如何获取书签的页码

Question

from typing import List
from PyPDF2 import PdfFileReader
from PyPDF2.generic import Destination

def get_outlines(pdf_filepath: str) -> List[Destination]:
    """Get the bookmarks of a PDF file."""
    with open(pdf_filepath, "rb") as fp:
        pdf_file_reader = PdfFileReader(fp)
        outlines = pdf_file_reader.getOutlines()
    return outlines

print(get_outlines("PDF-export-example.pdf"))

pyPdf.pdf.Destination有很多属性，但是我找不到那个书签的任何参考页码。我怎样才能得到书签的页码?

---

例如，outlines[1].page.idnum返回的数字大约是PDF文档中引用页码的3倍，我假设引用的是一些比页面小的对象，因为在整个PDF文档大纲上运行.page.idnum返回的数字数组甚至与"real"不线性相关;PDF文档的页码目的地，大约乘以~ 3

---

更新:这个问题和这个一样:根据大纲分割pdf，尽管我不明白作者在他自己的回答中做了什么。似乎太复杂了，我无法使用

Answer 1

正如@theta指出的"根据大纲分割pdf "包含提取页码所需的代码。如果你觉得这很复杂，我复制了一部分将页面id映射到页码的代码，并使其成为一个函数。下面是打印书签0[0]页码的工作示例:

from PyPDF2 import PdfFileReader

def _setup_page_id_to_num(pdf, pages=None, _result=None, _num_pages=None):
    if _result is None:
        _result = {}
    if pages is None:
        _num_pages = []
        pages = pdf.trailer["/Root"].getObject()["/Pages"].getObject()
    t = pages["/Type"]
    if t == "/Pages":
        for page in pages["/Kids"]:
            _result[page.idnum] = len(_num_pages)
            _setup_page_id_to_num(pdf, page.getObject(), _result, _num_pages)
    elif t == "/Page":
        _num_pages.append(1)
    return _result
# main
f = open('document.pdf','rb')
p = PdfFileReader(f)
# map page ids to page numbers
pg_id_num_map = _setup_page_id_to_num(p)
o = p.getOutlines()
pg_num = pg_id_num_map[o[0].page.idnum] + 1
print(pg_num)

对于@theta来说可能太晚了，但可能会对其他人有所帮助:)顺便说一句，我在stackoverflow上的第一篇文章，所以如果我没有遵循通常的格式，请原谅

进一步扩展:如果你正在寻找书签在页面上的确切位置，这将使你的工作更容易:

from PyPDF2 import PdfFileReader
import PyPDF2 as pyPdf
def _setup_page_id_to_num(pdf, pages=None, _result=None, _num_pages=None):
    if _result is None:
        _result = {}
    if pages is None:
        _num_pages = []
        pages = pdf.trailer["/Root"].getObject()["/Pages"].getObject()
    t = pages["/Type"]
    if t == "/Pages":
        for page in pages["/Kids"]:
            _result[page.idnum] = len(_num_pages)
            _setup_page_id_to_num(pdf, page.getObject(), _result, _num_pages)
    elif t == "/Page":
        _num_pages.append(1)
    return _result
def outlines_pg_zoom_info(outlines, pg_id_num_map, result=None):
    if result is None:
        result = dict()
    if type(outlines) == list:
        for outline in outlines:
            result = outlines_pg_zoom_info(outline, pg_id_num_map, result)
    elif type(outlines) == pyPdf.pdf.Destination:
        title = outlines['/Title']
        result[title.split()[0]] = dict(title=outlines['/Title'], top=outlines['/Top'], 
        left=outlines['/Left'], page=(pg_id_num_map[outlines.page.idnum]+1))
    return result
# main
pdf_name = 'document.pdf'
f = open(pdf_name,'rb')
pdf = PdfFileReader(f)
# map page ids to page numbers
pg_id_num_map = _setup_page_id_to_num(pdf)
outlines = pdf.getOutlines()
bookmarks_info = outlines_pg_zoom_info(outlines, pg_id_num_map)
print(bookmarks_info)

注意:我的书签是章节编号(例如:1.1介绍)，我正在将书签信息映射到章节编号。如果您的书签不同，请修改这部分代码:

    elif type(outlines) == pyPdf.pdf.Destination:
        title = outlines['/Title']
        result[title.split()[0]] = dict(title=outlines['/Title'], top=outlines['/Top'], 
        left=outlines['/Left'], page=(pg_id_num_map[outlines.page.idnum]+1))

Answer 2

使用vjayky和Giulio D建议递归管理书签

PyPDF2>= v1.25

from PyPDF2 import PdfFileReader
def printBookmarksPageNumbers(pdf):
    def review_and_print_bookmarks(bookmarks, lvl=0):
        for b in bookmarks:
            if type(b) == list:
                review_and_print_bookmarks(b, lvl + 4)
                continue
            pg_num = pdf.getDestinationPageNumber(b) + 1 #page count starts from 0
            print("%s%s: Page %s" %(" "*lvl, b.title, pg_num))
    review_and_print_bookmarks(pdf.getOutlines())
with open('document.pdf', "rb") as f:
    pdf = PdfFileReader(f)
    printBookmarksPageNumbers(pdf)

PyPDF2 & lt;v1.25

from PyPDF2 import PdfFileReader
def printBookmarksPageNumbers(pdf):
    # Map page ids to page numbers
    pg_id_to_num = {}
    for pg_num in range(0, pdf.getNumPages()):
        pg_id_to_num[pdf.getPage(pg_num).indirectRef.idnum] = pg_num
    def review_and_print_bookmarks(bookmarks, lvl=0):
        for b in bookmarks:
            if type(b) == list:
                review_and_print_bookmarks(b, lvl + 4)
                continue
            pg_num = pg_id_to_num[b.page.idnum] + 1 #page count starts from 0 
            print("%s%s: Page %s" %(" "*lvl, b.title, pg_num))
    review_and_print_bookmarks(pdf.getOutlines())
with open('document.pdf', "rb") as f:
    pdf = PdfFileReader(f)
    printBookmarksPageNumbers(pdf)

Answer 3

在2019年，对于那些对更快的方式感兴趣的人来说，可以使用:

from PyPDF2 import PdfFileReader
def printPageNumberFrom(filename):
    with open(filename, "rb") as f:
       pdf = PdfFileReader(f)
       bookmarks = pdf.getOutlines()
       for b in bookmarks:
           print(pdf.getDestinationPageNumber(b) + 1) #page count starts from 0 

Answer 4

我不确定，但根据pypdf的文档。书签的页码就是。页面。