Question

canActivate(route: ActivatedRouteSnapshot): Observable<boolean> 
{
   return concat(this.loadIfRequired(), this.hasBookInStore(route.params['id'])).pipe(skip(1));
}

private loadIfRequired(): Observable<any>
{
   return booksState$.pipe(
    map(state => state.loaded),
    take(1),
    tap(loaded=> {
      if (!loaded) {
        this.store.dispatch(new BooksActions.Load());
      }
    }));
}

private hasBookInStore(id: string): Obsevable<boolean>
{
  return this.booksState$.pipe(
    first(state => state.loaded),
    map(state => state.list.find(book => book.id === id)),
    map((book) => {
     if(!!book) return true;
     this.router.navigate(['/404']);
     return false;
   })
  );
}

private get booksState$(): Observable<BooksState>
{
 return this.store.pipe(
    select(state => state.books));
}

我尝试从字符串中提取“ **保险**：不适用”。但是我上面的代码不起作用。我该怎么做？

谢谢！

Answer 1

我会将内容视为（半结构化的）键值文件格式。

您可以将键值对与正则表达式进行匹配，如下所示：

(\*\*[a-zA-Y ?]+\*\*) : ((?:(?!\*\*).)*)(?= |$)

Demo

说明：

(\*\*[a-zA-Y ?]+\*\*)键：您可能需要调整字符范围
:带有空格的kv分隔符
((?:(?!\*\*).)*)会用一个经过调节的贪婪令牌捕获该值：除文字**之外的所有内容，其后是(?= |$)字符串$的结尾或空格。（？= | $）

Sample Code：

import re
regex = r"(\*\*[a-zA-Z ?]+\*\*) : ((?:(?!\*\*).)*)(?= |$)"
test_str = "**Amount** : $25k  **Name** : James **Excess** : None Returned \\n **In Suit?** : Y **Venue** : SF **Insurance** : N/A \\n **FTSA** : None listed"
matches = re.finditer(regex, test_str, re.MULTILINE)
for match in matches:
    if match.group(1) == "**Insurance**":
        print (match.group(2))

Python Regex-提取包含星号的句子

1 个答案: