🎨 ✨ find by name

Search on word/name, if partially found in resource it will return the resource

utils/pure.js

@@ -18,16 +18,35 @@ const Category = {
 }
 
 /// Functions
+// isNotEmpty [a] -> Bool
+const isNotEmpty = R.compose(R.not, R.isEmpty)
+
 // getAllResources :: [Category] -> [Resource]
 const getAllResources = R.compose(R.flatten, R.map(R.prop('resources')))
 
 // tagsNotEmpty :: Resource -> Bool
-const tagsNotEmpty = R.compose(R.not, R.isEmpty, R.prop('tags'))
+const tagsNotEmpty = R.compose(isNotEmpty, R.prop('tags'))
+
+// cleanString :: String -> String
+const cleanString = R.compose(R.toLower, R.trim)
 
 // true if list2 has element that appears in list1 else false
 // includesElOf([1, 2])([2]) -> true
 // includesElOf([1, 2], [3]) -> false
+// includesElOf(['a', 'b'], ['a']) -> true
+// includesElOf(['aa', 'b'], ['a']) -> false
 // includesElOf :: [a] -> [a] -> Bool
 const includesElOf = R.curry((list1, list2) => R.any(el => R.includes(el, list2), list1))
 
-export { getAllResources, tagsNotEmpty, includesElOf }
+// Similar to includesElOf, but partially included strings are also allowed
+// includesElOf(['a', 'b'])(['a']) -> true
+// includesElOf(['aa', 'b'], ['a']) -> true
+// includesElOf(['aa', 'b'], ['c']) -> false
+// includesElOf :: [String] -> [String] -> Bool
+const partiallyIncludesElOf = R.curry((list1, list2) => 
+  R.any(el2 =>
+    R.any(R.includes(el2), list1),
+  list2)
+)
+
+export { getAllResources, tagsNotEmpty, includesElOf, partiallyIncludesElOf, cleanString }