Php 来自$\u POST方法的JSON打印结果

当我以数组格式获取结果时，这会阻止我以所需格式发布值

我需要所需格式的结果（这些输入来自while循环）：

我得到的结果如下：

{
"delv_num":"0006020715",
"hu_id":"HU000000000000000225",
"box_scan_value":"68889",
"type":"box",
"hu_detail":{
"reference_id":["TD0000000000991","TD0000000000990","TD0000000000767"],
"qty":["2","0","0"]
}
}

但我需要以下格式的结果：

{
"delv_num":"0006020715",
"hu_id":"HU000000000000000225",
"box_scan_value":"68889",
"type":"box",
"hu_detail":[{
"reference_id":"TD0000000000991",
"qty":"2"
},
{
"reference_id":"TD0000000000990",
"qty":"0"
},
{
"reference_id":"TD0000000000767",
"qty":"0"
}
]
}

提前谢谢

像这样试试

$references =  $_POST['ref_id'];
$qtys = $_POST['qty'];

$data = array(
'delv_num' => $delv_num,
'hu_id' => $hu_id,
'box_scan_value' => $box_scan_value,
'type' => $type,
);

$huDetails = [];

foreach($references as $index => $ref){
   $huDetails[$index]['ref'] = $ref;
   $huDetails[$index]['qty'] = $qtys[$index]; //assuming you have `ref` and `qty` has same number of index, if not then you need to modify your html structure
}

$data['hu_details'] = $huDetails;

print_r(json_encode($data))

---

你应该可以自己做这件事。如果没有，您应该再次学习php基础知识：

$complete = [
    'delv_num'       => $delv_num,
    'hu_id'          => $hu_id,
    'box_scan_value' => $box_scan_value,
    'type'           => $type,
    'hu_detail' => []
];
foreach ($_POST['ref_id'] as $index => $refId) {
    $huDetail = [
        'reference_id'=> $refId,
        'qty' => $_POST['qty'][$index]
    ];
    $complete['hu_detail'][] = $huDetail;
}

echo json_encode($complete);

---

只需通过更改输入的名称来更改表单的结构

<input type="text" name="hu_detail[0][reference_id]" value="abc">
<input type="text" name="hu_detail[0][qty]" value="12">

试试这个：

$result = array();
$array_a = array("1","2","3");

$array_b = array("test1","test2","test3");

foreach($array_a as $key=> $value){
  $result[$key]['reference_id'] = $value;
  $result[$key]['qty'] = $array_b[$key];
}

$myJSONString = json_encode($result);

---

一句话中可能的解决方案：

$data1 = [
    'delv_num' => $delv_num,
    'hu_id' => $hu_id,
    'box_scan_value' => $box_scan_value,
    'type' => $type,
    'hu_detail' => array_map(
         function ($id, $qty) {
             return ['reference_id': $id, 'qty': $qty];
         },
         $_POST['ref_id'],
         $_POST['qty']
    ),
];

---

是什么阻止您构建所需的数据结构？你可能需要更多地了解。这是一个很长的页面，但它包含了你需要的所有答案。我读了，它对我的查询没有帮助。然后再读一遍。提示：用于迭代

$\u POST['ref\u id']

并构建列表以放入

'hu\u details'

，不要像在目标对象中一样放入

$\u POST['ref\u id']

。将3个数组创建为$data1、$data2、$data3，并传递相应的值，最后合并数组

$result = array();
$array_a = array("1","2","3");

$array_b = array("test1","test2","test3");

foreach($array_a as $key=> $value){
  $result[$key]['reference_id'] = $value;
  $result[$key]['qty'] = $array_b[$key];
}

$myJSONString = json_encode($result);

$data1 = [
    'delv_num' => $delv_num,
    'hu_id' => $hu_id,
    'box_scan_value' => $box_scan_value,
    'type' => $type,
    'hu_detail' => array_map(
         function ($id, $qty) {
             return ['reference_id': $id, 'qty': $qty];
         },
         $_POST['ref_id'],
         $_POST['qty']
    ),
];