php - if mysql any field is empty then show a word like null -


i need know that, if there empty field display null in retrieve form. mean,

name----------age----------country

xyz----------" "---------usa

so, form show

name: xyz

age: show null

country: usa

<?php $dbhost = ''; $dbuser = ''; $dbpass = '';   $conn = mysql_connect($dbhost, $dbuser, $dbpass);  if(! $conn ) {   die('could not connect: ' . mysql_error()); } $no=$_get['no']; $sql = "select * tablename name='$name'"; mysql_select_db('dbname'); $retval = mysql_query( $sql, $conn );  if(! $retval ) {   die('could not data: ' . mysql_error());  }  while($row = mysql_fetch_array($retval, mysql_assoc)) {     ?>    <table width="100%"><tr><td>  <table>  <tr>  <tr><td class="a" style="width:20%">name</td><td style="width:20%"  class="a"><a class="res" ><?php echo $row['name'];?></a></td></tr>  <tr><td class="a" style="width:20%">age</td><td style="width:20%"   class="a"><a class="res"><?php echo $row['age'];?></a></td></tr>  <tr><td class="a" style="width:20%">country</td><td style="width:20%"  class="a"><a class="res"><?php echo $row['country'];?></a></td></tr>  </tr>  </table>   <?php  }       mysql_close($conn);  ?>`

just test if result empty , if show null.

<?php if ($row['name'] != '') {     echo $row['name']; } else {     echo "null"; }?>

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