Question

您好我从服务器端返回了以下JSON：

{"command":"SELECT","rowCount":1,"oid":null,"rows":[{"username":"xxxx"}],"fields":[{"name":"username","tableID":34722,"columnID":3,"dataTypeID":1043,"dataTypeSize":-1,"dataTypeModifier":204,"format":"text"}],"_parsers":[null]}

我在javascript中解析json，如下所示：

 var ParsedJSONResponse = $.parseJSON(JSONResponseFromServerSide);

变量"ParsedJSONResponse"总是为null，我的JSON有效我在JSONLint中检查了它，那么结果是什么？

Answer 1

您应该解析字符串

var stringJson = '{"command":"SELECT","rowCount":1,"oid":null,"rows":[{"username":"xxxx"}],"fields":[{"name":"username","tableID":34722,"columnID":3,"dataTypeID":1043,"dataTypeSize":-1,"dataTypeModifier":204,"format":"text"}],"_parsers":[null]}';  
var ParsedJSONResponse = $.parseJSON(stringJson);

你只有一个Json（JSONResponseFromServerSide），没有理由解析它。

解析Json对象返回null。

$.parseJSON({}); // returns `null`

Answer 2

您的代码是正确的，请查看此JSFiddle。

var JSONResponseFromServerSide = '{"command":"SELECT","rowCount":1,"oid":null,"rows":[{"username":"xxxx"}],"fields":[{"name":"username","tableID":34722,"columnID":3,"dataTypeID":1043,"dataTypeSize":-1,"dataTypeModifier":204,"format":"text"}],"_parsers":[null]}'; //it's your json string
$.parseJSON(JSONResponseFromServerSide); //nothing wrong

如果JSONResponseFromServerSide已经是JavaScript对象，那么您不必执行parseJSON

JQuery解析JSON返回null

2 个答案: