JSON.parseObject("ok", String.class) 为什么总是报错，是啥原因呢

[ZuoxLiu]

com.alibaba.fastjson.JSONException: illegal input : o, offset 1, character o, line 1, column 1, fastjson-version 2.0.47 ok
at com.alibaba.fastjson.JSON.parseObject(JSON.java:525)

[MandarGharse]

Hi [ZuoxLiu],
What is the expected output ? The error is because "ok" is not a valid json string

[ZuoxLiu]

Hi [ZuoxLiu], What is the expected output ? The error is because "ok" is not a valid json string

请求接口时，统一返回{code: 200, msg: "success", data: "ok"}因为格式是统一的，data是动态的，当判断code是200后，需要将data转换为特定的对象，当是其它对象时或返回的字符串不是"ok"是都没有问题，而data是"ok"时，返回上面的报错，需要对此字符串特殊处理。