Question

我在PHP中创建JSON以从另一个应用程序读取，我想将JSON数据分类为

jobs:
  job:
    ID: 123
    name: Job Name 1
  job:
    ID: 321
    name: Job Name 2
  job:
    ID: 231
    name: Job Name 3

我得到的是

0:
  ID: 123
  name: Job Name 1
1:
  ID: 321
  name: Job Name 2
2:
  ID: 231
  name: Job Name 3

我用来创建JSON的PHP代码如下：

$rs = mysqli_query($dbc, "SELECT id, name FROM job");
while($obj = mysqli_fetch_object($rs)) {
    $arr[] = $obj;
}
echo json_encode($arr, JSON_FORCE_OBJECT);

如何将JSON中的每个项目归类为“作业”并将整个列表归类为“作业”，我需要做什么？

Answer 1

试试这个

$rs = mysqli_query($dbc, "SELECT id, name FROM job");
while($obj = mysqli_fetch_object($rs)) {
    $arr['job'][] = $obj;
}
echo json_encode($arr, JSON_FORCE_OBJECT);

Answer 2

试试这样的

while($obj = mysqli_fetch_object($rs)) {
    $arr['jobs'][]['job'] = $obj;
}
echo json_encode($arr);

更新：

$obj_user1 = new stdClass;
$obj_user1->id = "1";
$obj_user1->name = "a";

$obj_user2 = new stdClass;
$obj_user2->id = "2";
$obj_user2->name = "b";

$obj_user = array($obj_user1,$obj_user2);

foreach($obj_user as $a){
   $arr['jobs'][]['job'] = $a;
}

print_r(json_encode($arr));

输出 {"jobs":[{"job":{"id":"1","name":"a"}},{"job":{"id":"2","name":"b"}}]}

为我的JSON对象创建正确的密钥

2 个答案: