php - 无法显示查询结果

Question

我的查询在下面工作，但我无法显示应该为“2”的结果(非重复计数的结果)。相反，我得到“作者总数:ID”

//var_dump Result:

array(1) {
[0]=> object(stdClass)#349 (1) {
["COUNT(DISTINCT writing.ID)"]=> string(1) “2″
}
}

//Code:

$authors = $wpdb->get_results("SELECT COUNT(DISTINCT writing.ID)
FROM writing
LEFT JOIN stories on writing.SID = stories.SID
LEFT JOIN wp_users ON writing.ID = wp_users.ID        
WHERE (stories.SID = $the_SID)"
);

echo var_dump($authors);
print "Total Authors:" .$authors[0]->writing.ID ."<p>";

Answer 1

正如您从 var_dump 的输出中看到的，您需要的值被存储为对象的成员 COUNT(DISTINCT writing.ID)。您可以遵循一些解决方法。

1. 将索引存储为单独的变量，然后在打印时使用它。

   $t = "COUNT(DISTINCT writing.ID)";
   print "Total Authors:" .$authors[0]->$t . "<p>";
   

2. 在 MySQL 查询中使用别名。

   SELECT COUNT(DISTINCT writing.ID) AS writingID
   FROM writing
   LEFT JOIN stories ON writing.SID = stories.SID
   LEFT JOIN wp_users ON writing.ID = wp_users.ID        
   WHERE (stories.SID = $the_SID)
   

   然后，

   print "Total Authors:" .$authors[0]->writingID . "<p>";