Space or no-space in expressions

[beberlei]

It is a difference to have foo="bar" or foo = "bar" when evaluating this grammar. Is there a way to have both the same?

[Hywan]

Hello :-),

I don't think there is any difference. Both should be supported.

[beberlei]

@Hywan thants for the quick reply.

It renders different models var_dump((string)$model); from interpret():

$model->expression =
    $model->{'='}(
        $model->variable('method'),
        'POST'
    );

$model = new \Hoa\Ruler\Model();
$model->expression =
    $model->variable('method="POST"');

[beberlei]

@Hywan i should mention, we are not using ruler to execute, but we are visiting the AST ourselves to convert it into Elasticsearch query.

[Hywan]

Indeed, I can reproduce the bug. I don't have right now to fix it, but probably on Friday. Is it working for you?

[Hywan]

$ echo 'foo = "bar"' | vendor/bin/hoa compiler:pp vendor/hoa/ruler/Grammar.pp 0 -s
 #  namespace     token name            token value                     offset
-------------------------------------------------------------------------------
 0  default       identifier            foo                                  0
 1  default       identifier            =                                    4
 2  default       string                "bar"                                6
 3  default       EOF                                                       12

$ echo 'foo="bar"' | vendor/bin/hoa compiler:pp vendor/hoa/ruler/Grammar.pp 0 -s
 #  namespace     token name            token value                     offset
-------------------------------------------------------------------------------
 0  default       identifier            foo="bar"                            0
 1  default       EOF                                                       10

I suppose I know how to fix that.

[beberlei]

@Hywan don't worry or feel the need to commit to a time frame, i have a "workaround" for now :-)

[Hywan]

The trick is easy actually:

Ruler/Grammar.pp

Line 67 in 28174ca

%token identifier [^\s\(\)\[\],\.]+

We must change the token to be:

- %token  identifier    [^\s\(\)\[\],\.]+
+ %token  identifier    [^\s\(\)\[\],\.=]+

We should include also common operators. That's because an identifier can be anything.

[beberlei]

I tried to copy the grammar with the small change, but it leads to a new error:

Unrecognized token "=" at line 1 and column 9:
release = 1