Question

有没有办法可以将这种java.lang.object解析/转换为java.lang.Object的数组或List：

[
    {procedure_name_txt=nameText, procedure_name_ets_id=ID-1234, procedure_ets_desc=description_123}, 
    {procedure_name_txt=deafness, procedure_name_ets_id=ID-99022, procedure_ets_desc=description_31222}
]

已经尝试过使用Gson，但遗憾的是对象格式不是json格式。

Answer 1

由于以下几个原因，您提供的文件不是有效的JSON：

数组元素键未包含在引号"中。
数组元素键/值对不是以:分隔，而是以=分隔。
字符串值未包含在引号"中。

我几乎可以肯定该文档实际上是一个集合/数组的精美打印toString结果，其元素是地图或POJO，其中toString被覆盖。这样的输出从不意味着以编程方式解析。但是，Gson可以解析＆＃34; tostring＆＃34;结果是因为它的简单性违反了JSON格式语法并且太宽松了。

private static final String LOOKS_LIKE_JSON = "[\n" +
        "    {procedure_name_txt=nameText, procedure_name_ets_id=ID-1234, procedure_ets_desc=description_123}, \n" +
        "    {procedure_name_txt=deafness, procedure_name_ets_id=ID-99022, procedure_ets_desc=description_31222}\n" +
        "]";

//  private static final Type procedureListType = TypeToken.getParameterized(List.class, Procedure.class).getType();
private static final Type procedureListType = new TypeToken<List<Procedure>>() {
}.getType();

//  private static final Type procedureArrayType = Procedure[].class;
private static final Type procedureArrayType = new TypeToken<Procedure[]>() {
}.getType();

private static final Type stringToStringMapListType = new TypeToken<List<Map<String, String>>>() {
}.getType();

private static final Type stringToStringMapArrayType = new TypeToken<Map<String, String>[]>() {
}.getType();

private static final Gson gson = new Gson();
private static final JsonParser jsonParser = new JsonParser();

public static void main(final String... args) {
    final List<Procedure> procedureList = gson.fromJson(LOOKS_LIKE_JSON, procedureListType);
    System.out.println(procedureList);
    final Procedure[] procedureArray = gson.fromJson(LOOKS_LIKE_JSON, procedureArrayType);
    System.out.println(Arrays.toString(procedureArray));
    final List<Map<String, String>> mapList = gson.fromJson(LOOKS_LIKE_JSON, stringToStringMapListType);
    System.out.println(mapList);
    final Map<String, String>[] mapArray = gson.fromJson(LOOKS_LIKE_JSON, stringToStringMapArrayType);
    System.out.println(Arrays.toString(mapArray));
    final JsonElement jsonElement = jsonParser.parse(LOOKS_LIKE_JSON);
    System.out.println(jsonElement);
}

Procedure类如下：

final class Procedure {

    @SerializedName("procedure_name_txt")
    final String name = null;

    @SerializedName("procedure_name_ets_id")
    final String id = null;

    @SerializedName("procedure_ets_desc")
    final String description = null;

    @Override
    public String toString() {
        return new StringBuilder("Procedure{")
                .append("name='").append(name).append('\'')
                .append(", id='").append(id).append('\'')
                .append(", description='").append(description).append('\'')
                .append('}')
                .toString();
    }

}

输出：

IntelliJ IDEA生成的Procedure.toString：

[程序{name =＆＃39; nameText＆＃39;，id =＆＃39; ID-1234＆＃39;，description =＆＃39; description_123＆＃39;}，程序{name =＆＃39;耳聋＆＃39;，id =＆＃39; ID-99022＆＃39;，description =＆＃39; description_31222＆＃39;}]

字符串到字符串条目的映射列表（实际上这反映了给定的输入），JDK toString()：

[{procedure_name_txt = nameText，procedure_name_ets_id = ID-1234，procedure_ets_desc = description_123}，{procedure_name_txt = deafness，procedure_name_ets_id = ID-99022，procedure_ets_desc = description_31222}]

Gson JSON树模型toString()：

[{＆＃34; procedure_name_txt＆＃34;：＆＃34; nameText＆＃34;＆＃34; procedure_name_ets_id＆＃34;：＆＃34; ID-1234＆＃34;＆＃34; procedure_ets_desc＆＃34 ;：＆＃34; description_123＆＃34;}，{＆＃34; procedure_name_txt＆＃34;：＆＃34;耳聋＆＃34;＆＃34; procedure_name_ets_id＆＃34;：＆＃34; ID-99022＆＃34; ，＆＃34; procedure_ets_desc＆＃34;：＆＃34; description_31222＆＃34;}]

java.lang.Object到arraylist

1 个答案: