Help with is_null

[janey]

I am trying to use is_null() for the first time to ensure I only display a result if the field in my table is not empty (using mySQL), but I can’t seem to get it to work properly.

My query is:

  <?
$connection = mysql_connect($host, $user, $pass) or die ("Unable to connect!");
mysql_select_db($db) or die ("Unable to select database!");
$query = "SELECT * FROM dogs ";
$result = mysql_query($query) or die ("Error in query: $query. " . mysql_error());
if (mysql_num_rows($result) > 0)
{
	while($row = mysql_fetch_object($result))
	{
	?> 

Then later in my page, I would print the results with:
Breed Name = <? echo $row->breed; ?>

What I would like to do is only have the breed name line show up if that field in the table has data. So, I tried:

  <?
$connection = mysql_connect($host, $user, $pass) or die ("Unable to connect!");
mysql_select_db($db) or die ("Unable to select database!");
$query = "SELECT * FROM dogs ";
$result = mysql_query($query) or die ("Error in query: $query. " . mysql_error());
if (mysql_num_rows($result) > 0)
{
	while($row = mysql_fetch_object($result))
	{
if ( is_null( . $row["breed "] . ) ) {
echo("<P>");
} else {
echo( . $row["breed "] . );
?> 

I think I am close but this isn’t quite right. Any suggestions?

Thanks!!

[PHPThorsten]

sure it's NULL what in your DB stands? and not ""?
If there is an empty sting "" it isn't NULL.
you may use then empty() instead of is_null() for checking

[janey]

Hmm...I've tried it both ways and neither seem to work.

I'm wondering if there is first a problem with my syntax because I am getting the following error:

Parse error: parse error, unexpected '.', expecting ')' in query_dog.php4 on line 153

And line 153 is

if ( is_null( . $row["vol_link"] . ) ) {

I don't really see where the error is, however.

Thanks!

[PHPThorsten]

the dots are the error. you don't need them....

if (empty($row["vol_link"]) ) {

[janey]

Oops, thanks...that is better.

Unfortunately, I'm still not getting the right results. Maybe I am not conceptualizing this correctly.

Am I correct to put the if statement after my query and then later in my document I can echo with <? echo ($row->breed); ?> and if there is data in the breed field a result will show but if there isn’t data it will be left blank?

Here, let me show you my code again:

  <?
$connection = mysql_connect($host, $user, $pass) or die ("Unable to connect!");
mysql_select_db($db) or die ("Unable to select database!");
$query = "SELECT * FROM dogs ";
$result = mysql_query($query) or die ("Error in query: $query. " . mysql_error());
if (mysql_num_rows($result) > 0)
{
while($row = mysql_fetch_object($result))
	{
if ( empty( $row["breed"] ) ) {
echo("<P>");
} else {
echo( $row["breed "] );
	?>

<table width="100%" border="0" cellspacing="1" cellpadding="1">
<tr>
<td>
Breed Name: <? echo $row->breed; ?>

</td></tr></table>

The result I am hoping to get is to have the breed name display in the table cell only if data is present.

Thanks again!

[PHPThorsten]

ok, I've just installed a little DB for testing this.
so, this is working on my system:

$connection = mysql_connect($host, $user, $pass) or die ("Unable to connect!");
mysql_select_db($db) or die ("Unable to select database!");
$query = "SELECT * FROM dogs";
$result = mysql_query($query) or die ("Error in query: $query. " . mysql_error());
if (mysql_num_rows($result) > 0)
{
  while($row = mysql_fetch_object($result))
  {
    if(empty( $row->breed) ) 
    {
      echo("<P>");
    } 
    else 
    {
      echo $row->breed;
    }
  }
}

[janey]

Thanks so much, that helped a bunch!