我正在尝试创建一个在项目中包含任务的任务管理器。
但是我很难同时显示project.project(名称)和task.taks(任务名称)。
我的表具有以下字段:
project (project_id (PK) , project (name of project)
tasks(id,task(任务名称),project_id(PK的外键)。
我尝试过
$sql="
SELECT project.project
, tasks.task
, tasks.description
, tasks.Priority
, tasks.Due
FROM project
JOIN tasks
on project.project_id = tasks.project_id
WHERE project_id='{$_GET['project_id']}'
";
但是似乎有些问题。
谢谢
答案 0 :(得分:1)
尽管您没有指出任何错误(这将有助于您快速地回答问题),但您的陈述含糊不清。
$sql="
SELECT project.project
, tasks.task
, tasks.description
, tasks.Priority
, tasks.Due
FROM project
JOIN tasks
on project.project_id = tasks.project_id
WHERE project_id='{$_GET['project_id']}'
";
您的project_id
表和project
表中都有一个tasks
字段。如果不指定使用WHERE子句的对象,它将不起作用。
$sql="
SELECT project.project
, tasks.task
, tasks.description
, tasks.Priority
, tasks.Due
FROM project
JOIN tasks
on project.project_id = tasks.project_id
WHERE project.project_id='{$_GET['project_id']}'
";
还请确保对$ _GET ['project_id']进行了清理和正确准备。