no selecting data

[franknu]

Ok, I am having Some issues now, My main problem now is that it is not selecting from database. Meaning

it is not display the value from the database when i ask it to do so.


here is partial part of the code

PHP: Select all

 
$db = mysql_connect($host, $username, $password) or die(mysql_error());
mysql_select_db($database) or die(mysql_error());
 
 
 
$BusinessName = (isset($_POST['BusinessName']) ? $_POST['BusinessName'] : '');
$Slogan = (isset($_POST['Slogan']) ? $_POST['Slogan']:'');
$Business_Address = (isset($_POST['Business_Address']) ? $_POST['Business_Address']:'');
$Tel = (isset($_POST['Tel']) ? $_POST['Tel']:'');
$Website = (isset($_POST['Website']) ? $_POST['Website']:'');
$Email = (isset($_POST['Email']) ? $_POST['Email']:'');
$Member_Status = (isset($_POST['Member_Status']) ? $_POST['Member_Status']:'');
$Fax =(isset($_POST['Fax']) ? $_POST['Fax']:'');
$type = (isset($_POST['type']) ? $_POST['type']:'');
$make = (isset($_POST['make']) ? $_POST['make']:'');
$Categories = (isset($_POST['Categories']) ? $_POST['Categories']:'');
$Keyword = (isset($_POST['Keyword']) ? $_POST['Keyword']:'');
$Picture1 =  (isset($_POST['Picture1']) ? $_POST['Picture1']:'');
$Headline =  (isset($_POST['Headline']) ? $_POST['Headline']:'');
$Slogan2 = (isset($_POST['Slogan2']) ? $_POST['Slogan2']:'');
$Description1 = (isset($_POST['Description1']) ? $_POST['Description1']:'');
$Description2 = (isset($_POST['Description2']) ? $_POST['Description2']:'');
$Description3= (isset($_POST['Description3']) ? $_POST['Description3']:'');
$Contact2 = (isset($_POST['Contact2']) ? $_POST['Contact2']:'');
$Picture2 =  (isset($_POST['Picture2']) ? $_POST['Picture2']:'');
$Picture3 = (isset($_POST['Picture3']) ? $_POST['Picture3']:'');
$Picture4 =  (isset($_POST['Picture4']) ? $_POST['Picture4']:'');
$User_Name = (isset($_POST['User_Name']) ? $_POST['User_Name']:'');
$Password = (isset($_POST['Password']) ? $_POST['Password']: '');
$User_Name=strtolower($_POST['User_Name']);
$Password=strtolower($_POST['Password']);
 
if ( isset($_POST['User_Name']) && isset($_POST['Password']) ) {
 
 
$query = "SELECT * FROM business_info where User_Name='".$_POST['User_Name']."' AND `Password`='".$_POST['Password']."'";
 
 
}
 
  if ($result = mysql_query($query)) 
{
    if (mysql_num_rows($result))
 {
      $row = mysql_fetch_assoc($result);
 
      $BusinessName= $row['BusinessName'];
      $Keyword = $row['Keyword'];
      $Picture1 =  $row['Picture1'];
      $Headline = $row['Headline'];
      $Slogan =$row['Slogan'];
      $Slogan2 = $row['Slogan2'];
      $Description1 =$row['Description1'];
      $Description2 = $row['Description2'];
      $Description3= $row['Description3'];
      $Contact2 =  $row['Contact2'];
      $Picture2 = $row['Picture2'];
      $Picture3 = $row['Picture3'];
      $Categories=$row['Categories'];
      $Business_Address= $row['Business_Address'];
      $make=$row['make'];
      $type=$row['type'];
      $Tel= $row['Tel'];
      $Website=$row['Website'];
    } 
else 
{
      echo "<p><b>username and/or password not found. Try
again?
 
</b></p>";
exit;
    }
  } 
else 
{
    echo "Query failed<br />$query<br />". mysql_error();
exit;
  }
 
?>
 
 
 
 
 
 
<?
echo
<table width="504" border="1" cellpadding="0" cellspacing="0" bordercolor="#CCCCCC">
<tr>
<td width="494"><table width="497" border="0">
<tr>
<td width="487">Banner</td>
</tr>
<tr>
 
 
 
 
// like this line down here should display name from database INSTEAD Is
displayin (Welcome $BusinessName )
 
<td background="fondo2.jpg">Welcome $BusinessName </td>
 
</tr>
<tr>
<td><table width="480" border="0">
<tr>

[grahame]

Try printing out $query and seeing the SQL you're actually running. That should give you a clue as to what is wrong. I suspect the backtics around Password ...

[franknu]

ok print our $query and this is what is shows;

SELECT * FROM business_info WHERE User_Name='franklin' AND `Password`='franklin01'

i think that maybe the problem is in the way i am echoing the the data

like i am echoing the whole html codes in only one

example

<?
echo'

html codes';

?>

could that be the problem

[grahame]

You do NOT want the quote around the word password in the query, as I suspected in my previous answer. Take them out. As the code is written, you are testing if the literal string "password" is the same as the literal string "franklin01" which it never is. You probably want to test if the string in the column called password is equal t the string "franklin01" instead.

[franknu]

Well, i guess i was right it, is the was echo was set up for whatever reason it did not like it

thank u