[RESOLVED] first sql attempt

themanwhowas

<html>
<body>
<?PHP
DEFINE ('DB_USER', 'root');
DEFINE ('DB_PASS', 'pass');
DEFINE ('DB_HOST', '127.0.0.1');
DEFINE ('DB_NAME', 'site');


$con = mysql_connect(DB_HOST, DB_USER, DB_PASS) OR die('dead');
mysql_select_db(DB_NAME) OR die('still dead');

?>

<Form action="<?PHP $PHP_SELF ?>" method="post">
<input type="text" name="name1">
<input type="text" name="name2">
<input type="text" name="email">
<input type="text" name="password">
<input type="submit" name="submit" value="submit">

<?

if(isset($_POST['submit']))
{
$fn = $_POST['name1'];
$ln = $_POST['name2'];
$em = $_POST['email'];
$pass = $_POST['password'];


$query = "insert into users(first_name) values('$fn')";

mysql_query($query);
}

if($result)
{
echo'ok';
}
else
{
echo'no go';
}

?>

whats wrong with this? i have gone through the book and im stuck 🙁

bpat1434

where do you populate $result? If it's meant to be teh result of the mysql_query() call, you need to make it as such:

$result = mysql_query($query);

roscor

Try This basic form with validation.

<?php

if($_POST['submit']=="send") {
   $errcount = 0;
   $error1="<font face='Verdana' size='2'><span style='color:red'>";
   $error2="<font face='Verdana' size='2'><span style='color:red'>";

   if(preg_match('/^[a-zæøåÆØÅ0-9_-]{4,}$/i', $_POST['name1'])){
        $name1 = $_POST['name1'];     

    }
        else {
        $error1 .= "<font face='Verdana' size='1'>&nbsp;Required! min 4";
        $errcount++;  

         }   

       if(preg_match('/^[\w]+[-\w._]*@[\w]+[-\w]+(\.[\w]+[-\w]+)*\.[\w]{2,6}$/ ', $_POST['email'])){
        $email = $_POST['email'];
    }
        else {
        $error2 .= "<font face='Verdana' size='1'>Required!";
        $vaild = "<font face='Verdana' size='1' color='#FF0000'>That Email is not vaild, please try again";
        $errcount++;  

        }
     if ($errcount !=0) {   

     //displays errors in form boxes
}
    else {

    $host = "localhost";
$username = "root";
$password = "password";
$database = "site";

 // connect to db:
$conn = mysql_connect($host, $username, $password) or die(mysql_error());
mysql_select_db($database,$conn) or die(mysql_error());
 $query = "INSERT INTO users VALUES ('','$name1','$email',)";
 mysql_query($query);

//create a new page called newlogon.php 
//new logon welcome page
echo '<meta http-equiv="refresh" content="0;url=newlogon.php">';


}
}



?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" >


<table border="0" cellpadding="7" cellspacing="1" bordercolor="#111111" width="800" height="49" align="center">
 <tr>
<td width="1"  height="25"><font face="Verdana" size="2" color="#000000">name1:<font face="Verdana" size="2" color="#FF0000">*</font><font face="Verdana" size="2" color="#000000">	
</td>
    <td width="403" height="25" align="left">
    <input type="text" value="<? echo $name1; ?>" id="name1" name="name1"> <font face="Veranda" size="2"><? echo $error1; ?></font></td>  

     <tr>
    <td width="1"  height="26"><font face="Verdana" size="2" color="#000000">Email:<font face="Verdana" size="2" color="#FF0000">*</font><font face="Verdana" size="2" color="#000000"></font></td>

<td width="403" height="26"><font face="Verdana" size="2" color="#000000"><input type="text" name="email" size="20" value="<? echo $email; ?>"></font>&nbsp;<font face="Veranda" size="2"><? echo $error2;?></font></td></tr>

<tr>
    <td width="117" height="1"><font face="Verdana" size="2" color="#000000"><input name="submit" type="submit" value="send" /><input type='button' value='Retry' onClick='history.go(-1)'>
  </font></td>


</tr>
</table>

You will have to ensure that you have the database set up along with the a newlogon.php page. It is a guide only for you to work on.

roscor

bogu

To grab the results u should use [man]mysql_fetch_assoc[/man] or [man]mysql_fetch_array[/man] ...

bpat1434

But he's not grabbing results, he's just trying to make sure things were inserted. But even if they were grabbing results, they'd still need to store the results of the mysql_query() call into a variable first....

To ensure that rows have indeed been inserted, use mysql_affected_rows($result)

bogu

bpat1434 wrote:
But he's not grabbing results, he's just trying to make sure things were inserted...

U are right I didn't pay attention ...:p

themanwhowas

thanks guys. the first reply solved the problem. in the excitement i forgot to reply. oops 🙂