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\":[]}]}"------解决思路----------------------
不劳而获是大忌，给你写个示例，结合jq
<br><br>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\":[]}]}");<br> <br>var select="<select name='user'>";<br>$(a.userlist).each(function(i,data){<br><br>select+="<option value="+data.userid+">"+data.name+"</option>";<br>})<br><br>select+="</select>"; <br>$("body").append(select)<br>
