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[英]Getting the JSON string from a Twitter search result into Java to be parsed
[英]Outputting parsed JSON results from Google Custom Search in Java
我正在嘗試從Java命令行中的Google自定義搜索輸出一些結果(標題,URL)以進行測試,但是我一直收到java.io.EOFException錯誤。 編譯器列出了有問題的行,但是即使花了數小時來尋找答案,我也無法弄清楚要更改的內容。 我從Stack Overflow上現有的問題中提取了大部分代碼。 任何幫助表示贊賞。
package google.api.search;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.URL;
import com.google.gson.Gson;
class GSearch {
public static void main(String args[]) throws IOException {
String key = ""; //replace with API key
String qry = ""; // search key word
String cx = ""; //replace with cx
URL url = new URL ("https://www.googleapis.com/customsearch/v1?key=" +key+ "&cx=" +cx+ "&q=" +qry+ "&alt=json");
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.setRequestMethod("GET");
conn.setRequestProperty("Accept","application/json");
BufferedReader br = new BufferedReader(new InputStreamReader ( ( conn.getInputStream() ) ) );
String output;
{while ((output = br.readLine()) != null){
GResults results = new Gson().fromJson(output, GResults.class);
System.out.println(results);
}
conn.disconnect();
}
}
}
GResults類:
public class GResults {
String title;
String link;
public GResults(String title, String link) {
this.title = title;
this.link = link;
}
public String getTitle(){
return title;
}
public String getLink(){
return link;
}
public void setTitle(String title){
this.title = title;
}
public void setLink(String link){
this.link = link;
}
public String toString(){
return ("Title:%s, Link:%s", title, link);
}
}
錯誤行:
GResults results = new Gson().fromJson(output, GResults.class);
錯誤訊息:
Exception in thread "main" com.google.gson.JsonSyntaxException: java.io.EOFException: End of input at line 1 column 2
at com.google.gson.Gson.fromJson(Gson.java:813)
at com.google.gson.Gson.fromJson(Gson.java:768)
at com.google.gson.Gson.fromJson(Gson.java:717)
at com.google.gson.Gson.fromJson(Gson.java:689)
at google.api.search.GSearch.main(GSearch.java:26)
Caused by: java.io.EOFException: End of input at line 1 column 2
at com.google.gson.stream.JsonReader.nextNonWhitespace(JsonReader.java:1377)
at com.google.gson.stream.JsonReader.doPeek(JsonReader.java:483)
at com.google.gson.stream.JsonReader.hasNext(JsonReader.java:403)
at com.google.gson.internal.bind.ReflectiveTypeAdapterFactory$Adapter.read(ReflectiveTypeAdapterFactory.java:166)
at com.google.gson.Gson.fromJson(Gson.java:803)
... 4 more
我想我在注意到之前大約閱讀了20次,這是正確的代碼
final StringBuilder builder = new StringBuilder(50);
String output;
while ((output = br.readLine()) != null) {
builder.append(output);
}
final GResults results = new Gson().fromJson(builder.toString(), GResults.class);
Gson拋出了適當的異常,因為您正在逐行閱讀並將該行交給gson進行反序列化。 例如,第一行是{或[或“ message”:{,並且這不是有效的json JsonSyntax。
請享用 :)
經過原始的JSON輸出后,我終於找出了主要缺陷。 Google的JSON通過[]
符號返回了一個“項目” 數組 ,因此在GResults類中添加了一個List之后:
import java.util.List;
public class GResults {
public String link;
public List<GResults> items;
public String getLink(){
return link;
}
public List<GResults> getItems(){
return items;
}
public void setLink(String link){
this.link = link;
}
public void setGroups(List<GResults> items){
this.items = items;
}
public void getThing (int i){
System.out.println(items.get(i));
}
public String toString(){
return String.format("%s", link);
}
}
我可以在主GSearch類中使用以下命令返回一系列鏈接:
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.setRequestMethod("GET");
conn.setRequestProperty("Accept","application/json");
BufferedReader br = new BufferedReader(new InputStreamReader ( ( conn.getInputStream() ) ) );
GResults results = new Gson().fromJson(br, GResults.class);
for (int i=0; i<10; i++)
results.getThing(i);
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