-1

我的 SD 卡中有一个包含 json 数组的文本文件。比如说文件是这样的：

[
{"CountryID" : "1","CountryName" : "Australia"},
{"CountryID" : "2","CountryName" : "Japan"},
{"CountryID" : "3","CountryName" : "China"},
{"CountryID" : "4","CountryName" : "India"},
{"CountryID" : "5","CountryName" : "Holland"}
]

现在我想根据 Country ID 获取数据。前任。我想通过 ID = 2 并且我只得到一个对象。我可以在字符串变量中获取整个文件并遍历每个对象以查找我的数据。但我不认为这是最好的做法。因为在我的真实文件中，我可能有超过 1000 个我不想循环的对象。

谢谢

4

2 回答 2

0

如果将有 1000 条记录，请改用 sqllite。Json 并不意味着是数据库结构。

于 2012-07-20T07:18:42.593 回答
0

您可以使用以下代码示例来完成您的任务

    Type listType = new TypeToken<List<Country>>() {}.getType();
    Gson gson = new Gson();
    String json = "["
            + "{\"CountryID\" : \"3\",\"CountryName\" : \"China\"},"
            + "{\"CountryID\" : \"2\",\"CountryName\" : \"Japan\"},"
            + "{\"CountryID\" : \"1\",\"CountryName\" : \"Australia\"},"
            + "{\"CountryID\" : \"4\",\"CountryName\" : \"India\"},"
            + "{\"CountryID\" : \"5\",\"CountryName\" : \"Holland\"}" 
            + "]";
    // parsing the JSON array
    ArrayList<Country> countries = gson.fromJson(json, listType);
    System.out.println(countries);

    // sorting the country list based on CountryID. Ensure list is sorted, since we do binary search next
    Collections.sort(countries);
    System.out.println(countries);

    Country searchKey = new Country(3,"");
    // searching for country with ID 3
    int index = Collections.binarySearch(countries, searchKey);
    if(index < 0 || index >= countries.size() ) System.out.println("Country with ID "+searchKey.getCountryID()+ " not found in results");
    else 
    {
        System.out.println("Found Country with ID : " + searchKey.getCountryID()+ " @ index : " + index);
        Country searchResult = countries.get(index);
        System.out.println("Searched result : "+searchResult);
    }
  • 这里我使用GSON来解析Country列表
  • 为此，我使用了一个名为的数据持有者类Country

找到班级Country

public class Country implements Comparable<Country> {


    public Country(int countryID, String countryName) {
        super();
        CountryID = countryID;
        CountryName = countryName;
    }
    //"CountryID" : "1","CountryName" : "Australia"

    private int CountryID;
    private String CountryName;
    /**
     * Gets the countryID.
     * 
     * @return <tt> the countryID.</tt>
     */
    public int getCountryID() {
        return CountryID;
    }
    /**
     * Sets the countryID.
     *
     * @param countryID <tt> the countryID to set.</tt>
     */
    public void setCountryID(int countryID) {
        CountryID = countryID;
    }
    /**
     * Gets the countryName.
     * 
     * @return <tt> the countryName.</tt>
     */
    public String getCountryName() {
        return CountryName;
    }
    /**
     * Sets the countryName.
     *
     * @param countryName <tt> the countryName to set.</tt>
     */
    public void setCountryName(String countryName) {
        CountryName = countryName;
    }
    /* (non-Javadoc)
     * @see java.lang.Object#toString()
     */
    @Override
    public String toString() {
        return "Country [CountryID=" + CountryID + ", CountryName="
                + CountryName + "]";
    }
    /* (non-Javadoc)
     * @see java.lang.Comparable#compareTo(java.lang.Object)
     */
    @Override
    public int compareTo(Country o) {
        // TODO Auto-generated method stub
        return this.getCountryID()-o.getCountryID();
    }


}
  • Country类实现Comparable接口来执行Collections'二进制搜索
  • 解析成功后，我对Country列表进行了排序
  • 然后使用binarySearchofCollections类找到你的索引search-key-country
  • 现在只需使用返回index的来检索您的search-result-country对象
于 2012-07-20T12:24:36.970 回答