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Posted by Luigi Donatello Asero on 10/13/21 11:22
"Hilarion" <hilarion@SPAM.op.SMIECI.pl> skrev i meddelandet
news:dc5eod$73f$1@news.onet.pl...
> Luigi Donatello Asero wrote:
> >> > The page https://www.scaiecat-spa-gigi.com/sv/form.php shows a form
but
> Something tells me that you havent checked the obvious possible causes.
> Check if your table is named "Formular" or "Formulδr" (those are
> totally different words for MySQL).
I have already done it.
But I do not see the data in the form.
Please do not answer, if you do not want to.
The code follows:
<?php
if(isset($_POST['add']) && ($_POST['add']=="Skicka")) {
$fornamn = $_POST['fornamn'];
$efternamn = $_POST['efternamn'];
$db = mysql_connect("local host", "user", "password")
or die('I could not connect');
mysql_select_db("scaiecat_?", $db);
$query="INSERT INTO Formular('Fornamn','Efternamn')
VALUES($Fornamn','$Efternamn')";
mysql_query($query) or die('Error, insert query failed');
}?>
<html>
<body>
<form method="post">
<?php
echo
"<LABEL FOR='Fornamn'>Fφrnamn</LABEL>";
?>
<INPUT TYPE="TEXT" NAME='Fnamn' id='Fornamn'><br>
<?php
echo
"<LABEL FOR='Efternamn'>Efternamn</LABEL>";
?>
<INPUT TYPE="TEXT" NAME='Efternamn' id='Efternamn'><br>
<INPUT TYPE=SUBMIT VALUE=Skicka>
</FORM> </form>
</body>
</html>
--
Luigi Donatello (un italiano che vive in Svezia)
https://www.scaiecat-spa-gigi.com/sv/francavilla-di-sicilia.php
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