Json获取数据值

时间:2017-10-07 06:56:32

标签: php jquery json

我只需要获取id数据

{"register":{"id":"1","members_count":null,"name":"Neha","user_name":"Neha","email":"help.eduexpression.com@gmail.com","profileId":"8866","maritialstatus_id":"1","religion_id":"1","caste_id":null,"mothertongue_id":"4","employed_id":"6","country_id":"103"}}

需要做jquery json decode

o / p:register.id

2 个答案:

答案 0 :(得分:3)

var jsondata='{"register":{"id":"1","members_count":null,"name":"Neha","user_name":"Neha","email":"help.eduexpression.com@gmail.com","profileId":"8866","maritialstatus_id":"1","religion_id":"1","caste_id":null,"mothertongue_id":"4","employed_id":"6","country_id":"103"}}';

var result=$.parseJSON(jsondata);

// OR

var result=JSON.parse(jsondata);

console.log(result.register.id); //result will be 1

enter image description here

答案 1 :(得分:0)

请尝试这个:

var obj ='{"register":{"id":"1","members_count":null,"name":"Neha","user_name":"Neha","email":"help.eduexpression.com@gmail.com","profileId":"8866","maritialstatus_id":"1","religion_id":"1","caste_id":null,"mothertongue_id":"4","employed_id":"6","country_id":"103"}}';

var json = JSON.parse(obj);

alert(json.register['id']);