Question

我阅读了关于未定义索引错误的所有答案，但对我来说没有帮助，因为我已经使用isset函数来检查plz如何解决这个问题..

<?php
$con=mysqli_connect("localhost","root","","contact");
if (mysqli_connect_errno())
{
    echo "failed".mysqli_connect_error();
    }

检查提交的数据

if(isset($_POST['submit']))
        {
        $name=$_POST['name']; 
        $website=$_POST['website']; 
        $gender=$_POST['gender']; 
        $comment=$_POST['comment']; 
        }
        $sql="insert into form(name,website,gender,comment) Values('$_POST[name]','$_POST[website]','$_POST[gender]','$_POST[comment]')";

        if(!mysqli_query($con,$sql))
            {
                die('error:'.mysqli_error($con));
            }
        else "added";

        mysqli_close($con);
        ?>
    <html>
        <body>
            <form method=post action="<?php echo $_SERVER['PHP_SELF']; ?>">
                Name: <input type="text" name="name"><br>
                E-mail: <input type="text" name="email"><br>
                Website: <input type="text" name="website"><br>
                <input type="radio" name="gender" value="female">Female
                <input type="radio" name="gender" value="male">Male<br>
                Comment: <textarea name="comment" rows="5" cols="40"></textarea>
                 <input type=submit name="submit"><br>
            </form>
        </body>

这些错误来自

Notice: Undefined index: name in H:\Wamp\Xamp\htdocs\form.php on line 15

Notice: Undefined index: website in H:\Wamp\Xamp\htdocs\form.php on line 15

Notice: Undefined index: gender in H:\Wamp\Xamp\htdocs\form.php on line 15

Notice: Undefined index: comment in H:\Wamp\Xamp\htdocs\form.php on line 15

Answer 1

更新此插入查询

$sql="insert into form(name,website,gender,comment) values('". $name ."','". $website ."','". $gender ."','". $comment ."')";

希望这对你有帮助！

Answer 2

请尝试以下更正的代码：

if(isset($_POST['submit']))
        {
        $name=isset($_POST['name']) ? $_POST['name'] : ''; 
        $website=isset($_POST['website']) ? $_POST['website'] : ''; 
        $gender=isset($_POST['gender']) ? $_POST['gender'] : ''; 
        $comment=isset($_POST['comment']) ? $_POST['comment'] : ''; 
        $sql="insert into form(name,website,gender,comment) Values('$name','$website','$gender','$comment')";

        // Open the database connection here
        // aka, mysqli_connect()


        if(!mysqli_query($con,$sql))
            {
                die('error:'.mysqli_error($con));
            }
        else "added";

        mysqli_close($con);
        }
        ?>
    <html>
        <body>
            <form method=post action="<?php echo $_SERVER['PHP_SELF']; ?>">
                Name: <input type="text" name="name"><br>
                E-mail: <input type="text" name="email"><br>
                Website: <input type="text" name="website"><br>
                <input type="radio" name="gender" value="female">Female
                <input type="radio" name="gender" value="male">Male<br>
                Comment: <textarea name="comment" rows="5" cols="40"></textarea>
                 <input type=submit name="submit"><br>
            </form>
        </body>

我所做的是添加验证来检查这些字段是否已设置，如果是，则设置值，如果没有，则将变量（又名$ name）设置为''。如果必填字段=''（等于空白），您应该添加一些进一步的验证。

我还调整了你的查询以不使用$ _POST变量，而是使用你分配$ _POST值的变量，所以你知道它们确实存在。

最后，我将mysql连接代码和查询本身移动到if（isset（submit））语句中，因此它不会尝试处理那些尚未提交表单的常规页面加载。

Answer 3

更改

if(isset($_POST['submit']))
        {
        $name=$_POST['name']; 
        $website=$_POST['website']; 
        $gender=$_POST['gender']; 
        $comment=$_POST['comment']; 
        }
        $sql="insert into form(name,website,gender,comment) Values('$_POST[name]','$_POST[website]','$_POST[gender]','$_POST[comment]')";
        if(!mysqli_query($con,$sql))
        {
            die('error:'.mysqli_error($con));
        }
       else "added";

       mysqli_close($con);
       ?>

到

if(isset($_POST['submit']))
{
        $name=$_POST['name']; 
        $website=$_POST['website']; 
        $gender=$_POST['gender']; 
        $comment=$_POST['comment']; 
        $sql="insert into form(name,website,gender,comment) values ('$name','$website','$gender','$comment')";
        if(!mysqli_query($con,$sql))
        {
            die('error:'.mysqli_error($con));
        }
        else "added";
        mysqli_close($con);

  }?>

Answer 4

比朋友问题更难解决怎么了？

if(isset($_POST['submit']))
        {
        $name=$_POST['name']; 
        $website=$_POST['website']; 
        $gender=$_POST['gender']; 
        $comment=$_POST['comment']; 


        $sql="insert into form(name,website,gender,comment) Values('". $name . "','" . $website . "','" . $gender . "','" . $comment . "')";

//这些也在if（isset（））Block

中

if(!mysqli_query($con,$sql))
            {
                die('error:'.mysqli_error($con));
            }
        else "added";
        }

thanx to all

未定义的索引表单错误

4 个答案: