Question

我有一个没有返回正确数据的查询过滤器... 这是为网站主页列表提取新闻文章。我有一个名为HideFromHome（boolean）的字段，我正在尝试过滤，以及发布和删除日期：

$today=date('Y-m-d');   
$filter = "HideFromHome != 1 AND ";
$filter .= "((PublishDate <= '".$today."') AND (RemoveDate > '".$today."')) OR ";
$filter .= "((PublishDate IS NULL AND RemoveDate > '".$today."')) OR "; 
$filter .= "((PublishDate <= '".$today."' AND RemoveDate IS NULL)) OR ";
$filter .= "((PublishDate IS NULL AND RemoveDate IS NULL))";

一切正常，除非它返回所有文章，即使选择了HideFromHome ......有人能发现问题吗？

Answer 1

您需要更多括号。每个'OR'与第一个'AND'处于同一水平尝试：

$today=date('Y-m-d');   
$filter = "HideFromHome != 1 AND (";
$filter .= "((PublishDate <= '".$today."') AND (RemoveDate > '".$today."')) OR ";
$filter .= "((PublishDate IS NULL AND RemoveDate > '".$today."')) OR "; 
$filter .= "((PublishDate <= '".$today."' AND RemoveDate IS NULL)) OR ";
$filter .= "((PublishDate IS NULL AND RemoveDate IS NULL)))";

正如H-Man2的回答FALSE AND TRUE OR TRUE所解释的那样TRUE。

Answer 2

您可能需要围绕部分或部分使用括号，因为false and true or true为true。

Answer 3

也许你试试HideFromHome <> 1 对于MySQL，!=（“不等于”）是<>（“更少或更大”）。

SQL查询过滤器问题

3 个答案: