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Fix [is[variable]] operator doesn't work for "fake" variables #6303 (#6996)

* Add tests for [is[variable]] and "faked" variables

See GH #6303

* Make is[variable] and variables[] operators resilient to fake widgets

Co-authored-by: jeremy@jermolene.com <jeremy@jermolene.com>
This commit is contained in:
Rob Hoelz
2022-10-18 11:08:04 -05:00
committed by GitHub
parent 7b408c7adf
commit 24dbf69180
3 changed files with 38 additions and 5 deletions

View File

@@ -19,13 +19,13 @@ exports.variable = function(source,prefix,options) {
var results = [];
if(prefix === "!") {
source(function(tiddler,title) {
if(!(title in options.widget.variables)) {
if(options.widget.getVariable(title) === undefined) {
results.push(title);
}
});
} else {
source(function(tiddler,title) {
if(title in options.widget.variables) {
if(options.widget.getVariable(title) !== undefined) {
results.push(title);
}
});

View File

@@ -16,9 +16,15 @@ Filter operator for returning the names of the active variables
Export our filter function
*/
exports.variables = function(source,operator,options) {
var names = [];
for(var variable in options.widget.variables) {
names.push(variable);
var names = [],
widget = options.widget;
while(widget && !widget.hasOwnProperty("variables")) {
widget = widget.parentWidget;
}
if(widget && widget.variables) {
for(var variable in widget.variables) {
names.push(variable);
}
}
return names.sort();
};