Can I control the HTTP headers sent by window.open
(cross browser)?
If not, can I somehow window.open
a page that then issues my request with custom headers inside its popped-up window?
I need some cunning hacks.
This question is related to
javascript
window.open
As the best anwser have writed using XMLHttpResponse
except window.open
, and I make the abstracts-anwser as a instance.
The main Js file is download.js
Download-JS
// var download_url = window.BASE_URL+ "/waf/p1/download_rules";
var download_url = window.BASE_URL+ "/waf/p1/download_logs_by_dt";
function download33() {
var sender_data = {"start_time":"2018-10-9", "end_time":"2018-10-17"};
var x=new XMLHttpRequest();
x.open("POST", download_url, true);
x.setRequestHeader("Content-type","application/json");
// x.setRequestHeader("Access-Control-Allow-Origin", "*");
x.setRequestHeader("Authorization", "JWT " + localStorage.token );
x.responseType = 'blob';
x.onload=function(e){download(x.response, "test211.zip", "application/zip" ); }
x.send( JSON.stringify(sender_data) ); // post-data
}
Although it is easy to construct a GET query using window.open()
, it's a bad idea (see below). One workaround is to create a form that submits a POST request. Like so:
<form id="helper" action="###/your_page###" style="display:none">
<inputtype="hidden" name="headerData" value="(default)">
</form>
<input type="button" onclick="loadNnextPage()" value="Click me!">
<script>
function loadNnextPage() {
document.getElementById("helper").headerData.value = "New";
document.getElementById("helper").submit();
}
</script>
Of course you will need something on the server side to handle this; as others have suggested you could create a "proxy" script that sends headers on your behalf and returns the results.
Problems with GET
You can't directly add custom headers with window.open() in popup window but to work that we have two possible solutions
- Write Ajax method to call that particular URL with headers in a separate HTML file and use that HTML as url in
<i>window.open()</i>
here is abc.html
$.ajax({
url: "ORIGIONAL_URL",
type: 'GET',
dataType: 'json',
headers: {
Authorization : 'Bearer ' + data.id_token,
AuthorizationCheck : 'AccessCode ' +data.checkSum ,
ContentType :'application/json'
},
success: function (result) {
console.log(result);
},
error: function (error) {
} });
call html
window.open('*\abc.html')
here CORS policy can block the request if CORS is not enabled in requested URL.
- You can request a URL that triggers a server-side program which makes the request with custom headers and then returns the response redirecting to that particular url.
Suppose in Java Servlet(/requestURL) we'll make this request
`
String[] responseHeader= new String[2];
responseHeader[0] = "Bearer " + id_token;
responseHeader[1] = "AccessCode " + checkSum;
String url = "ORIGIONAL_URL";
URL obj = new URL(url);
HttpURLConnection urlConnection = (HttpURLConnection) obj.openConnection();
urlConnection.setRequestMethod("GET");
urlConnection.setDoInput(true);
urlConnection.setDoOutput(true);
urlConnection.setRequestProperty("Content-Type", "application/json");
urlConnection.setRequestProperty("Accept", "application/json");
urlConnection.setRequestProperty("Authorization", responseHeader[0]);
urlConnection.setRequestProperty("AuthorizationCheck", responseHeader[1]);
int responseCode = urlConnection.getResponseCode();
if (responseCode == HttpURLConnection.HTTP_OK) {
BufferedReader in = new BufferedReader(new
InputStreamReader(urlConnection.getInputStream()));
String inputLine;
StringBuffer response1 = new StringBuffer();
while ((inputLine = in.readLine()) != null) {
response1.append(inputLine);
}
in.close();
response.sendRedirect(response1.toString());
// print result
System.out.println(response1.toString());
} else {
System.out.println("GET request not worked");
}
`
call servlet in window.open('/requestURL')
If you are in control of server side, it might be possible to set header value in query string and send it like that? That way you could parse it from query string if it's not found in the headers.
Just an idea... And you asked for a cunning hack :)
Source: Stackoverflow.com