" varchar":" string"类型org.json.jsonobject无法转换为jsonarray

时间:2015-10-17 13:12:51

标签: android json

我在运行项目时遇到问题看到这个错误:" bookname":" farnaz"类型org.json.jsonobject无法转换为jsonarray。我搜索了这个错误,但没有找到这个问题。 这是我的代码:

  public class Books {
  long bookid;
  String bookname;
  String author;
  String publisher;
  double cost;
  int likenumber;
  int readedtimes;
  public long getBookid() {
    return bookid;
 }
 public void setBookid(long bookid) {
    this.bookid = bookid;
 }
 public String getBookname() {
    return bookname;
 }
 public void setBookname(String bookname) {
    this.bookname = bookname;
 }
 public String getAuthor() {
    return author;
 }
 public void setAuthor(String author) {
    this.author = author;
 }
 public String getPublisher() {
    return publisher;
 }
 public void setPublisher(String publisher) {
    this.publisher = publisher;
 }
 public double getCost() {
    return cost;
 }
 public void setCost(double cost) {
    this.cost = cost;
 }
 public int getLikenumber() {
    return likenumber;
 }
 public void setLikenumber(int likenumber) {
    this.likenumber = likenumber;
 }
 public int getReadedtimes() {
    return readedtimes;
 }
 public void setReadedtimes(int readedtimes) {
    this.readedtimes = readedtimes;
}
}
public class BooksJSONParser {
public static List<Books> parseFeed(String content) {
    try {
        JSONArray ar = new JSONArray(content);
        List<Books> blist = new ArrayList<Books>();
        for (int i = 0; i < ar.length(); i++) {
            JSONObject obj = ar.getJSONObject(i);
            Books b = new Books();
            b.setBookname(obj.getString("bookname"));               
            b.setAuthor(obj.getString("author"));
            b.setPublisher(obj.getString("publisher"));
            b.setCost(obj.getDouble("cost"));`enter code here`
            b.setBookid(obj.getLong("id"));
            b.setLikenumber(obj.getInt("likes"));
            b.setReadedtimes(obj.getInt("readedtimes"));`enter code here`   
            blist.add(b);
        return blist;
    } catch (JSONException e) {
        e.printStackTrace();
        return null;
    }

 }

0 个答案:

没有答案