dongshao5573 2013-12-13 11:03
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PHP错误。 “Undefined index”表单中的数据不会插入到mysql数据库中

it is a error in the VALUES in the php. It says it is a undefined index in every single value. I have tried for a long time now, but i can't figure out what the problem is. is it in my database or inside the code? Please help me :)

<html>
<body>
<h1>A small example page to insert some data in to the MySQL database using PHP</h1>
<form action="insert.php" method="post">
Firstname: <input type="text" name="navn" /><br><br>
Lastname: <input type="text" name="adresse" /><br><br>

<input type="submit" />
</form>

<?php
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

mysql_select_db("forum", $con);

$sql="INSERT INTO bruker (navn, adresse)
VALUES
('$_POST[navn]',
'$_POST[adresse]')";

if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
echo "1 record added";

mysql_close($con)
?>

</body>
</html>
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  • duanchensou8685 2013-12-13 11:08
    关注

    your post value will work once submit button is clicked

    <form action="" method="post">

<input type="submit" name="submit" />

<?php
if(isset($_POST['submit']))
{
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

mysql_select_db("forum", $con);

$sql="INSERT INTO bruker (navn, adresse)
VALUES
('$_POST[navn]',
'$_POST[adresse]')";

if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
echo "1 record added";

mysql_close($con)
}
?>

    you can also hide your warning by using

    error_reporting(0);

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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