如何将我的结果集转换成一个分组的多维数组，然后是一个json字符串？
如何将我的结果集转换成一个分组的多维数组，然后是一个json字符串？
提问于 2018-08-15 21:21:48
我的数据库中有一个表，其结构和数据如下所示：
---------------------------------------------
id | type | name | age
---------------------------------------------
1 F-A jon 24
2 F-A roy 25
3 F-E robert 26
4 F-E sina 25我希望根据json值对数据进行分组，以生成一个特定的结构化数组，以生成以下type：
{
type:{
"F-A": [
{
"name": "jon",
"age": "24"
"id": "1"
},
{
"name": "roy",
"age": "25",
"id": "2"
}
],
"F-E": [
{
"name": "robert",
"age": "26",
"id": "3"
},
{
"name": "sina",
"age": "25",
"id": "4"
},
]
}
}请注意，每个唯一类型的子数组都包含与该值关联的行数据的集合。
我的模型：
public function prescriptionData(){
$table = 'prescription';
$this->db->select('*');
$this->db->from($table);
$query=$this->db->get();
$locations = [];
//===
$val4=[];
$this->db->select('type,name,age,id');
$this->db->from($table);
$query2=$this->db->get();
Foreach($query2->result() as $k=>$va13){
$val4[$k]= $va13;
}
foreach($query2->result_array() as $val){
if(isset($val['type'])){
$locations[type][$val['type']]=$val4;
}else{
$locations[type][$val['type']]= $val4;
}
}
return $locations;
}我的控制器：
public function prescription(){
$userCount['result'] = $userCount1 = $this->Querydata->prescriptionData();
if($userCount['result']>0){
$validation = array(
"responseCode" => $this->res = 200,
"responseMessage" => $this->login = 'Training programs details successfully added',
"data" => $this->data = $userCount['result'] );
echo json_encode($validation);这是我错误的json响应：
"type": {
"F-A": [
{
"type": "F-A",
"name": "Jon",
"age": "24",
"id": "1"
},
{
"type": " F-A",
"name": "roy",
"age": "25",
"id": "2"
},
{
"type": "F-E",
"name": "robert",
"age": "26",
"id": "3"
},
{
"type": " F-E",
"name": "sina",
"age": "25",
"id": "4"
}
],
" F-A": [
{
"type": "F-A",
"name": "Jon",
"age": "24",
"id": "1"
},
// etc.回答 1
Stack Overflow用户
发布于 2018-08-16 04:44:56
假设您正在对type列进行预排序，您只需要对键type和来自type列的迭代值进行硬编码，然后将剩余的行数据作为子数组进行推送。循环结束后，对输出数组进行编码。
我将使用pretty_print使输出更容易阅读，但您不一定需要这样做。
代码：(Demo)
$resultset = [
['id' => 1, 'type' => 'F-A', 'name' => 'jon', 'age' => 24],
['id' => 2, 'type' => 'F-A', 'name' => 'roy', 'age' => 25],
['id' => 3, 'type' => 'F-E', 'name' => 'robert', 'age' => 26],
['id' => 4, 'type' => 'F-E', 'name' => 'sina', 'age' => 25]
];
foreach ($resultset as $row) {
$output['type'][$row['type']][] = ['name' => $row['name'], 'age' => $row['age'], 'id' => $row['id']];
}
echo json_encode($output, JSON_PRETTY_PRINT);输出：
{
"type": {
"F-A": [
{
"name": "jon",
"age": 24,
"id": 1
},
{
"name": "roy",
"age": 25,
"id": 2
}
],
"F-E": [
{
"name": "robert",
"age": 26,
"id": 3
},
{
"name": "sina",
"age": 25,
"id": 4
}
]
}
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/51859543
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