本文介绍了jQuery的AJAX GET返回405不允许的方法的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我试图访问要求认证的API。这是我现在用的是code，但我不断收到405不允许的方法错误。有什么想法吗？ （我的用户名和密码是正确的）
函数basic_auth（user和pass）{
VAR TOK =用户+：+通过;
VAR哈希= $ .base64.en code（TOK）;
返回基本+散;
}
VAR AUTH = basic_auth（用户名，密码）;
VAR releaseName1 =11.6.3;
VAR releaseName2 =11.6.3确认;
$阿贾克斯（{
键入：GET，
网址： "https://www10.v1host.com/Company/rest-1.v1/Data/Story?sel=Description,Number,Name,Timebox.Name,Parent,AssetState&where=Custom_Release2.Name='"+releaseName1+"','"+releaseName2+"';AssetState='64'",
beforeSend：功能（XHR）{
xhr.setRequestHeader（'授权'，验证）;
}，
数据类型：XML，
异步：假的，
成功：parseXml
}）;
功能parseXml（XML）{
$（XML）.find（项目）;
}
解决方案
您不能进行的JavaScript / AJAX跨域调用（不包括一些严重kludging）。我们已经为这个在过去做的是创建一个本地jsp的代理，我们与我们的JavaScript调用，但只是一个传递到远程URL。
下面是一些例子JSP code是接近我已经用打一个SOLR实例返回JSON。
最后字符串编码=UTF-8; //这是默认的，除非（Tomcat的server.xml中）另有规定
//看到http://tomcat.apache.org/tomcat-6.0-doc/config/http.html#Common_Attributes和
// http://wiki.apache.org/tomcat/FAQ/CharacterEncoding#Q2
最后弦乐URI_ENCODING =ISO-8859-1;
HashMap的＆LT;字符串，字符串＆GT; defaultSettings =新的HashMap＆LT;字符串，字符串＆GT;（）{
{
把（重量，JSON）;
把（行，10）;
放（开始，0）;
}
};
BufferedReader中searchResponse = NULL;
OutputStreamWriter forwardedSearchRequest = NULL;
// TODO：在那里，我们需要传递的任何头？
//简单的传递请求到搜索服务器和返回任何结果
尝试 {
URL searchURL =新的URL（http://yourdestinationurlhere.com）;
HttpURLConnection的康恩=（HttpURLConnection类）searchURL.openConnection（）;
//读取请求数据并将其发送的POST数据（不变）
conn.setDoOutput（真正的）;
forwardedSearchRequest =新OutputStreamWriter（conn.getOutputStream（））;
//至少为Tomcat 6.0，默认是真的ISO-8859-1，虽然它被报告为UTF-8
//所以，我们将明确地将它设置为URI_ENCODING在这两个地方
request.setCharacterEncoding（URI_ENCODING）;
查询字符串=（字符串）的request.getParameter（Q）;
如果（（查询=空）及！＆安培;！（.equals（query.trim（））））{
查询= URLEn coder.en code（查询，request.getCharacterEncoding（））; //我们必须使用相同的设置作为容器的URI编码
forwardedSearchRequest.write（与＆amp; Q =）;
forwardedSearchRequest.write（查询）;
} 其他 {
//空的查询可能返回所有结果，让我们规避
forwardedSearchRequest.write（＆放大器; Q =帮助）;
}
对于（字符串键：defaultSettings.keySet（））{
字符串与resultType =（字符串）的request.getParameter（密钥）;
如果（（与resultType == NULL）||.equals（resultType.trim（）））与resultType = defaultSettings.get（密钥）;
forwardedSearchRequest.write（与＆amp;+键+=）;
forwardedSearchRequest.write（与resultType）;
}
forwardedSearchRequest.flush（）;
//读取并转发响应
//重置，可能已被发送为止的任何
out.clearBuffer（）;
//做到这一点，只有当我们有一个200响应code
如果（conn.getResponse code（）== HttpURLConnection.HTTP_OK）{
// Web服务器可能正在运行的窗口1252，让我们确保我们拥有正确的CS
searchResponse =新的BufferedReader（新的InputStreamReader（conn.getInputStream（），编码））;
字符串的contentType = conn.getHeaderField（内容类型）;
如果（（的contentType = NULL）及！及（.equals（的contentType））response.setHeader！）（内容类型，则contentType）;
字符串缓冲区;
而（（缓冲液= searchResponse.readLine（））！= NULL）通过out.println（缓冲液）;
} 其他 {
//盘出一个模拟的Solr JSON的响应，包括状态和错误
response.setHeader（内容类型，应用/ JSON）;
通过out.println（{responseHeader：{状态：-1，响应code：+ conn.getResponse code（）+
，responseMessage：\+ conn.getResponseMessage（）+\}}）;
}
}赶上（例外五）{
抛出新的ServletException异常（异常 - + e.getClass（）的getName（），E。）;
} 最后 {
如果（forwardedSearchRequest！= NULL）forwardedSearchRequest.close（）;
如果（！searchResponse = NULL）searchResponse.close（）;
}
I am trying to access an API that requires authentication. This is the code that i am using but i keep getting a 405 Method Not Allowed error. Any thoughts? (my username and password are correct)
function basic_auth(user, pass){
var tok = user + ':' + pass;
var hash = $.base64.encode(tok);
return "Basic " + hash;
}
var auth = basic_auth('username','password');
var releaseName1 = "11.6.3";
var releaseName2 = "11.6.3 Confirmed";
$.ajax({
type: "GET",
url: "https://www10.v1host.com/Company/rest-1.v1/Data/Story?sel=Description,Number,Name,Timebox.Name,Parent,AssetState&where=Custom_Release2.Name='"+releaseName1+"','"+releaseName2+"';AssetState='64'",
beforeSend: function(xhr){
xhr.setRequestHeader('Authorization', auth);
},
dataType: "xml",
async: false,
success: parseXml
});
function parseXml(xml){
$(xml).find("item");
}
解决方案
You can't make javascript/Ajax calls across domains (without some serious kludging). What we've done for this in the past is to create a local jsp proxy that we call with our javascript but is just a pass-through to the remote URL.
Here is some example JSP code that is close to what I've used to hit a SOLR instance returning JSON.
final String ENCODING = "UTF-8"; // this is the default unless specified otherwise (in server.xml for Tomcat)
// see http://tomcat.apache.org/tomcat-6.0-doc/config/http.html#Common_Attributes and
// http://wiki.apache.org/tomcat/FAQ/CharacterEncoding#Q2
final String URI_ENCODING = "ISO-8859-1";
HashMap<String, String> defaultSettings = new HashMap<String, String>() {
{
put("wt", "json");
put("rows", "10");
put("start", "0");
}
};
BufferedReader searchResponse = null;
OutputStreamWriter forwardedSearchRequest = null;
// TODO: are there any headers that we need to pass on?
// simply pass on the request to the search server and return any results
try {
URL searchURL = new URL("http://yourdestinationurlhere.com");
HttpURLConnection conn = (HttpURLConnection) searchURL.openConnection();
// read the request data and send it as POST data (unchanged)
conn.setDoOutput(true);
forwardedSearchRequest = new OutputStreamWriter(conn.getOutputStream());
// at least for Tomcat 6.0, the default is really iso-8859-1, although it is reported as UTF-8
// so, we will explicitly set it to URI_ENCODING in both places
request.setCharacterEncoding(URI_ENCODING);
String query = (String) request.getParameter("q");
if ((query != null) && (! "".equals(query.trim()))) {
query = URLEncoder.encode(query, request.getCharacterEncoding()); // we must use the same setting as the container for URI-encoding
forwardedSearchRequest.write("&q=");
forwardedSearchRequest.write(query);
} else {
// empty queries may return all results, so let's circumvent that
forwardedSearchRequest.write("&q=help");
}
for(String key:defaultSettings.keySet()) {
String resultType = (String) request.getParameter(key);
if ((resultType == null) || "".equals(resultType.trim())) resultType = defaultSettings.get(key);
forwardedSearchRequest.write("&"+key+"=");
forwardedSearchRequest.write(resultType);
}
forwardedSearchRequest.flush();
// read and forward the response
// reset anything that may have been sent so far
out.clearBuffer();
// do this only if we have a 200 response code
if (conn.getResponseCode() == HttpURLConnection.HTTP_OK) {
// the web server may be running as windows-1252, so let's ensure we have the right CS
searchResponse = new BufferedReader(new InputStreamReader(conn.getInputStream(), ENCODING));
String contentType = conn.getHeaderField("Content-Type");
if ((contentType != null) && (! "".equals(contentType))) response.setHeader("Content-Type", contentType);
String buffer;
while ((buffer = searchResponse.readLine()) != null) out.println(buffer);
} else {
// dish out a mock-Solr-JSON response that includes a status and an error
response.setHeader("Content-Type", "application/json");
out.println("{ responseHeader: {status: -1, responseCode: " + conn.getResponseCode() +
", responseMessage: \"" + conn.getResponseMessage() + "\" } }");
}
} catch (Exception e) {
throw new ServletException("Exception - " + e.getClass().getName(), e);
} finally {
if (forwardedSearchRequest != null) forwardedSearchRequest.close();
if (searchResponse != null) searchResponse.close();
}
这篇关于jQuery的AJAX GET返回405不允许的方法的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！