Combine const and enum errors into a single message

[ShivaGupta-14]

Description

When multiple const or enum constraints fail (e.g., via allOf), produce a single message with the most restrictive constraint instead of multiple redundant messages.

• Compute the intersection of allowed values across all constraints
• If intersection has 1 value: use const message
• If intersection has multiple values: use enum message
• If intersection is empty (contradictory): use boolean schema message
• Fixes: Combine const and enum handlers #132

Checklist

• npm test
• npm run lint
• npm run type-check

[jdesrosiers]

This approach isn't going to work. Try,

{
  "allOf": [
    { "const": "a" },
    { "const": "b" }
  ]
}

[ShivaGupta-14]

This approach isn't going to work. Try,

{
  "allOf": [
    { "const": "a" },
    { "const": "b" }
  ]
}

Hi @jdesrosiers,

thanks for pointing out the issue! I understood the problem:

Issue: with a schema like above one -> allOf: [{ const: "a" }, { const: "b" }], if the instance is "a"

• const: "a" passes (not in errors)
• Only const: "b" fails
• My original code only saw failing constraints -> couldn't detect the contradiction
• My Solution: I updated the handler to collect ALL constraints (both passing and failing), then compute the intersection:

const passed = normalizedErrors[...][schemaLocation] === true;
if (!passed) hasFailure = true;
// always collect
constraints.push({ allowedValues: [...], schemaLocation });

now:

• Both ["a"] and ["b"] are collected
• Intersection = [] (empty)
• Returns getBooleanSchemaErrorMessage()

Tests Added:

• contradictory const - empty intersection (instance matches one const)
• contradictory enum - empty intersection (disjoint enum sets)

All tests pass. Is this approach good, or does it need further improvements?
Thanks!

[jdesrosiers]

That should work. Just a couple small things I'd like to see.

[jdesrosiers]

Let's add a test where that would fail if we didn't use the deterministic stringify approach.

Also, as a style rule, I like to put helper functions at the bottom.

[jdesrosiers]

All of this could be avoided if you used a Set and Set operations to determine the intersection.