爲什麼這段代碼不輸出結果？

Question

我得到一個不會有任何錯誤或警告。我的表稱爲產品，該字段稱爲名稱。我想找到第一和第二列之間的關係。

爲什麼我在這裏失蹤？

<?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

。在你的查詢中沒有T2，但在你的選擇你有t2.name as second。

其實，你沒有一個FROM條款在所有...

編輯：

這是在你想達到什麼樣的一個隨機猜測......我不噸真的知道，但你應該能夠根據需要編輯查詢。

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

Answer 2

也許你應該檢查$ query2是否不爲FALSE，否則顯示mysql_error（）錯誤。另請參見http://www.php.net/manual/en/function.mysql-query.php上的示例。