Make `Compose.parseWhereClause(where)` a public method

[justinleebehnke]

• [x] Feature
• [ ] Bug Request

  Description

  I might be a random edge case but for me personally, I need the WHERE clause as a string. I know I could traverse the awesome tree structure that you have built in for the WHERE clause, it really is awesome, but I am 99% sure that's exactly what's going on in the parseWhereClause() method. Just a thought. I really appreciate your repo it's very nice.

[paustint]

@justinleebehnke - This makes sense to me and I think that it should be available. I am working on it now and writing some tests to make sure I did not break anything as well as updating the README with examples.

You will need to create a new instance of the Compose class and pass in a new option { autoCompose: false } - this ensures that the passed in query data will not be autocomposed into a soql query string - since for your use-case you will only need to compose one part of the parsed query.

I should have a new version published today.

Here is an example of how it will be used:

  it(`Should compose the where clause properly`, () => {
    const soql = `SELECT Id FROM Account WHERE Name = 'Foo'`;
    const parsedQuery = parseQuery(soql);
    const composer = new Compose(parsedQuery, { autoCompose: false });
    const whereClause = composer.parseWhereClause(parsedQuery.where);
    expect(whereClause).to.equal(`Name = 'Foo'`);
  });
  it(`Should compose the where clause properly with semi-join`, () => {
    const soql = `SELECT Id FROM Account WHERE Id IN (SELECT AccountId FROM Contact WHERE Name LIKE '%foo%')`;
    const parsedQuery = parseQuery(soql);
    const composer = new Compose(parsedQuery, { autoCompose: false });
    const whereClause = composer.parseWhereClause(parsedQuery.where);
    expect(whereClause).to.equal(`Id IN (SELECT AccountId FROM Contact WHERE Name LIKE '%foo%')`);
  });