Php 将数据表单插入mysql表_Php_Mysql_Forms

Php 将数据表单插入mysql表

php mysql forms

Php 将数据表单插入mysql表,php,mysql,forms,Php,Mysql,Forms,谁来告诉我哪里错了 html代码是： <form action="insert.php" method="POST"> <label>Firstname:</label> <input type="text" name="firstname" value="first name" /><br /> <label>Lastname:</label> <input type="text" name="lastn

谁来告诉我哪里错了

html代码是：

<form action="insert.php" method="POST">
<label>Firstname:</label> <input type="text" name="firstname" value="first name" /><br />
<label>Lastname:</label> <input type="text" name="lastname" value="last name" /><br />
<label>email:</label> <input type="text" name ="email" value="email" /><br />


名字：

姓氏：

电子邮件：

php代码是：

<?php
 if(isset($_POST['submit'])!='')
    {
    $fname=$_POST ['firstname'];
    $lname=$_POST ['lastname'];
    $email=$_POST ['email'];
    mysql_connect ("localhost","root","mydatabase") or die (mysql_error ());
    mysql_select_db ("my_db") or die (mysql_error ());
    mysql_query ("ISERT INTO userinfo (`firstname`,`lastname`,`email`)
    VALUES ('$fname','$lname','$email')");
    echo "successfully updated" ;
    }

它是INSERT
而不是ISERT
。。你有个打字错误
mysql_query ("ISERT INTO userinfo (`firstname`,`lastname`,`email`)
              ^^^^^ 

修改代码。。
试试这个
if ($_POST) {
    $fname = $_POST['firstname'];
    $lname = $_POST['lastname'];
    $email = $_POST['email'];
    mysql_connect ("localhost","root","mydatabase") or die (mysql_error ());
    mysql_select_db ("my_db") or die (mysql_error ());
    mysql_query ("INSERT INTO userinfo (firstname,lastname,email)
    VALUES ('$fname','$lname','$email')");
    echo "successfully updated" ;
}

在您的示例中，我没有看到$\u POST['submit']
，您在INSERT
中犯了错误，在“INSERT”中，代码是SQL注入的成熟目标，我建议你尽快阅读这篇文章-你能告诉我如何切换到preparedstatements吗？我是否也应该在html代码中编辑以切换到preparedstatements？不，你不必在html中编辑。下面是如何使用准备好的语句的链接。如果你有任何问题，请回来。
if ($_POST) {
    $fname = $_POST['firstname'];
    $lname = $_POST['lastname'];
    $email = $_POST['email'];
    mysql_connect ("localhost","root","mydatabase") or die (mysql_error ());
    mysql_select_db ("my_db") or die (mysql_error ());
    mysql_query ("INSERT INTO userinfo (firstname,lastname,email)
    VALUES ('$fname','$lname','$email')");
    echo "successfully updated" ;
}