我们可以在 PHP 脚本中使用 if...else 条件来准备基于 NULL 值的查询。为了说明这一点，我们使用以下示例 -

示例

在此示例中，我们使用名为 'tcount_tbl' 的表，其中包含以下数据 -

mysql> SELECT * from tcount_tbl;
+-----------------+----------------+
| tutorial_author | tutorial_count |
+-----------------+----------------+
|      mahran     |       20       |
|      mahnaz     |      NULL      |
|       Jen       |      NULL      |
|      Gill       |       20       |
+-----------------+----------------+
4 rows in set (0.00 sec)

现在，以下是一个PHP脚本，它从外部获取 ‘tutorial_count’的值，并将其与字段中可用的值进行比较。

<?php
   $dbhost = 'localhost:3036';
   $dbuser = 'root';
   $dbpass = 'rootpassword';
   $conn = mysql_connect($dbhost, $dbuser, $dbpass);

   if(! $conn ) {
      die('Could not connect: ' . mysql_error());
   }

   if( isset($tutorial_count )) {
      $sql = 'SELECT tutorial_author, tutorial_count FROM tcount_tbl
         WHERE tutorial_count = $tutorial_count';
   } else {
      $sql = 'SELECT tutorial_author, tutorial_count FROM tcount_tbl
         WHERE tutorial_count IS $tutorial_count';
   }

   mysql_select_db('TUTORIALS');
   $retval = mysql_query( $sql, $conn );

   if(! $retval ) {
      die('Could not get data: ' . mysql_error());
   }

   while($row = mysql_fetch_array($retval, MYSQL_ASSOC)) {
      echo "Author:{$row['tutorial_author']} <br> ".
         "Count: {$row['tutorial_count']} <br> ".
         "--------------------------------<br>";
   }
   echo "Fetched data successfully<p>";
   mysql_close($conn);
?></p>