Files
online_sys/frontend_大健康/find_syntax_issue.py
KQL a7242f0c69 Initial commit: 教务系统在线平台
- 包含4个产业方向的前端项目:智能开发、智能制造、大健康、财经商贸
- 已清理node_modules、.yoyo等大文件,项目大小从2.6GB优化至631MB
- 配置完善的.gitignore文件

🤖 Generated with [Claude Code](https://claude.com/claude-code)

Co-Authored-By: Claude Sonnet 4.5 <noreply@anthropic.com>
2025-12-12 18:16:55 +08:00

42 lines
1.5 KiB
Python

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
print("查找语法问题...")
with open('/Users/apple/Documents/cursor/教务系统/frontend_大健康/src/mocks/resumeInterviewMock.js', 'r', encoding='utf-8') as f:
lines = f.readlines()
# 跟踪括号栈
stack = []
line_num = 0
for i, line in enumerate(lines, 1):
for j, char in enumerate(line):
if char in '{[':
stack.append((char, i, j))
elif char in '}]':
if not stack:
print(f"{i}行第{j}列: 多余的关闭符号 '{char}'")
print(f" 内容: {line.strip()}")
else:
opener, open_line, open_col = stack.pop()
expected = '}' if opener == '{' else ']'
if char != expected:
print(f"{i}行第{j}列: 不匹配的关闭符号")
print(f" 期望 '{expected}' 但得到 '{char}'")
print(f" 开始于第{open_line}行第{open_col}")
print(f" 内容: {line.strip()}")
# 检查未关闭的括号
if stack:
print("\n未关闭的括号:")
for char, line_num, col in stack[-5:]: # 显示最后5个未关闭的
print(f"{line_num}行第{col}列: '{char}' 未关闭")
if line_num <= len(lines):
print(f" 内容: {lines[line_num-1].strip()}")
# 检查第6518行附近
print("\n第6518行附近的内容:")
for i in range(max(0, 6515), min(len(lines), 6525)):
if i < len(lines):
print(f"{i+1}: {lines[i].rstrip()}")