Question

我的查询在下面工作，除了我无法显示应为“2”的结果（非重复计数的结果）。相反，我得到“Total Authors：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

从var_dump的输出中可以看出，您所需的值将存储为对象的成员COUNT(DISTINCT writing.ID)。您可以遵循一些解决方法。

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>";

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>";

无法获取要显示的查询结果

1 个答案: