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feat($compile): '=?' makes '=' binding optional

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If you bind using '=' to a non-existant parent property, the compiler
will throw a NON_ASSIGNABLE_MODEL_EXPRESSION exception, which is right
because the model doesn't exist.

This enhancement allow to specify that a binding is optional so it
won't complain if the parent property is not defined. In order to mantain
backward compability, the new behaviour must be specified using '=?' instead
of '='. The local property will be undefined is these cases.

Closes #909
Closes #1435
This commit is contained in:
Luis Ramón López
2013-01-26 20:15:06 +01:00
committed by Igor Minar
parent 30162b769c
commit ac899d0da5
3 changed files with 39 additions and 3 deletions
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4
docs/content/guide/directive.ngdoc
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@@ -336,7 +336,9 @@ compiler}. The attributes are:
Given `<widget my-attr="parentModel">` and widget definition of
`scope: { localModel:'=myAttr' }`, then widget scope property `localModel` will reflect the
value of `parentModel` on the parent scope. Any changes to `parentModel` will be reflected
in `localModel` and any changes in `localModel` will reflect in `parentModel`.
in `localModel` and any changes in `localModel` will reflect in `parentModel`. If the parent
scope property doesn't exist, it will throw a NON_ASSIGNABLE_MODEL_EXPRESSION exception. You
can avoid this behavior using `=?` or `=?attr` in order to flag the property as optional.
* `&` or `&attr` - provides a way to execute an expression in the context of the parent scope.
If no `attr` name is specified then the attribute name is assumed to be the same as the