如何返回其中包含 json 数组的 json 对象？答案

【问题标题】：How to return json object with json array inside of it?如何返回其中包含 json 数组的 json 对象？
【发布时间】：2021-05-04 08:02:10
【问题描述】：

我有多对多的关系像这样：服务器版本：10.4.17-MariaDB

表格颜色（ID、名称）。
表格项目（id、title....）。
表 item_color(id,items_id,color_id)。

我的查询是这样的：

SELECT items.*,colors.name FROM items,item_color,colors
 where
 items.id = item_color.item_id
 and
 colors.id = item_color.color_id

我使用 php 函数 json_encode()。 如何退货：

{
    "id": "22",
    "title": "my products 515151",
    "descreption": "5454545455",
    "price": "0.05",
    "quantity": "2",
    "date_added": "2021-01-29 14:37:24",
    "primary_image": "http://localhost/ecomerce/uploads/1611927444hat.jpg",
    "color": [
      0 : "pink",
      1 : "white"
      ]

  }

如果这样：

    "id": "22",
    "title": "my products 515151",
    "descreption": "5454545455",
    "price": "0.05",
    "quantity": "2",
    "date_added": "2021-01-29 14:37:24",
    "primary_image": "http://localhost/ecomerce/uploads/1611927444hat.jpg",
    "color": "pink"
  },
  {
    "id": "22",
    "title": "my products 515151",
    "descreption": "5454545455",
    "price": "0.05",
    "quantity": "2",
    "date_added": "2021-01-29 14:37:24",
    "primary_image": "http://localhost/ecomerce/uploads/1611927444hat.jpg",
    "color": "red"
  },

【问题讨论】：

按颜色聚合并按所有其他列分组

标签： php mysql sql json mariadb

【解决方案1】：

有两种方法：

如果要保留查询，必须先解析数据，然后再将其解析为 json。

...
function regroup_color($data = array()){
  $ret = array();
  foreach($data as $d){
    $id = $d['id'];
    if(!isset($ret[$id])){
      $color = $d['color'];
      unset($d['color']);
      $ret[$id] = $d;
      $ret[$id]['color'][] = $color;
    }else{
      $ret[$id]['color'][] = $d['color'];
    }
  }
  return $ret;
}
$data = regroup_color($data);
echo json_encode($data)
...

或者你可以……

查询 2 部分，第一个是获取所有 items，第二个是获取 colors

...
$query = "SELECT * FROM items";
// get the data here using query above
$data = {{result of query}};

foreach($data as $i => $d){
  $id = $d['id'];
  $query = "SELECT * FROM item_color JOIN colors ON item_color.color_id = colors.id";
  // get the data here using query above
  $colors = {{result of query}};
  foreach($colors as color){
    $data[$i]['color'][] = $color['name'];
  }
}
echo json_encode($data)
...

【讨论】：

【解决方案2】：

我还想添加类别，所以这就是我带来的

function regroup_color($data = array(),$data2 = array()){
  // array that will hold our data and we will return it later
$ret = array();
// fetch the data as d
foreach($data as $d){
  //save the id 
  $id = $d['id'];
  //make sure no id was set
  if(!isset($ret[$id])){
    // save the name of the color
    $color = $d['color'];
    unset($d['color']);
    //save all the item data
    $ret[$id] = $d;
    // add color to color array
    $ret[$id]['color'][] = $color;
  }else{
    // if wa alredy did all the above things just keep adding colors to color array
    $ret[$id]['color'][] = $d['color'];
  }
}
  
  
foreach($data2 as $d){

    //save the id 
      $id = $d['id'];
      //make sure no id was set
      if(!isset($ret[$id])){
        // save the name of the color
        $category = $d['category'];
        unset($d['category']);
        $ret[$id] = $d;
        $ret[$id]['category'][] = $category;
      }else{
        $ret[$id]['category'][] = $d['category'];
      }
    }

返回 $ret; }

结果：

"22": {
"id": "22",
"title": "my products 515151",
"descreption": "5454545455",
"price": "0.05",
"quantity": "2",
"date_added": "2021-01-29 14:37:24",
"primary_image": "http://localhost/ecomerce/uploads/1611927444hat.jpg",
"color": [
  "black",
  "white",
  "pink",
  "red",
  "yellow"
],
"category": [
  "buttom",
  "hats",
  "shoes"
]

}

【讨论】：