聚合JSON输出

Question

我在PHP上有一个JSON输出，如：

[{"Title":"Message","count":"380","Number":"600"}]

如何获得如下结果？

"Column 'examplecolumn' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause."

Answer 1

在php中：

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

参考：example