PHP获取JSON生成select下拉选框问题

这两天搞微信企业号接口，获取了一段JSON，想在PHP里通过userlist中的usrid和name内容生成相关的select下拉选框，应该怎样写好，好似只能用AJAX来搞吧？

"{\"errcode\":0,\"errmsg\":\"ok\",\"userlist\":[{\"userid\":\"ersuo\",\"name\":\"\u6881\u51ef\u6b23\",\"department\":[]},{\"userid\":\"sabrina\",\"name\":\"\u8d75\u5b9d\u83b9\",\"department\":[]},{\"userid\":\"kelly\",\"name\":\"\u9648\u70ab\u534e\",\"department\":[]},{\"userid\":\"eva\",\"name\":\"eva\",\"department\":[]},{\"userid\":\"zhongzhong\",\"name\":\"\u949f\u548f\u6bb7\",\"department\":[]}]}"

回复讨论(解决方案)

就用ajax实现

求代码

不劳而获是大忌，给你写个示例，结合JQ

var a=JSON.parse("{\"errcode\":0,\"errmsg\":\"ok\",\"userlist\":[{\"userid\":\"ersuo\",\"name\":\"\u6881\u51ef\u6b23\",\"department\":[]},{\"userid\":\"sabrina\",\"name\":\"\u8d75\u5b9d\u83b9\",\"department\":[]},{\"userid\":\"kelly\",\"name\":\"\u9648\u70ab\u534e\",\"department\":[]},{\"userid\":\"eva\",\"name\":\"eva\",\"department\":[]},{\"userid\":\"zhongzhong\",\"name\":\"\u949f\u548f\u6bb7\",\"department\":[]}]}"); var select="";$(a.userlist).each(function(i,data){select+=""+data.name+"";})select+=""; $("body").append(select)