According to the code provided.
Changes
- Change
<input type="button" value="Ilmoita">
to <input type="submit" name="Ilmoita" value="Ilmoita">
- Change
if(isset( $_POST["tapahtuma"]) ) {
to if(isset( $_POST["Ilmoita"]) ) {
- Change
$link = mysql_connect("localhost","root","","db");
to $link = mysqli_connect("localhost","root","","db");
- Put
$auts = mysql_real_escape_string($_POST['auts']);
before mysqli_query
Updated Code
<form method="post" name="tapahtuma">
<input type="text" name="auts">
Submit: <input type="submit" name="Ilmoita" value="Ilmoita">
</form >
<?php
if(isset( $_POST["Ilmoita"]) ) {
$link = mysqli_connect("localhost","root","","db");
$auts = mysql_real_escape_string($_POST['auts']);
mysqli_query($link, "INSERT INTO table (column) VALUES ('".$auts."')");
mysqli_close($link);
}?>
I will suggest you to use prepared statements to avoid SQL Injections
<form method="post" name="tapahtuma">
<input type="text" name="auts">
Submit: <input type="submit" name="Ilmoita" value="Ilmoita">
</form>
<?php
if(isset($_POST["Ilmoita"])) {
$link = mysqli_connect("localhost","root","","db") or die("connection failed: " . mysqli_connect_error());
$result = mysqli_prepare($link, "INSERT INTO `table` (`column`) VALUES (?)");
mysqli_stmt_bind_param($result, "s", $_POST['auts']);
mysqli_stmt_execute($result);
mysqli_close($link);
}?>
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