将包含对象数组的 JSON 转换为正确的 Java 对象

Question

我有一个简单的 Spring Boot 项目，其中调度程序定期使用 RESTful API 并转换传入的 JSON 文件。JSON 文件实际上是具有一些键和值的对象数组：[{"CoID":1,"CoName":"کشاورزی و دامپروری مگسال","CoNameEnglish":"MagsalAgriculture & Animal Husbandry Co.","CompanySymbol":"MAGS","CoTSESymbol":"زمگسا","GroupID":1,"GroupName":"كشاورزی و دامپروری","IndustryID":1,"IndustryName":"كشاورزی، دامپروری و خدمات وابسته به آن","InstCode":"5054819322815158","TseCIsinCode":"IRO1MAGS0006","TseSIsinCode":"IRO1MAGS0001","MarketID":1,"MarketName":"بورس"},...]我有一个名为的类，其Company字段与 JSON 文件中数组中的对象之一类似：@JsonIgnoreProperties(ignoreUnknown = true)public class Company {private int CoID;private String CoName;private String CoNameEnglish;private String CompanySymbl;private String CoTSESymbl;private int GroupID;private String GroupName;private int IndustryID;private String IndustryName;private String IndustryCode;private String TseCIsinCode;private String TseSIsinCode;private int MarketID;private String MarketName;    // And proper getters, setters and constructor //我还创建了一个名为的包装类CompanyList：public class CompanyList {private ArrayList<Company> companyList;public ArrayList<Company> getCompanyList() {    return companyList;}public void setCompanyList(ArrayList<Company> companyList) {    this.companyList = companyList;}public CompanyList() {}@Overridepublic String toString() {    return "CompanyList [companyList=" + companyList + "]";}}

Answer 1

第二种和第三种方法应该可以正常工作。您需要检查您的 json 响应结构。

您可以使用以下 json 进行测试（它们适用于您的代码）：

第二种方法：

[{"tseCIsinCode":null,"tseSIsinCode":null,"coName":"n1","industryID":0,"coID":0,"coNameEnglish":null,"companySymbl":null,"coTSESymbl":null,"groupID":0,"groupName":null,"industryName":null,"industryCode":null,"marketID":0,"marketName":null},{"tseCIsinCode":null,"tseSIsinCode":null,"coName":"n2","industryID":0,"coID":0,"coNameEnglish":null,"companySymbl":null,"coTSESymbl":null,"groupID":0,"groupName":null,"industryName":null,"industryCode":null,"marketID":0,"marketName":null}]

第三：

{"companyList":[{"coName":"n1","coID":0,"coNameEnglish":null,"companySymbl":null,"coTSESymbl":null,"groupID":0,"groupName":null,"industryID":0,"industryName":null,"industryCode":null,"tseCIsinCode":null,"tseSIsinCode":null,"marketID":0,"marketName":null},{"coName":"n2","coID":0,"coNameEnglish":null,"companySymbl":null,"coTSESymbl":null,"groupID":0,"groupName":null,"industryID":0,"industryName":null,"industryCode":null,"tseCIsinCode":null,"tseSIsinCode":null,"marketID":0,"marketName":null}]}

更新：

第二种方法修复：更改 json 字段名称 -“CoName”->“coName”、“CoID”->“coID”等。更改后它将完美运行。

第三种方法修复：用第二种方法包装你的 json"{\"companyList\":[...] 并更改字段名称

第二次更新 如果您无法更改响应中的 json。您可以在 Company 类中使用映射

@JsonProperty("CoName")
private String CoName;