Question

我尝试传递的参数是“Submitted_By：test2 OR Submitted_By：test”。如果我在错误消息中复制并粘贴确切的URL，我可以正常访问该URL。当传递诸如“Submitted_By：test2”之类的参数时，它工作正常，但是当添加“OR”时，它会抛出错误。有什么想法吗？

12:51:53,246 ERROR [stderr] (default task-1) java.io.IOException: Server returned HTTP response code: 400 for URL: http://localhost:1234/solr/STINGRA/select/?q=Submitted_By:test2 OR Submitted_By:test&fl=Submission_ID,Submitted_By,File_Name,TOT&version=2.2&start=0&rows=50&indent=on&wt=xml&callback=?&json.wrf=on_data
12:51:53,247 ERROR [stderr] (default task-1)    at sun.net.www.protocol.http.HttpURLConnection.getInputStream0(HttpURLConnection.java:1839)
12:51:53,247 ERROR [stderr] (default task-1)    at sun.net.www.protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:1440)
12:51:53,247 ERROR [stderr] (default task-1)    at java.net.URL.openStream(URL.java:1038)
12:51:53,247 ERROR [stderr] (default task-1)    at webapp.Search.getResultsBeta(Search.java:144)
12:51:53,247 ERROR [stderr] (default task-1)    at webapp.WebappController.searchResultsBeta(WebappController.java:87)
12:51:53,248 ERROR [stderr] (default task-1)    at webapp.WebappController$$FastClassBySpringCGLIB$$b0db6f6a.invoke(<generated>)

它引用的第144行是提供的代码片段中的最后一行。 Search.getResultsBeta方法：

public static Integer[] getResultsBeta(String query) {
    String inputLine;
    ArrayList<Integer> subId = new ArrayList<>();
    int numOfResults = 0;
    try {
        URL temp;
        temp = new URL("http://localhost:1234/solr/STINGRA/select/?q=" + query + "&fl=Submission_ID,Submitted_By,File_Name,TOT&version=2.2&start=0&rows=50&indent=on&wt=xml&callback=?&json.wrf=on_data");
        BufferedReader in;
        InputStreamReader sr = new InputStreamReader(temp.openStream());

Answer 1

您应该使用url = url.replaceAll(" ", "%20");

将空格编码为'％20'

Answer 2

通过编码查询变量来解决问题。

更改我的代码：

URL temp;
temp = new URL("http://localhost:1234/solr/STINGRA/select/?q=" + query + "&fl=Submission_ID,Submitted_By,File_Name,TOT&version=2.2&start=0&rows=50&indent=on&wt=xml&callback=?&json.wrf=on_data");

对此：

URL temp;
query = URLEncoder.encode(query);
temp = new URL("http://localhost:1234/solr/STINGRA/select/?q=" + query + "&fl=Submission_ID,Submitted_By,File_Name,TOT&version=2.2&start=0&rows=50&indent=on&wt=xml&callback=?&json.wrf=on_data");

获取java.io.IOException：服务器返回HTTP响应代码：400用于URL但可以在浏览器中

2 个答案: