Question

无法在android中运行sencha / phonegap应用程序.. 获得异常＆＃34; cmd：命令失败，退出代码为ENOENT 3＆＃34;这是我安装的版本..

D:\testapp>sencha -version   
Sencha Cmd v4.0.1.45
[ERR] No such property : version
 at com.sencha.cli.AbstractCommand$Properties.getProperty(AbstractCommand.ja
va:417)
 at com.sencha.cli.AbstractCommand.parseConfigParam(AbstractCommand.java:441
)
 at com.sencha.cli.AbstractCommand.processConfigParam(AbstractCommand.java:4
91)
 at com.sencha.cli.AbstractCommand.configure(AbstractCommand.java:114)
 at com.sencha.command.Sencha.main(Sencha.java:136)


D:\testapp>phonegap -version  
3.4.0-0.19.7

D:\testapp>cordova -version  
3.4.0-0.1.3

D:\testapp>ant -version  
Apache Ant(TM) version 1.9.3 compiled on December 23 2013

D:\testapp>phonegap run android  
[phonegap] detecting Android SDK environment...  
[phonegap] using the local environment  
[phonegap] adding the Android platform...  
[error] cmd: Command failed with exit code ENOENT  

D:\testapp>phonegap build android  
[phonegap] detecting Android SDK environment...  
[phonegap] using the local environment  
[phonegap] adding the Android platform...   
[error] cmd: Command failed with exit code ENOENT  

Thanks in advance

Answer 1

检查以下步骤：

cordova create testapp com.example.testapp TestApp
cd testapp
cordova platform add android
cordova build android

在此之后，您应该复制项目并将其粘贴到cordova项目的直接www文件夹中。然后运行

cordova run android

从上面的代码段中，您的系统中似乎已安装cordova。注意：以上CLI步骤适用于cordova版本3.x

无法解决Phone Gap [error] cmd：命令失败，退出代码为ENOENT 3

1 个答案: