在 PHP 中合并两个 json
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Merging two json in PHP
提问by Calipso
I have two json's
我有两个json
First one is
第一个是
[{"COLUMN_NAME":"ORDER_NO","COLUMN_TITLE":"Order Number"}
,{"COLUMN_NAME":"CUSTOMER_NO","COLUMN_TITLE":"Customer Number"}]
Second one is
第二个是
[{"COLUMN_NAME":"ORDER_NO","DEFAULT_VALUE":"1521"},
{"COLUMN_NAME":"CUSTOMER_NO","DEFAULT_VALUEE":"C1435"}]
I want to merge them and have a json like
我想合并它们并有一个像
[{"COLUMN_NAME":"ORDER_NO","COLUMN_TITLE":"Order Number","DEFAULT_VALUE":"1521"}
,{"COLUMN_NAME":"CUSTOMER_NO","COLUMN_TITLE":"Customer Number","DEFAULT_VALUEE":"C1435"}]
is there a way to merge them? It is also OK for me if a stucture change in JSON is required
有没有办法合并它们？如果需要改变 JSON 的结构，我也可以
thanks.
谢谢。
回答by DarkSide
Something like json_encode(array_merge(json_decode($a, true),json_decode($b, true)))should work.
类似的东西json_encode(array_merge(json_decode($a, true),json_decode($b, true)))应该工作。
array_mergein official PHP documentation
PHP 官方文档中的array_merge
json_decodein official PHP documentation
PHP 官方文档中的json_decode
EDIT:try adding trueas second parameter to json_decode. That'll convert objects to associative arrays.
编辑：尝试将true第二个参数添加到 json_decode。这会将对象转换为关联数组。
EDIT 2: try array-merge-recursiveand see my comment below. Sorry have to log out now :(
This looks like a full correct solution: https://stackoverflow.com/a/20286594/1466341
编辑 2：尝试array-merge-recursive在下面查看我的评论。抱歉现在必须注销:(这看起来是一个完整正确的解决方案：https: //stackoverflow.com/a/20286594/1466341
回答by Ben Fortune
Managed to throw this together. There is most likely a better solution, but this is the closest I got.
设法把它放在一起。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);
returns,
返回，
[{"COLUMN_NAME":"ORDER_NO","DEFAULT_VALUE":"1521","COLUMN_TITLE":"Order Number"},
{"COLUMN_NAME":"CUSTOMER_NO","DEFAULT_VALUEE":"C1435","COLUMN_TITLE":"Customer Number"}]
回答by bobthenob
This works like a charm for me
这对我来说就像一个魅力
json_encode(array_merge(json_decode($a, true),json_decode($b, true)))
here is a full example
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)));
回答by Welisson Carlos Dias
This worked to me! $dados1 = json_encode(arrray('data'=>'1')); $dados2 = json_encode(arrray('data2'=>'2')); echo json_encode(array_merge(json_decode($dados1, true),json_decode($dados2, true)));
这对我有用！$dados1 = json_encode(arrray('data'=>'1')); $dados2 = json_encode(arrray('data2'=>'2')); 回声 json_encode(array_merge(json_decode($dados1, true),json_decode($dados2, true)));
