CODEWITHSUNDEEP
javaurimalformedurlexception
MorkPork
MorkPork

Reputation: 884

java.net.MalformedURLException: no protocol on URL based on a string modified with URLEncoder

So I was attempting to use this String in a URL :- 

http://site-test.com/Meetings/IC/DownloadDocument?meetingId=c21c905c-8359-4bd6-b864-844709e05754&itemId=a4b724d1-282e-4b36-9d16-d619a807ba67&file=\\s604132shvw140\Test-Documents\c21c905c-8359-4bd6-b864-844709e05754_attachments\7e89c3cb-ce53-4a04-a9ee-1a584e157987\myDoc.pdf

In this code: -

String fileToDownloadLocation = //The above string
URL fileToDownload = new URL(fileToDownloadLocation);
HttpGet httpget = new HttpGet(fileToDownload.toURI());

But at this point I get the error: -

java.net.URISyntaxException: Illegal character in query at index 169:Blahblahblah

I realised with a bit of googling this was due to the characters in the URL (guessing the &), so I then added in some code so it now looks like so: -

String fileToDownloadLocation = //The above string
fileToDownloadLocation = URLEncoder.encode(fileToDownloadLocation, "UTF-8");
URL fileToDownload = new URL(fileToDownloadLocation);
HttpGet httpget = new HttpGet(fileToDownload.toURI());

However, when I try and run this I get an error when I try and create the URL, the error then reads: -

java.net.MalformedURLException: no protocol: http%3A%2F%2Fsite-test.testsite.com%2FMeetings%2FIC%2FDownloadDocument%3FmeetingId%3Dc21c905c-8359-4bd6-b864-844709e05754%26itemId%3Da4b724d1-282e-4b36-9d16-d619a807ba67%26file%3D%5C%5Cs604132shvw140%5CTest-Documents%5Cc21c905c-8359-4bd6-b864-844709e05754_attachments%5C7e89c3cb-ce53-4a04-a9ee-1a584e157987%myDoc.pdf

It looks like I can't do the encoding until after I've created the URL else it replaces slashes and things which it shouldn't, but I can't see how I can create the URL with the string and then format it so its suitable for use. I'm not particularly familiar with all this and was hoping someone might be able to point out to me what I'm missing to get string A into a suitably formatted URL to then use with the correct characters replaced?

Any suggestions greatly appreciated!

Upvotes: 43

Views: 375422

Answers (7)

Praveen Dasare
Praveen Dasare

Reputation: 51

I was adding the value in the double quotes (") when I changed the key-value pairs from my .properties file the issue was resolved key-value pair should not be enclosed under double quotes

Upvotes: 0

fge
fge

Reputation: 121840

You want to use URI templates. Look carefully at the README of this project: URLEncoder.encode() does NOT work for URIs.

Let us take your original URL:

http://site-test.test.com/Meetings/IC/DownloadDocument?meetingId=c21c905c-8359-4bd6-b864-844709e05754&itemId=a4b724d1-282e-4b36-9d16-d619a807ba67&file=\\s604132shvw140\Test-Documents\c21c905c-8359-4bd6-b864-844709e05754_attachments\7e89c3cb-ce53-4a04-a9ee-1a584e157987\myDoc.pdf

And convert it to a URI template with two variables (on multiple lines for clarity):

http://site-test.test.com/Meetings/IC/DownloadDocument
    ?meetingId={meetingID}&itemId={itemID}&file={file}

Now let us build a variable map with these three variables using the library mentioned in the link:

final VariableMap = VariableMap.newBuilder()
    .addScalarValue("meetingID", "c21c905c-8359-4bd6-b864-844709e05754")
    .addScalarValue("itemID", "a4b724d1-282e-4b36-9d16-d619a807ba67e")
    .addScalarValue("file", "\\\\s604132shvw140\\Test-Documents"
        + "\\c21c905c-8359-4bd6-b864-844709e05754_attachments"
        + "\\7e89c3cb-ce53-4a04-a9ee-1a584e157987\\myDoc.pdf")
    .build();

final URITemplate template
    = new URITemplate("http://site-test.test.com/Meetings/IC/DownloadDocument"
        + "meetingId={meetingID}&itemId={itemID}&file={file}");

// Generate URL as a String
final String theURL = template.expand(vars);

This is GUARANTEED to return a fully functional URL!

Upvotes: 2

Anton Cavanaugh
Anton Cavanaugh

Reputation: 1049

Very simple fix:

String encodedURL = UriUtils.encodePath(request.getUrl(), "UTF-8");

No extra functionality needed.

Upvotes: 0

Ajay
Ajay

Reputation: 740

This code worked for me

public static void main(String[] args) {
    try {
        java.net.URL url = new java.net.URL("http://path");
        System.out.println("Instantiated new URL: " + url);
    }
    catch (MalformedURLException e) {
        e.printStackTrace();
    }
}

Instantiated new URL: http://path

Upvotes: 1

Dave
Dave

Reputation: 3311

Thanks to Erhun's answer I finally realised that my JSON mapper was returning the quotation marks around my data too! I needed to use "asText()" instead of "toString()"

It's not an uncommon issue - one's brain doesn't see anything wrong with the correct data, surrounded by quotes!

discoveryJson.path("some_endpoint").toString();
"https://what.the.com/heck"

discoveryJson.path("some_endpoint").asText();
https://what.the.com/heck

Upvotes: 1

erhun
erhun

Reputation: 3647

I have the same problem, i read the url with an properties file:

String configFile = System.getenv("system.Environment");
        if (configFile == null || "".equalsIgnoreCase(configFile.trim())) {
            configFile = "dev.properties";
        }
        // Load properties 
        Properties properties = new Properties();
        properties.load(getClass().getResourceAsStream("/" + configFile));
       //read url from file
        apiUrl = properties.getProperty("url").trim();
            URL url = new URL(apiUrl);
            //throw exception here
    URLConnection conn = url.openConnection();

dev.properties

url = "https://myDevServer.com/dev/api/gate"

it should be

dev.properties

url = https://myDevServer.com/dev/api/gate

without "" and my problem is solved.

According to oracle documentation

  • Thrown to indicate that a malformed URL has occurred. Either no legal protocol could be found in a specification string or the string could not be parsed.

So it means it is not parsed inside the string.

Upvotes: 11

m-szalik
m-szalik

Reputation: 3654

You need to encode your parameter's values before concatenating them to URL.
Backslash \ is special character which have to be escaped as %5C

Escaping example:

String paramValue = "param\\with\\backslash";
String yourURLStr = "http://host.com?param=" + java.net.URLEncoder.encode(paramValue, "UTF-8");
java.net.URL url = new java.net.URL(yourURLStr);

The result is http://host.com?param=param%5Cwith%5Cbackslash which is properly formatted url string.

Upvotes: 32

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