PHP获取JSON生成select下拉选框问题

Question

phpjson微信企业

Answer 1

不劳而获是大忌，给你写个示例，结合JQvar 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="<select name='user'>";
$(a.userlist).each(function(i,data){select+="<option value="+data.userid+">"+data.name+"</option>";
})select+="</select>"; 
$("body").append(select)