SocketTimeoutException while opening New Spring St

Question

I'm using the ZIP distribution of STS 3.5.1 for win32. I start STS with this cmd script:

set JAVA_HOME=d:\jdk\jdk1.7.0_21
set PATH=%JAVA_HOME%\bin;%PATH%;
start .\sts-3.5.1.RELEASE\sts.exe

Then, when I do "New > Spring Starter Project", I get this error:

If I replace Java 7 with Java 6 by changing the first line of my cmd script, everything works fine: the wizard starts.

set JAVA_HOME=d:\jdk\jdk1.6.0_31

Is there something special to do if I want to use Java 7 ?

Answer 1

You probably want to get a more recent version of the JDK 1.7 (e.g. 1.7.0_51) as there were quite a few unpleasant bugs in earlier releases. Otherwise I don't have any suggestions. Can you see the service (http://start.spring.io) from your browser?