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我有以下问题：这是我的表的结构在此处输入图像描述

在此处输入图像描述

这是我的代码：

            $SQL_NEXT_PROJECT_ID ="SELECT id  FROM projects WHERE id < '".$id_project."' and project_types_id='".$project_type_id."' ORDER BY id DESC LIMIT 1";
            $conn->query($SQL_NEXT_PROJECT_ID);
            foreach ($conn->query($SQL_NEXT_PROJECT_ID) as $rowProjectNext) {
                $next_project_id = $rowProjectNext['id'];
            }

            $SQL_PREVIOUS_PROJECT_ID ="SELECT id  FROM projects WHERE id > '".$id_project."' and project_types_id='".$project_type_id."' ORDER BY id ASC LIMIT 1";
            $conn->query($SQL_PREVIOUS_PROJECT_ID);
            foreach ($conn->query($SQL_PREVIOUS_PROJECT_ID) as $rowProjectPrevious) {
                $previous_project_id = $rowProjectPrevious['id'];
            }

问题是我需要使用“图像”表中的附加条件进行查询，我必须评估主字段 = '1'，我一直在尝试使用内部连接但它似乎不起作用，在这里我附上我一直在尝试的查询

$SQL_NEXT_PROJECT_ID ="SELECT projects.id  FROM projects INNER JOIN images WHERE projects.id > '".$id_project."' and project_types_id='1' and main='1' ORDER BY projects.id DESC LIMIT 1";

基本上我需要做的是根据项目的当前 id ($id_project) 创建下一个/上一个指向特定类型项目 (project_types_id) 的链接，同时考虑到项目应该有一个具有主字段的图像这一事实等于“1”

希望我能很好地解释自己...

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2 回答 2

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好吧，你什么都没有加入……你需要一个ON条款。

SELECT pr.id  FROM projects pr 
INNER JOIN images img ON (pr.id = img.projects_id AND img.main = 1)
WHERE pr.id > ? 
AND pr.project_types_id = 1 
ORDER BY pr.id DESC LIMIT 1
于 2012-11-19T22:10:56.040 回答
0

最后我这样做了

$_nextPrevious = array();

            $SQL_SEARCH_VALUES="SELECT images.projects_id AS projectId FROM projects INNER JOIN images ON ( projects.id = images.projects_id ) AND project_types_id =  '".$project_type_id."' AND images.main =  '1' LIMIT 0 , 30";
            $conn->query($SQL_SEARCH_VALUES);
            foreach($conn->query($SQL_SEARCH_VALUES) as $rowNextPrevious) {
                $value = $rowNextPrevious['projectId'];                    
                array_push($_nextPrevious, $value);
            }


            $currentValue = array_search($id_project, $_nextPrevious);

            $next = $currentValue + 1;

            $previous = $currentValue - 1;

            $next_project_id = $_nextPrevious[$next];

            $previous_project_id = $_nextPrevious[$previous];

            return array($project_name,$project_description,$project_path, $_images, $next_project_id, $previous_project_id);

使用数组而不是两次搜索...

于 2012-11-24T06:45:00.880 回答