0
我目前有一个硬编码的json文件。这里是一个片段:使用mysql查询创建嵌套的json对象
{
"name": "Projects",
"children": [
{
"name":"Project_Category(example:MobileApps)",
"description":"category",
"children": [{
"name":"Sub_category(example:MusicPlayer)",
"description":"some_description(example:The music app will play music from an android device...)",
"children":[
{//child 1 of the MusicPlayer subcategory
"name": "Actual_project_name(example:JukeBox)",
"description":"Actual_project_description",
"children":[
{"name":"projectGroupMember1", "email":"[email protected]"},
{"name":"projectGroupMemeber2", "email":"[email protected]"}
]},
{ //child 2 of the MusicPlayer subcategory
"name": "another_project_title",
"description":"another_project_description",
"children":[
{"name":"projectGroupMember1", "email":"[email protected]"},
{"name":"projectGroupMember2", "email":"[email protected]"}
]
}
]//end of MusicPlayer's children
}, //end of MobileApps children
...
拥有所有这些数据存储在数据库中,我一直在试图使用PHP和嵌套的MySQL查询生成这个输出,我将用于其他任务的JSON文件(数据视觉)。我的目标是在每次打开Web应用程序时生成该文件,以便从数据库获得最新的版本。但是,我一直在使用嵌套mysql查询来创建这个输出有困难。所以我的问题是,鉴于我的json文件的结构是可行的吗?我应该做更好的方法吗?任何建议都会有帮助。
为什么要在嵌套的SQL查询中做到这一点?分而治之,并建立通过一步一步的JSON。并非一切都需要一步做 – MatthiasLaug 2013-04-21 12:58:35
您需要提供更多的信息。你的数据库结构是什么样的?你有什么尝试? – alexn 2013-04-21 13:02:47