为什么这段代码不输出结果？

Question

我得到一个不会有任何错误或警告。我的表称为产品，该字段称为名称。我想找到第一和第二列之间的关系。

为什么我在这里失踪？

<?php 

mysql_connect("localhost"," "," "); 
mysql_select_db(" "); 
mysql_query('set names utf8'); 

$query = mysql_query("SELECT t1.name AS first FROM products t1"); 
$row = mysql_fetch_assoc($query); 
if (!$query) { 
    die('Invalid query: ' . mysql_error()); 
} 

do { 
    $first = $row['first']; 
    $query2 = mysql_query("SELECT t1.name AS first, 
           t2.name AS second, 
           (MATCH (t2.name) AGAINST ('$first')) AS relevance 
FROM products t1, products t2 
          WHERE t1.name = '$first' 
          ORDER BY relevance DESC"); 
if (!$query2) { 
    die('Invalid query: ' . mysql_error()); 
} 
    $most_similar = mysql_fetch_assoc($query2); 
} while ($row = mysql_fetch_assoc($query)); 
?> 

Answer 1

。在你的查询中没有T2，但在你的选择你有t2.name as second。

其实，你没有一个FROM条款在所有...

编辑：

这是在你想达到什么样的一个随机猜测......我不吨真的知道，但你应该能够根据需要编辑查询。

$query2 = mysql_query("SELECT '$first' AS first, 
          t.name AS second, 
          (MATCH (t.name) AGAINST ('$first')) AS relevance 
         FROM products t 
         ORDER BY relevance DESC"); 

Answer 2

也许你应该检查$ query2是否不为FALSE，否则显示mysql_error（）错误。另请参见http://www.php.net/manual/en/function.mysql-query.php上的示例。