Application startup task failed with exit code 1 when Deploy on azure

Question

I'm newbie on azure.I try to following article from : http://msdn.microsoft.com/en-us/library/windowsazure/hh690944(VS.103).aspx

first as beginner to create helloworld app with JSP

I have uploading both of file to blob storage using azure storage explorer :

Service Configuration.cscfg
WindowsAzurePackage.cspkg

Then i try to create new hosting services and inclue these file inside.

But on progress like stop on status :

Application startup task failed with exit code 1. [2012-06-05T14:51:58Z]

for long time i waiting, firstly it saya initializating, then Recovering role, but still same.progress is stop.

What is that error ? is there my configuration missing ?

Martin Sawicki · Accepted Answer · 2012-08-06 23:50:07Z

1

As of the June 2012 update of the Windows Azure Plugin for Eclipse with Java, the basic deployment of a Java app server + some apps got much simpler -- now it's just part of the new Windows Azure project wizard UI. The "old" way of copying startup scripts still works, and especially more advanced users may still find that preferable, but if you're just getting started with Java on Windows Azure, you will find the wizard-based approach just much simpler, quicker and with less opportunity for error.

Easiest to start with the 5 min demo at: http://channel9.msdn.com/Blogs/Interoperability/Java-Applications-in-Windows-Azure-Cloud-Services-using-Eclipse

answered Aug 6, 2012 at 23:50

Martin Sawicki

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David Makogon · Accepted Answer · 2012-06-05 15:07:07Z

0

Just curious if you got this to run correctly locally.

One thing you must do, if you haven't, is modify the startup.cmd to properly match the version of Tomcat you're deploying with your app. Notice that the startup script ships with tomcat's folder being apache-tomcat-7.0.22. If you pulled Tomcat today, you'd get an updated build number, so you'd need to update this name in the startup script. Otherwise you'll see your initialization fail.

answered Jun 5, 2012 at 15:07

David Makogon

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3 Comments

troya_adromeda Over a year ago

i have modify my startup cmd as my apache tomcat.I modify this variable SET SERVER_DIR_NAME=apache-tomcat-7.0.27 this is my tomcat version.But on zip file my file name is tomcat7.zip.How about this ?

David Makogon Over a year ago

Same thing with zip file: The startup script unzips it, so the name of the zip needs to match the name in the startup script.

troya_adromeda Over a year ago

but with this file name acutally i can deploy on azure emulator on local.Should it indicate that my app should work on azure cloud, isn't it ? i can running on azure emulator with tomcat7.zip as filename.

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