[RESOLVED] View my emial address's in php

I'm having trouple viewing my emial address that have been saved in to mysql database how can i sort this here is my code there is no error showing in this code.

http://www.prksoft.com/testview.php

mysql info

PHP Code:

---

<?php
$dbhost='localhost';
$dbuser='USERNAME';
$dbpass='PASSWORD';

$conn=mysql_connect($dbhost, $dbuser, $dbpass)
    or die('Could not connect to mysql');

$dbname='prksoft_newsletter';
mysql_select_db($dbname);
?>

---

This is the textview code

PHP Code:

---

<?php
include "db_connect.php";

$conn=mysql_connect($dbhost, $dbuser, $dbpass)
    or die('Could not connect to mysql');


$sql = "select * from $newsletter";
$result = mysql_query($sql);

while($row=mysql_fetch_array($result)){
?>

<table width="400" border="1" align="center">
 <tr>
  <td width="63"><strong>Email Address</strong></td>
  <td width="8"><strong>:</strong></td>
  <td width="307"><? echo $rows['address']; ?></td>
</tr>
</table>

<p>&nbsp;       </p>

---

Well for starters, you don't have a close brace for your while loop.

Place

Code:

---

ERROR_REPORTING(E_ALL);

---

At the top of your code, chances are the server you're running off of has error reporting turned off.

here is the proper testview.php code.

I put ERROR_REPORTING(E_ALL); at the top and it still shows no errors.

PHP Code:

---

<?php
include "db_connect.php";

$conn=mysql_connect($dbhost, $dbuser, $dbpass)
    or die('Could not connect to mysql');


$sql = "select * from $newsletter";
$result = mysql_query($sql);

while($row=mysql_fetch_array($result)){
?>

<table width="400" border="1" align="center">
 <tr>
  <td width="63"><strong>Email Address</strong></td>
  <td width="8"><strong>:</strong></td>
  <td width="307"><? echo $rows['address']; ?></td>
</tr>
</table>

<p>&nbsp;       </p>

<?php
}
mysql_close($conn);
?>

---

What does $newsletter contain? have you assigned it a proper table name?

-tg

I sorted that problem now it wont show the email address's?

Code:

---

<?php

include "db_connect.php";

$conn=mysql_connect($dbhost, $dbuser, $dbpass)
        or die('Could not connect to mysql');


$sql = "select * from newsletter";
$result = mysql_query($sql);

while($row=mysql_fetch_array($result)){
?>

<table width="400" border="1" align="center">
 <tr>
  <td width="63"><strong>Email Address</strong></td>
  <td width="8"><strong>:</strong></td>
  <td width="307"><? echo $rows['address']; ?></td>
</tr>
</table>

<p>&nbsp;      </p>

<?php
}
mysql_close($conn);
?>

---

You're doing

Code:

---

echo $rows['address'];

---

When your variable is row, not rows.

Also are you selecting the database? You have a select statement in your original post, but I don't see it in the testview.php code.

If db_connect.php is connecting and selecting, you may be overwriting your connection object when you do it again in testview.php

Code:

---

include "db_connect.php";

$conn=mysql_connect($dbhost, $dbuser, $dbpass)
        or die('Could not connect to mysql');

---

samba - yes... presumably these lines here take care of that:

Code:

---

include "db_connect.php";

$conn=mysql_connect($dbhost, $dbuser, $dbpass)
        or die('Could not connect to mysql');

---

-tg

tg - Sorry, edited my post for more clarity on what I meant... db_connect.php is making $conn, selecting a db, but then testview.php makes a new connection right after that into the same variable $conn... If you put the included code of db_connect.php into testview.php, you get this:

Code:

---

$dbhost='localhost';
$dbuser='USERNAME';
$dbpass='PASSWORD';

$conn=mysql_connect($dbhost, $dbuser, $dbpass)
    or die('Could not connect to mysql');

$dbname='prksoft_newsletter';
mysql_select_db($dbname);

$conn=mysql_connect($dbhost, $dbuser, $dbpass)
        or die('Could not connect to mysql');

---

mysql_connect is happening twice, and the second time isn't selecting a database.

Now that i got that sorted when i try to signup it dosent save the address in to the mysql database this is the code?

address http://www.prksoft.com/signup.php

Code:

---

<html>
<head></head>
<body>
<form action="signup.php" method="POST">
<input type="text" size="35" name="address">  <input type="submit" value="Get Special Offers!" name="submit">
</form>

<?php
if ($submit)
{
include "db_connect.php";
$address = $_POST['address'];
mysql_query("INSERT INTO newsletter (address)
                        VALUES ('$address')");
echo "$address has been added.";
}
?>

---

That'll almost work. Just change this...

Code:

---

if ($submit)

---

...to this...

Code:

---

if (isset($_POST['submit']))

---

But you should also never insert to a database without sanitizing the input first - look up "SQL injection" for the troubles that can cause. So, do something like this here:

Code:

---

$address = mysql_real_escape_string(stripslashes($_POST['address']));

---

You can look up both of these - mysql_real_escape_string() and stripslashes() - on PHP.net for more info.

How would i do error checking for duplicate email or blank text box?