在PHP中合并两个json

Question

我有两个json的

第一个是

    [{"COLUMN_NAME":"ORDER_NO","COLUMN_TITLE":"Order Number"}
,{"COLUMN_NAME":"CUSTOMER_NO","COLUMN_TITLE":"Customer Number"}]

第二个是

[{"COLUMN_NAME":"ORDER_NO","DEFAULT_VALUE":"1521"},
{"COLUMN_NAME":"CUSTOMER_NO","DEFAULT_VALUEE":"C1435"}]

我想合并他们,并有一个像json

[{"COLUMN_NAME":"ORDER_NO","COLUMN_TITLE":"Order Number","DEFAULT_VALUE":"1521"}
,{"COLUMN_NAME":"CUSTOMER_NO","COLUMN_TITLE":"Customer Number","DEFAULT_VALUEE":"C1435"}]

有没有办法合并它们？如果需要更改JSON的结构,我也可以

谢谢.

Answer 1

有点像json_encode(array_merge(json_decode($a, true),json_decode($b, true)))应该工作.

在官方PHP文档中的 array_merge

官方PHP文档中的 json_decode

编辑:尝试添加true第二个参数到json_decode.那会将对象转换为关联数组.

编辑2:尝试array-merge-recursive看下面的评论.抱歉,现在必须退出:(这看起来像一个完整正确的解决方案:https://stackoverflow.com/a/20286594/1466341

Answer 2

管理把它扔在一起.最有可能是更好的解决方案,但这是我得到的最接近的解决方案.

$a = '[{"COLUMN_NAME":"ORDER_NO","COLUMN_TITLE":"Order Number"},{"COLUMN_NAME":"CUSTOMER_NO","COLUMN_TITLE":"Customer Number"}]';
$b = '[{"COLUMN_NAME":"ORDER_NO","DEFAULT_VALUE":"1521"},{"COLUMN_NAME":"CUSTOMER_NO","DEFAULT_VALUEE":"C1435"}]';
$r = [];
foreach(json_decode($a, true) as $key => $array){
 $r[$key] = array_merge(json_decode($b, true)[$key],$array);
}
echo json_encode($r);

回报,

[{"COLUMN_NAME":"ORDER_NO","DEFAULT_VALUE":"1521","COLUMN_TITLE":"Order Number"},
{"COLUMN_NAME":"CUSTOMER_NO","DEFAULT_VALUEE":"C1435","COLUMN_TITLE":"Customer Number"}]

Answer 3

这对我来说就像一个魅力

json_encode(array_merge(json_decode($a, true),json_decode($b, true)))

这是一个完整的例子

$query="SELECT * FROM `customer` where patient_id='1111118'";

$mysql_result = mysql_query($query);

$rows = array();
while($r = mysql_fetch_assoc($mysql_result)) {
    $rows[] = $r;
}
$json_personal_information=json_encode($rows);
//echo $json_personal_information;




$query="SELECT * FROM `doctor` where patient_id='1111118'";

$mysql_result = mysql_query($query);

$rows = array();
while($r = mysql_fetch_assoc($mysql_result)) {
    $rows[] = $r;
}
$json_doctor_information=json_encode($rows);
//echo $json_doctor_information;

echo $merger=json_encode(array_merge(json_decode($json_personal_information, true),json_decode($json_doctor_information, true)));