在PHP中合并两个json

这样的东西应该可以工作：

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

或者与一个衬垫相同：

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

array_merge在官方PHP文档中

json_decode在官方PHP文档中

编辑：尝试将true作为第二个参数添加到json_decode。这将把对象转换为关联数组。

编辑2：尝试array-merge-recursive并查看下面的评论。抱歉，现在必须注销：(这看起来是一个完全正确的解决方案：https://stackoverflow.com/a/20286594/1466341

设法把这些放在一起。很可能有更好的解决方案，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);

退货，

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

这对我来说就像一个魅力

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

这是一个完整的示例

$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)));

这对我很管用！

$dados1 = json_encode(array('data'=>'1')); 
$dados2 = json_encode(array('data2'=>'2')); 
echo json_encode(array_merge(json_decode($dados1, true),json_decode($dados2, true)));