我有两个从
JSON转换的数据对象.两者都非常复杂,我想以类似的方式将它们合并到jQuery将如何使用extends合并两个对象.
例
JSON 1：
{
...
"blah":
{
"params":
{
"foo":
{
"default": "bar","misc": "0",...
},...
},...
},...
}
JSON 2：
{
...
"blah":
{
"params":
{
"foo":
{
"value": "val","misc": "1",...
}
合并成
{
...
"blah":
{
"params":
{
"foo":
{
"default": "bar","value": "val",...
}
使用PHP对象来处理这个问题的最好方法是什么？
将每个JSON字符串解码为关联数组,合并结果并进行重新编码
$a1 = json_decode( $json1,true ); $a2 = json_decode( $json2,true ); $res = array_merge_recursive( $a1,$a2 ); $resJson = json_encode( $res );
更新：
如果您具有特定的合并要求,则需要编写自己的合并功能.
我在下面写了一个符合您的要求的问题.
如果您还没有提到其他要求,则可能需要进行调整.
<?PHP
$json1 = '
{
"blah":
{
"params":
{
"foo":
{
"default": "bar","misc": "0"
}
},"lost":
{
"one": "hat","two": "cat"
}
}
}';
$json2 = '
{
"blah":
{
"lost": "gone","params":
{
"foo":
{
"value": "val","misc": "1"
}
}
},"num_array": [12,52,38]
}';
$a1 = json_decode( $json1,true );
/*
* Recursive function that merges two associative arrays
* - Unlike array_merge_recursive,a differing value for a key
* overwrites that key rather than creating an array with both values
* - A scalar value will overwrite an array value
*/
function my_merge( $arr1,$arr2 )
{
$keys = array_keys( $arr2 );
foreach( $keys as $key ) {
if( isset( $arr1[$key] )
&& is_array( $arr1[$key] )
&& is_array( $arr2[$key] )
) {
$arr1[$key] = my_merge( $arr1[$key],$arr2[$key] );
} else {
$arr1[$key] = $arr2[$key];
}
}
return $arr1;
}
$a3 = my_merge( $a1,$a2);
$json3 = json_encode( $a3 );
echo( $json3 );
/*
{
"blah":
{
"params":
{
"foo":
{
"default": "bar","misc": "1","value": "val"
}
},"lost": "gone"
},38]
}
*/