List<Foo> f = (List<Foo>) JSON.deserialize(jsonString, List<Foo>.class);
-
Recent Posts
Archives
Categories
Tags
- Adapter Pattern
- Administration
- Apex
- Apex Method of the Day
- Approvals
- Batch Apex
- Careers
- Chrome
- Controller
- Customer Portal
- Dreamforce
- Dynamic Binding
- Field Set
- FinancialForce
- FinancialForce.com
- Force.com Site
- Heroku
- Hierarchical Custom Setting
- Identity Provider
- Interface
- ISV
- JSON
- Locking
- Managed Package
- message-driven-architecture
- OAuth
- Orizuru
- Plugin
- Productivity
- Security
- SObject
- SObject Secret Life
- Triggers
- Visualforce
Top Posts & Pages
- Apex Method of the Day – JSON.serialize(Object)
- Apex Method of the Day - String myString.split(String regExp)
- Apex Method of the Day - JSON.deserialize(jsonString, List.class)
- Apex Method of the Day - String.format(String value, List<String> args)
- Salesforce as an Identity Provider for Customer Portal
RSS
Thank you, this saved me some research !
LikeLike