Question

This is my first post so please bear with me with inputting the code into here. Im trying to output some images to a PDF and need to create a if statement that looks for data with in a row.

$connection = mysql_connect("localhost", "testdb", "********")
    or die ("Unable to connect!");    

// select database  
mysql_select_db("testdb") or die ("Unable to select database!");      

// Select all the rows in the test table
$query = "SELECT * FROM test2 WHERE testid=89";
$result = mysql_query($query);
if (!$result) {
    die('Invalid query: ' . mysql_error());
}

while ($row= mysql_fetch_array($result)) {
    $image = $row[1]; 
    $text = $row[2];
}

That’s what I have so far and basically I need something along the line of this:

If (data in row 1) {
    print $image;
} else {
    print $text;
}

Answer 1

It’s hard to say exactly what you’re looking for since it isn’t very clear, but I think what you’re wanting to do is check to see if $image has a value, and if so, display it. If not, display $text instead.

If this is the case use empty(). It will tell you if a variable is empty or not.

if (!empty($image))
 {
   print $image;
 }
else
 {
  print $text;
 }

The following things are considered to be empty:

• “” (an empty string)
• 0 (0 as an integer)
• 0.0 (0 as a float)
• “0” (0 as a string)
• NULL
• FALSE
• array() (an empty array)
• $var; (a variable declared, but without a value)