我正在构建一个Sencha-Touch 2应用程序,从服务器端(mysql DB)恢复数据时遇到了一些麻烦.
I'm building a Sencha-Touch 2 app and I have some trouble with the recuperation of my data from server side (mysql DB).
这是我的数据模型:
Table1 : ID:int description:varchar(100) Table2 : ID:int description:varchar(100) table1_ID:int Table3 : ID:int name:varchar(100) info:varchar(100) table2_ID:int表1以一对多关系连接到表2,并且在表2和表3之间相同.
Table1 is join to Table2 with a one-to-many relationship and same between Table2 and Table3.
我要从服务器获取的是一个嵌套的JSON,它看起来像这样:
What I want from server is a nested JSON who looks like this :
[ Table1_object1_ID: 'id' : { Table1_object1_description: 'description', Table2_Objects : [ 'Table2_object1': { Table2_object1_id : 'id', Table2_object1_description : 'description' Table3_Objects : [ table3_object1: { Table3_object1_name : 'name', Table3_object1_info : 'info', }, table3_object2: { Table3_object2_name : 'name', Table3_object2_info : 'info', }, table3_object3: { Table3_object3_name : 'name', Table3_object3_info : 'info', }, etc... ], }, 'Table2_object2': { Table2_object2_id : 'id', Table2_object2_description : 'description' Table3_Objects : [ ... ] }, etc.... ] }, Table1_object2_ID: 'id' : { etc.... ]在我的App中,我为每个表使用3个模型,理想情况下,我想将数据保存在3个商店中,但这将是另一个问题;-)
In my App, I use 3 Models for each Table, and ideally I want to save my data in 3 Stores, but that will be an other problem ;-)
第一个商店(基于Table1中的模型)执行JsonP请求以获取嵌套JSON.
The first Store (based on the Model from Table1) do a JsonP request to get the Nested JSON.
实际上,我在PHP文件中的SQL请求很简单:
Actually my SQL request in the PHP file is simple :
SELECT * FROM Table1 INNER JOIN Table2 ON Table1.ID = Table2.table1_ID INNER JOIN Table3 ON Table2.ID = Table3.table2_ID;我试图根据SQL结果在PHP中创建一个数组,但无法获得预期的结果. 我还尝试使用GROUP BY和GROUP_CONCAT更改我的SQL,但是在这里相同,无法获取我想要的JSON.
I tried to make an array in PHP from my SQL results but cannot get the expect result. I also try to change my SQL with GROUP BY and GROUP_CONCAT but same here, cannot get the JSON I want.
一些帮助将不胜感激.
推荐答案带有一些示例数据的可运行代码: codepad /2Xsbdu23
Runnable code with some sample data: codepad/2Xsbdu23
我使用了3个不同的SELECT来避免不必要的重复. 当然,您必须自定义$result数组以完全符合所需的JSON格式,但是我认为这并不难.
I used 3 distinct SELECTs to avoid the unnecessary repetitions. Of course you have to customize the $result array to your exact desired JSON format, but I think it is not that hard.
// assume $t1/2/3 will be arrays of objects $t1 = SELECT Table1.* FROM Table1 WHERE Table1.ID = 111 $t2 = SELECT Table2.* FROM Table2 WHERE Table2.table1_ID = 111 $t3 = SELECT Table3.* FROM Table2 INNER JOIN Table3 ON Table2.ID = Table3.table2_ID WHERE Table2.table1_ID = 111 function array_group_by( $array, $id ){ $groups = array(); foreach( $array as $row ) $groups[ $row -> $id ][] = $row; return $groups; } // group rows from table2/table3 by their parent IDs $p2 = array_group_by( $t2, 'table1_ID' ); $p3 = array_group_by( $t3, 'table2_ID' ); // let's combine results: $result = array(); foreach( $t1 as $row1 ){ $row1 -> Table2_Objects = isset( $p2[ $row1 -> ID ]) ? $p2[ $row1 -> ID ] : array(); foreach( $row1 -> Table2_Objects as $row2 ) $row2 -> Table3_Objects = isset( $p3[ $row2 -> ID ]) ? $p3[ $row2 -> ID ] : array(); $result[] = $row1; } echo json_encode( $result );更多推荐
来自3个一对多表的嵌套JSON
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