prometheus/web/ui/module/codemirror-promql/src/parser/path-finder.ts
Marijn Haverbeke 8dbb2eaf0b
Clean up the Lezer grammar output tree (#11333)
Signed-off-by: Marijn Haverbeke <marijn@haverbeke.nl>
2022-09-29 22:26:46 +02:00

59 lines
1.9 KiB
TypeScript

// Copyright 2021 The Prometheus Authors
// Licensed under the Apache License, Version 2.0 (the "License");
// you may not use this file except in compliance with the License.
// You may obtain a copy of the License at
//
// http://www.apache.org/licenses/LICENSE-2.0
//
// Unless required by applicable law or agreed to in writing, software
// distributed under the License is distributed on an "AS IS" BASIS,
// WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
// See the License for the specific language governing permissions and
// limitations under the License.
import { SyntaxNode } from '@lezer/common';
// walkBackward will iterate other the tree from the leaf to the root until it founds the given `exit` node.
// It returns null if the exit is not found.
export function walkBackward(node: SyntaxNode | null, exit: number): SyntaxNode | null {
for (;;) {
if (!node || node.type.id === exit) {
return node;
}
node = node.parent;
}
return null;
}
export function containsAtLeastOneChild(node: SyntaxNode, ...child: (number | string)[]): boolean {
const cursor = node.cursor();
if (!cursor.next()) {
// let's try to move directly to the children level and
// return false immediately if the current node doesn't have any child
return false;
}
let result = false;
do {
result = child.some((n) => cursor.type.id === n || cursor.type.name === n);
} while (!result && cursor.nextSibling());
return result;
}
export function containsChild(node: SyntaxNode, ...child: (number | string)[]): boolean {
const cursor = node.cursor();
if (!cursor.next()) {
// let's try to move directly to the children level and
// return false immediately if the current node doesn't have any child
return false;
}
let i = 0;
do {
if (cursor.type.is(child[i])) {
i++;
}
} while (i < child.length && cursor.nextSibling());
return i >= child.length;
}