| Display Duplicates
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Gilead
msg:4397941 | 7:13 pm on Dec 14, 2011 (gmt 0) |
I am creating a new function for the project. Need to display all members that have the same account_number and show how many of them there are. I did something wrong because I'm getting errors along with the actual account numbers that have more than one.
error_reporting(E_ALL);
session_start();
include('config.php');
$table_name=(USER);
$query = "SELECT account_number,COUNT(*) AS Count
FROM $table_name
GROUP BY account_number
HAVING Count > 1; ";
$result = mysql_query($query) or die("No Account Numbers Present!" . mysql_error());
$sql = "SELECT * FROM $table_name";
$data = mysql_fetch_array($sql);
$firstname=mysql_real_escape_string($data['firstname']);
$organization=mysql_real_escape_string($data['organization_title']);
$lastname=mysql_real_escape_string($data['lastname']);
$username=mysql_real_escape_string($data['member_login']);
$password=mysql_real_escape_string($data['member_password']);
$email=mysql_real_escape_string($data['email']);
echo "<table>";
while($row = mysql_fetch_array($result))
{
echo '<tr>';
echo '<td>' .$row["account_number"] .'</td>';
echo '<td>' .$organization .'</td>';
echo '<td>' .$firstname .'</td>';
echo '<td>' .$lastname .'</td>';
echo '<td>' .$username .'</td>';
echo '<td>' .$password .'</td>';
echo '<td>' .$email .'</td>';
echo '<td>' .$row['COUNT(account_number)'] .'</td>';
echo '</tr>';
}
echo "</table>";
Output is as follows:
Warning: mysql_fetch_array() expects parameter 1 to be resource, string given in path\to.accounts.php on line 30
Notice: Undefined index: COUNT(account_number) in path\to.accounts.php on line 49
Notice: Undefined index: COUNT(account_number) in path\to.accounts.php on line 49
Notice: Undefined index: COUNT(account_number) in path\to.accounts.php on line 49
Notice: Undefined index: COUNT(account_number) in path\to.accounts.php on line 49
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This part is right. It's likely a very simplle fix, but I cannot see it right now.
Thanks for the help!
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enigma1
msg:4398225 | 2:39 pm on Dec 15, 2011 (gmt 0) |
$sql = "SELECT * FROM $table_name";
$data = mysql_fetch_array($sql); |
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You're pulling rows without making a query first.
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Gilead
msg:4398237 | 3:16 pm on Dec 15, 2011 (gmt 0) |
I knew it was something simple. How do I get the COUNT to work right. I wanted it to show how many users there were for the one account. Or am I going about it the wrong way?
Thanks!
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