undefine index

[franknu]

ok, I want to create a link to another page  within the database same table( all the info is on the same row)
the problem that i am having is
undefine index:

[code=php:0]
$BusinessName = ($_POST['BusinessName']);
$Keyword =($_POST['Keyword']);
$Picture1 =  ($_POST['Picture1']);
$Headline = ($_POST['Headline']);
$Slogan2 = ($_POST['Slogan2']);
$Description1 =($_POST['Description1']);
$Description2 = ($_POST['Description2']);
$Description3= ($_POST['Description3']);
$Contact2 =  ($_POST['Contact2']);
$Picture2 = ($_POST['Picture2']);
$Picture3 = ($_POST['Picture3']);
[/code]

if i add isset it would work but i was told that isset would only return true or false value. when i add isset works but dont return any value..

[code=php:0]
Here is my link:$bn = $row['BusinessName'];
echo "<a href=\"bizwebpage2.php?BusinessName=$bn\">$bn</a>";
[/code]

here is the page where it takes me

[code=php:0]
<?php

$host = "localhost";
$username = "localhost";
$password = "abc123";
$database = "contacts";


$db = mysql_connect($host, $username, $password) or die(mysql_error());
mysql_select_db($database) or die(mysql_error());

$BusinessName = ($_POST['BusinessName']);
$Keyword =($_POST['Keyword']);
$Picture1 =  ($_POST['Picture1']);
$Headline = ($_POST['Headline']);
$Slogan2 = ($_POST['Slogan2']);
$Description1 =($_POST['Description1']);
$Description2 = ($_POST['Description2']);
$Description3= ($_POST['Description3']);
$Contact2 =  ($_POST['Contact2']);
$Picture2 = ($_POST['Picture2']);
$Picture3 = ($_POST['Picture3']);

if($BusinessName)
{
$query = "SELECT * FROM business_info WHERE `BusinessName`= '$BusinessName' ";
$result = mysql_query($query) or die (mysql_error());
}
?>
<table>
  <tr>
    <td>
      <table>
        <tr>
          <td valign="top">
            <table>
              <tr>
                <td valign="top">
                  <table>
                    <tr> 
                      <td><?php echo"$Logo"; ?></td>
                    </tr>
                    <tr>
                      <td valign="top"><h2><?php echo "<h2>$BusinessName</h2>"; ?></h2></td>
                    </tr>
                    <tr>
                      <td valign="top"><?php echo "$Description1"; ?></td>
                    </tr>
                    <tr>
                      <td valign="top"><?php echo "$Description2"; ?></td>
                    </tr>
<tr>
                      <td valign="top"><?php echo "$Description3"; ?></td>
                    </tr>
                    <tr>
                      <td valign="top"><?php echo "$Contact2"; ?></td>
                    </tr>
                  </table>
                </td>
              </tr>
            </table>
          </td>
          <td valign="top">
            <table>
              <tr>
                <td>&nbsp;</td>
              </tr>
              <tr>
                <td valign="top"><?php echo"<img src='$Picture2' width='200' height='250'>"; ?>

</td>
              </tr>
              <tr>
                <td valign="top"> <?php echo "<img src='$Picture3'  width='200' height='250'>"; ?>  </td>
              </tr>
            </table>
          </td>
        </tr>
      </table>
      <table border='1'>
        <tr>
          <td>&nbsp;</td>
        </tr>
      </table>
    </td>
  </tr>
</table>
<?php

?>
[/code]

[Jenk]

It means the index does not exist, ergo the POST field was not submitted.

Use an if statement to check your requested fields are there.

[code]<?php

if (!isset($_POST['field'])) die('You must provide a value for \'field');

?>[/code]

[CheesierAngel]

if you do not want to test if they all are set and you dont wanna quit your script you could also use:

[code]
<?php

$BusinessName = @$_POST['BusinessName'];
$Keyword =@$_POST['Keyword'];
$Picture1 =  @$_POST['Picture1'];
$Headline = @$_POST['Headline'];
$Slogan2 = @$_POST['Slogan2'];
$Description1 =@$_POST['Description1'];
$Description2 = @$_POST['Description2'];
$Description3= @$_POST['Description3'];
$Contact2 =  @$_POST['Contact2'];
$Picture2 = @$_POST['Picture2'];
$Picture3 = @$_POST['Picture3'];

...
?>
[/code]

If the var was not submitted and/or an error occurs then the @$var will be null, else the $var value will be assigned


MODIFICATION:

You can work even alot shorter if you want:

[code]
<?php
$submittedVars = array();
foreach($_POST as $key => $value) {
  $submittedVars[$key] = $value;
}
?>
[/code]

Now you can print your table something like:

[code]
...
<td><?php echo @$submittedVars['DescriptionA']; ?></td>
<td><?php echo @$submittedVars['DescriptionB']; ?></td>
...
[/code]

[Jenk]

'Angel, you first problem is poor programming (sweeping errors under the carpet does not make them go away.) and the second is only hindering.. if you are going to transfer from $_POST to $submittedVars you can just do $submittedVars = $_POST, or just plain use @$_POST on every line, which leads me back to my first point.

[CheesierAngel]

It's indeed sweeping the errors under the carpet but alot shorter to write.
Using the @ in your code means you know there can be an error and if so use null instead of terminating the script.

The transfer from $_POST to $submittedVars is indeed unusefull in this example. But when the vars are hoovering around different functions and classes it's more obvious to make a copy of the $_POST variables. Therefor i mostly copy the array even when i don't use these in vars in other functions and or classes. (kind of a 'bad' habbit)

[franknu]

make sense now, i am not submitting the these variables, i have a page call bizwebpage2.php

when the user submit  their search criteria a list of business  display if they want to know more about that particular business they clink on the link  [code=php:0] $bn = $row['BusinessName'];
echo "<a href=\"bizwebpage2.php?BusinessName=$bn\">$bn</a>"; [/code], maybe i am doing the whole thing wrong

[franknu]

i already tried

$BusinessName = @$_POST['BusinessName'];

and i got the same problem. i just realized, is because i am not submitting the variables, they are in another page i just need to use them to display the data tha it is in those colums in another page any advise please

[Jenk]

Advice: Be more clear and descriptive with your questions..

[franknu]

Ok, when the user click on the link  which is on a page call business_display.php

[code=php:0]

this is the link on page business_display.php
$bn = $row['BusinessName'];
echo "<a href=\"bizwebpage2.php?BusinessName=$bn\">$bn</a>";

[/code]

user click on the link and it takes them to page

bizwebpage2.php

here is the code for bizwebpage2.php

[code=php:0]

<?php

$host = "localhost";
$username = "localhost";
$password = "abc123";
$database = "contacts";


$db = mysql_connect($host, $username, $password) or die(mysql_error());
mysql_select_db($database) or die(mysql_error());

$BusinessName = ($_POST['BusinessName']);
$Keyword =($_POST['Keyword']);
$Picture1 =  ($_POST['Picture1']);
$Headline = ($_POST['Headline']);
$Slogan2 = ($_POST['Slogan2']);
$Description1 =($_POST['Description1']);
$Description2 = ($_POST['Description2']);
$Description3= ($_POST['Description3']);
$Contact2 =  ($_POST['Contact2']);
$Picture2 = ($_POST['Picture2']);
$Picture3 = ($_POST['Picture3']);

if($BusinessName)
{
$query = "SELECT * FROM business_info WHERE `BusinessName`= '$BusinessName' ";
$result = mysql_query($query) or die (mysql_error());
}
?>
<table>
  <tr>
    <td>
      <table>
        <tr>
          <td valign="top">
            <table>
              <tr>
                <td valign="top">
                  <table>
                    <tr> 
                      <td><?php echo"$Logo"; ?></td>
                    </tr>
                    <tr>
                      <td valign="top"><h2><?php echo "<h2>$BusinessName</h2>"; ?></h2></td>
                    </tr>
                    <tr>
                      <td valign="top"><?php echo "$Description1"; ?></td>
                    </tr>
                    <tr>
                      <td valign="top"><?php echo "$Description2"; ?></td>
                    </tr>
<tr>
                      <td valign="top"><?php echo "$Description3"; ?></td>
                    </tr>
                    <tr>
                      <td valign="top"><?php echo "$Contact2"; ?></td>
                    </tr>
                  </table>
                </td>
              </tr>
            </table>
          </td>
          <td valign="top">
            <table>
              <tr>
                <td>&nbsp;</td>
              </tr>
              <tr>
                <td valign="top"><?php echo"<img src='$Picture2' width='200' height='250'>"; ?>

</td>
              </tr>
              <tr>
                <td valign="top"> <?php echo "<img src='$Picture3'  width='200' height='250'>"; ?>  </td>
              </tr>
            </table>
          </td>
        </tr>
      </table>
      <table border='1'>
        <tr>
          <td>&nbsp;</td>
        </tr>
      </table>
    </td>
  </tr>
</table>
<?php

?>
[/code]

so how can i display the info that  it is on Description1, Description2, ect that matched BusinessName which is on the same row on the table name business_info

[Jenk]

$_GET['BusinessName']