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 amp itemId a4b724d1-282e-4b36-9d16-d619a807ba67 amp 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  amp    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

User · Answer

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 amp itemId a4b724d1-282e-4b36-9d16-d619a807ba67 amp 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  amp itemId  itemID  amp 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  amp itemId  itemID  amp file  file         Generate URL as a String final String theURL   template expand vars     This is GUARANTEED to return a fully functional URL

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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

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

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