PHP擷取JSON產生select下拉選框問題

這兩天搞企業號介面，擷取了一段JSON，想在PHP裡通過userlist中的usrid和name內容產生相關的select下拉選框，應該怎樣寫好，好似只能用AJAX來搞吧？

"{\"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\":[]}]}"

回複討論(解決方案)

就用ajax實現

求代碼

不勞而獲是大忌，給你寫個樣本，結合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)

