JSON對象和JSON字串,json對象字串
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html xmlns="http://www.w3.org/1999/xhtml"><head><title>JSON.parse()</title><script type="text/javascript">//樣本1:此樣本使用 JSON.parse 將 JSON 字串轉換為對象var jsontext = '{"firstname":"Jesper","surname":"Aaberg","phone":["555-0100","555-0120"]}';//JSON 字串var contact = JSON.parse(jsontext);document.write(contact.surname + ", " + contact.firstname + ", "+ contact.phone);//樣本2:和執行個體1是一樣的效果var jsontext2 = {"firstname":"Jesper","surname":"Aaberg","phone":["555-0100","555-0120"]};//JSON 對象//var contact2 = JSON.parse(jsontext2);document.write("<br /><br />"+jsontext2.surname + ", " + jsontext2.firstname + ", "+ jsontext2.phone);</script></head><body></body></html>
輸出:
Aaberg, Jesper, 555-0100,555-0120
Aaberg, Jesper, 555-0100,555-0120
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前端頁面接收JSON對象的執行個體:
<script>sendRecord('1');function sendRecord(record){var req = {user_id:<?php echo $userId;?>,record:record,}$.ajax({url: "/3G/wall/ajax_send_record/",type:"post",data:req,dataType:"JSON", //返回資料格式為JSON對象success: function(res){if(res.result==1){ //因為傳遞過來是JSON對象,所以不用JSON.parse()解析alert('11');}else if(res.result==2){alert('22');}else if(res.result==3){alert('33');}},error: function(){alert('error000');console.log(this);}});}</script>
<?php function ajax_send_record() { $record = $_POST('record'); if ($record==1) { $json['result'] = 1; }elseif($record==2){$json['result'] = 2;}elseif(elseif($record==3){$json['result'] = 3;}$json = json_encode($json);echo $json; }?>
將json字串用jsonnet轉成對象
根據你的json資料,可以像下面方式進行轉換:
調用Custom 類中的 DeJson方法 傳入json字串 返回對象
java:json字串轉化為json對象
你好!!
import java.io.*; import org.json.*; public class Demo { public static void main(String[] args) throws Exception { String str = "{\"brand_no\":\"jycy,sy\",\"unit_rank\":\"2\",\"package\":\"2\"}"; JSONObject obj = new JSONObject(str); System.out.println(obj); System.out.println(obj.get("brand_no")); // "jycy,sy" } }