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Caused by java net malformedurlexception protocol not found

The problem is that you' re passing in the XML content itself - but DocumentBuilder. parse( String) accepts a URL to load the XML from - not the content itself. You probably want to use DocumentBuilder. parse( InputSource). Your URI is not a URI. There is no protocol component. It needs http: / / or whatever other protocol you intend. Hence change the line : InputStream is = new URL( source). like InputStream is = new URL( source. replace( " & quot; ", " \ " " ) ). The issue is that you are escaping the " before the start of url and its considering it to be part of Url. since " http is not a valid protocol, valid values are http https ftp etc. final String FORECAST_ BASE_ URL. Add http: / / before the or public String URL = w3schools.

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  • Video:Found caused malformedurlexception

    Found protocol java

    com/ webservices/ tempconvert. up vote 2 down vote. you missed the protocol http: / / public String URL = " http: / / www. NOTE: If you just need to convert from Celsius to Fahrenhiet ot vice- versa you could try this:. 273: E/ getImage: java. MalformedURLException: Protocol not found: > https% 3A% 2F% 2Fs3. com% 2Fknuser9% 2Fae2f40a6- c563- 4f9d- ade4- beb8d373ebc3. I am a little losted as to. I don' t know of any logger that will add enclosing quotes to a given log message. As such, I' m going to assume that your url String is literally. I think that the problem is that you are calling the URL constructor with an invalid URL string. Indeed, the exception message implies that the URL string starts with " 9: ".

    ( The ' protocol' component is the sequence of characters.