问题描述
我将STS 3.5.1的ZIP发行版用于Win32.我使用以下cmd脚本启动STS:
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
然后，当我执行新建> Spring Starter项目"时，出现此错误:
Then, when I do "New > Spring Starter Project", I get this error:
如果通过更改cmd脚本的第一行将Java 7替换为Java 7，一切正常:向导启动.
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
如果我想使用Java 7，有什么特别的事情要做吗?
Is there something special to do if I want to use Java 7 ?
推荐答案
您可能希望获得JDK 1.7的最新版本(例如1.7.0_51)，因为早期版本中存在许多令人不愉快的错误.否则我没有任何建议.您可以从浏览器中看到该服务( http://start.spring.io )吗?
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?
这篇关于在运行Java 7的STS上打开New Spring Starter Project向导时，出现SocketTimeoutException的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！