How to initialize and always-true BooleanExpression?

[Marcel Stör]

We dynamically build expressions based on a number of member variables.
This looks kind of like this:

public BooleanExpression getFilterExpression() {
BooleanExpression expression = <QEntity>.isNotNull();

if (A) {
BooleanBuilder builder = new BooleanBuilder();
<some-operation-on-the-builder>
expression = expression.and(builder);
}
if (B) {
BooleanBuilder builder = new BooleanBuilder();
<more-builder-stuff>
expression = expression.and(builder);
}
if (C) {
expression = expression.and(<another-predicate>);
}
if (D) {
expression = expression.and(<final-predicate>);
}
return expression;
}

After the first line we know for sure 'expression' is not null, that
makes the code more legible.
However, we'd much rather drop the superfluous always-true query
condition on the first line. Is there any other way we could initialize
the BooleanExpression?

Cheers,
Marcel

--
Marcel St�r, http://www.frightanic.com
Couchsurfing: http://www.couchsurfing.com/people/marcelstoer
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[Timo Westkämper]

Hi.

BooleanBuilder is mutable builder for BooleanExpression. Your code looks like you expect it to be immutable.

Your code should look like this

public BooleanExpression getFilterExpression() {

    BooleanBuilder builder = new BooleanBuilder();

    if (A) {
        builder.and(<some expression>);
    }
    if (B) {
        builder.and(<some expression>);
    }
    ...
    return builder;
}

Br,
Timo 

On Mon, May 14, 2012 at 10:25 PM, Querydsl on behalf of Marcel Stör <quer...@googlegroups.com> wrote:

We dynamically build expressions based on a number of member variables. This looks kind of like this:

public BooleanExpression getFilterExpression() {
 BooleanExpression expression = <QEntity>.isNotNull();

 if (A) {
   BooleanBuilder builder = new BooleanBuilder();
   <some-operation-on-the-builder>
   expression = expression.and(builder);
 }
 if (B) {
   BooleanBuilder builder = new BooleanBuilder();
   <more-builder-stuff>
   expression = expression.and(builder);
 }
 if (C) {
   expression = expression.and(<another-predicate>);
 }
 if (D) {
   expression = expression.and(<final-predicate>);
 }
 return expression;
}

After the first line we know for sure 'expression' is not null, that makes the code more legible.
However, we'd much rather drop the superfluous always-true query condition on the first line. Is there any other way we could initialize the BooleanExpression?

Cheers,
Marcel

--
Marcel Stör, http://www.frightanic.com

Couchsurfing: http://www.couchsurfing.com/people/marcelstoer
O< ascii ribbon campaign - stop html mail - www.asciiribbon.org

--
Timo Westkämper
Mysema Oy
+358 (0)40 591 2172
www.mysema.com

[Matthew Adams]

I'm doing something similar, but this fails because a BooleanBuilder is not type compatible with BooleanExpression.  The method that you advise Marcel to write won't compile, because it returns a BooleanExpression, and a BooleanBuilder is not type compatible with BooleanExpression.

How do you convert a BooleanBuilder to a BooleanExpression?

[Timo Westkämper]

Hi.

On Wed, May 23, 2012 at 4:47 PM, Querydsl on behalf of Matthew Adams <quer...@googlegroups.com> wrote:

I'm doing something similar, but this fails because a BooleanBuilder is not type compatible with BooleanExpression.  

Yes, that's true. I overlooked it. Both implement the Predicate interface and that is the type that should be returned. BooleanExpression provides an immutable DSL for predicate construction and BooleanBuilder provides provides a similar, but mutable DSL.

Conversion is not directly possible and shouldn't be necessary.

Br,
Timo
 

[Matthew Adams]

Ok.  Would it be bad design for BooleanBuilder to be changed to offer a toBooleanExpression() method?  Predicate doesn't have methods like and, or, etc.

[Timo Westkämper]

We tried to keep the core Expression model independent of the DSL types. Predicate is a core DSL type and BooleanBuilder and BooleanExpression are independent DSL types. I believe that providing direct conversion from BooleanBuilder to BooleanExpression will be misleading, since the signature is similar, but the semantics are different.

Also I don't see many case where BooleanBuilder to BooleanExpression would be useful.

Timo

[Matthew Adams]

Ok, no problem.  I defer to your expertise.  :)