Question

我在文件中有一些文字：

<form [formGroup]="myFormGroup" *ngIf="myFormGroup" (ngSubmit)="onSubmit()">
  <div>
    <label for="name">Name:</label>
    <input type="text" name="name" id="name" formControlName="name" />
  </div>
  <div>
    <label for="myCheckbox">My Checkbox:</label>
    <input type="checkbox" name="myCheckbox" id="myCheckbox" formControlName="myCheckbox" />
  </div>
  <div>
    <button type="submit" [disabled]="!myFormGroup.valid">Submit!</button>
  </div>

  <div *ngIf="myFormGroup.errors">
    {{myFormGroup.errors | json}}
  </div>
</form>

我只想找到这部分：

import { Pipe, PipeTransform } from '@angular/core';
import { DatePipe } from '@angular/common';

@Pipe({
  name: 'dateFormat'
})
export class DateFormatPipe extends DatePipe implements PipeTransform {

  transform(value: any, args?: any): any {

    switch (args) {
      case 'en': {
        return super.transform(value, 'MM/dd/yyyy h:mm a');
      }

      case 'fr': {
        return super.transform(value, 'dd/MM/yyyy HH:mm'); // <--- find proper separator
      }

      case 'de': {
        return super.transform(value, 'dd.MM.yyyy HH:mm');
      }

      case 'es': {
        return super.transform(value, 'yyyy/MM/dd hh:MM');
      }
    }
  }
}

我知道我可以使用此正则表达式：

class GroupsWizard(SessionWizardView):

    def get_template_names(self):
        return 'index.html'

    def get_context_data(self, form, **kwargs):
        context = super(GroupsWizard, self).get_context_data(form = form, **kwargs)
        context_data = Groups.objects.all()
        context.update({'context_data': context_data})
        return context

    def done(self, form_list, **kwargs):
        data = {k: v for form in form_list for k, v in form.cleaned_data.items()}
        instance = Groups.objects.create(**data)
        return render(self.request, 'done.html', {
            'form_data': [form.cleaned_data for form in form_list],
        })

我想做的是用逗号25968.254.000.1234.784.000000000000.85 8547968.82 0 874968.21 5896314.42替换第一部分中的句号25968.254.000.1234.784.000000000000.85 8547968.82，最后得到以下内容：

^\d{5}\.(\d{3})\.\d{3}\.(\d{4})\.\d{3}\.(\d{12})\.\d{2}\s([0-9.]*)\s

我该怎么做？

谢谢

Answer 1

您需要使用正则表达式capture-groups。

您将每个数字组包装在（）中，然后使用\ N引用它们，其中N是第N个组。

将此用于查找：

create table subject as
  select 1 id, 'Humanities' name from dual union all
  select 2   , 'Science'         from dual union all
  select 3   , 'Math'            from dual
;

create table course_semester as
  select 1 id_subject, 201801 semester, 1002 as id_profesor from dual union all
  select 2           , 201702         , 1002 as id_profesor from dual union all
  select 3           , 201801         , 1002 as id_profesor from dual
;

CREATE OR REPLACE FUNCTION listar_cursos(prof NUMBER)  RETURN VARCHAR IS
 test VARCHAR(500);
BEGIN
FOR item IN 
(
  SELECT subject.name AS name FROM subject
      INNER JOIN course_semester 
  ON subject.id = course_semester.id_subject
  WHERE course_semester.id_profesor = prof
  ORDER BY subject.name
)
LOOP
  test:= test || ',' || item.name;
END LOOP;
RETURN ltrim(test, ',');
END;
/

select listar_cursos(1002) from dual;

LISTAR_CURSOS(1002)
-----------------------
Humanities,Math,Science

然后将其替换：

^(\d{5}).(\d{3}).(\d{3}).(\d{4}).(\d{3}).(\d{12}).(\d{2})\s([0-9.]*)\s

Answer 2

如果仅在某些数字之后，则将其替换为逗号，从行的开头一直到找到一个非数字字符为止。它适用于任意数量的数字点。

Ctrl + H
查找内容：(?:^|\G)(\d+)\.
替换为：$1,
检查环绕
检查正则表达式
全部替换

说明：

(?:         : start non capture group
    ^       : beginning of line
  |         : OR
    \G      : restart from last match
)           : end group
(\d+)       : group 1, 1 or more digits
\.          : a dot

替换：

$1          : content of group 1 (i.e. the digits)
,           : a comma

给定示例的结果

25968,254,000,1234,784,000000000000,85  8547968.82  0   874968.21   5896314.42

Notepad ++使用正则表达式查找并用逗号替换点

2 个答案: