,Html,Mysql,Apache,Easyphp,运行这些脚本时，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"/&g
<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>
马修·吉格的留言簿
姓名：
电邮：
信息：
<?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,$message,date())";
?>
“已连接”您没有在任何地方定义
mysqlimysqli\u query（）$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 ."')");
很可能是格式问题，请按如下方式尝试：
<?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 . ']');
}
?>
您没有正确实例化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();
}
$name = mysqli_real_escape_string($mysqli, $_POST['Name']);
$email = mysqli_real_escape_string($mysqli, $_POST['Email']);
$message = mysqli_real_escape_string($mysqli, $_POST['Message']);
$mysqli->query("insert into contacts (Name,Email,Message,Timestamp) values ('$name','$email','$message',".date().")";
如果你不逃避用户的输入，用一个简单的方法就很容易破解你的数据库。试着通过If条件来调试它。成功了！非常感谢你！对不起，我不能投赞成票。。我没有足够的声誉。@MattG也看看Schlaus的答案，以防止出现安全问题（sql注入）。我在向数据库添加第二条记录时仍然遇到问题。似乎不允许超过1条记录。有什么想法吗？@MattG嗯，你有没有设置一个
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