Function ParseExprWithInt

pkg/sql/parser/parse.go:392–401

ParseExprWithInt parses a SQL scalar expression, using the given type when INT is used as type name in the SQL syntax. The caller is responsible for ensuring that the input is, in fact, a valid SQL scalar expression — the results are undefined if the string contains invalid SQL syntax.

(sql string, nakedIntType *types.T)

// scalar expression — the results are undefined if the string
// contains invalid SQL syntax.
func ParseExprWithInt(sql string, nakedIntType *types.T) (tree.Expr, error) {
	exprs, err := parseExprsWithInt([]string{sql}, nakedIntType)
	if err != nil {
		return nil, err
	}
	if len(exprs) != 1 {
		return nil, errors.AssertionFailedf("expected 1 expression, found %d", len(exprs))
	}
	return exprs[0], nil
}

// GetTypeReferenceFromName turns a type name into a type
// reference. This supports only “simple” (single-identifier)