oracle - data

Teodotko

Registered
Код:
<form id="reg" name="reg" method="post" action=""> 
Username<br />
<input name="user" type="text" class="txt" id="path" size="30"
<br /><br />
password<br />
<input name="pass" type="text" class="txt" id="author" size="30" />

<input name="submit" type="submit" class="btn" id="button" value="Reg" /> 
</form> 


<?php 

if($_POST['submit']){ 
if($_POST['user'] == ""){ 
echo "<script type=\"text/javascript\"> 
alert('Empty field!'); 
</script>";} 
else { 
$date = date("d.m.Y"); 
$sql = "INSERT INTO `account` (`user`, `pass`, `date`) VALUES (NULL,  '$_POST[user]', '$_POST[pass] ', '$_POST[date]', '$date');"; 
$result = mysql_query($sql) or die(mysql_error()); 
} 
} 
?>

Как същото нещо да се направи да работи с oracle база данни ? :)
 
Направих го така, но нещо не записва данните в базата данни :

Код:
<?php
$db_username = "alef";
$db_password = "password";
$db = "oci:dbname=archlord";
$conn = new PDO($db,$db_username,$db_password);
?>
<form id="reg" name="reg" method="post" action=""> 
Username<br />
<input name="user" type="text" class="txt" id="path" size="30"
<br /><br />
password<br />
<input name="pass" type="text" class="txt" id="author" size="30" />

<input name="submit" type="submit" class="btn" id="button" value="Reg" /> 
</form>
<?php 

if($_POST['submit']){ 
if($_POST['user'] == ""){ 
echo "<script type=\"text/javascript\"> 
alert('Empty field!'); 
</script>";} 
else { 
$date = date("d.m.Y"); 
$stmt = $conn->exec("INSERT INTO `AMT_ACCOUNT` (`ACCOUNTID`, `PASSWORD`) VALUES ('$_POST[user]', '$_POST[pass] ')");
} 
} 
?>
 
http://stackoverflow.com/questions/2119145/inserting-data-in-oracle-database-using-php

Прегледай това, има няколко варианта
 

Back
Горе