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

Question

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

http://site-test.com/Meetings/IC/DownloadDocument?meetingId=c21c905c-8359-4bd6-b864-844709e05754&item         


        

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

This code worked for me

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

Instantiated new URL: http://path

Answer 5

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!