[SOLVED] omitting an output if value is NULL.

[vomitbomb]

Put this code together so it omits entries if a value hasn't been put in.

It's not working and I'm getting:

Parse error: syntax error, unexpected '.' in C:\htdocs\showorders.php on line 19

 

$sql = "SELECT * FROM orders WHERE lname='$search'";
$result = mysql_query($sql, $conn);
$row = mysql_fetch_array( $result );

if(.$row['fname'] != NULL){
echo "Name: ".$row['fname'];
}

 

Anyone know another way I can do this?

[stopblackholes]

.$row['fname'] 

 

should be $row['fname']

[The Little Guy]

that is because you have period before the $row

 

Try this:

$sql = "SELECT * FROM orders WHERE lname='$search'";
$result = mysql_query($sql, $conn);
$row = mysql_fetch_array( $result );

if($row['fname'] != NULL){
echo "Name: ".$row['fname'];
}

[vomitbomb]

cool thanks for that.

 

just for reference? why is the dot in .$row?

[The Little Guy]

because who ever made it made a mistake... basically it shouldn't be their.

[vomitbomb]

lol I meant on:

 

echo "Name: ".$row['fname'];

[corbin]

It's simply PHP's concatenation character.

 

PHP:

"String" . "String" makes StringString

Javascript:

"String" + "String" makes StringString

 

And so on...  It's just the symbol in PHP that tells it to stick two strings together.  Kind of ironic now that I think about it.  A symbol which usually means end a sentence being used to stick two things together....