解压.docx文件实现提取图片
前言
.docx文件其实也就是一个压缩文件，当我们将一个.docx文件直接解压后可以看到如下目录
其中我们要找的图片就在word/media目录内，如图
所以，要提取word内的图片就需要将.docx文件解压，再从media文件内取得图片，然后将解压后的文件删除
代码实现
import
os
import
shutil
import
zipfile
def
get_pictures(word_path, result_path):
"""
获取word内的所有图片
:param word_path: word文件
:param result_path: 结果目录，无需手动创建
:return: None or generator，None：word内没有图片，generator：每个图片的路径
"""
tmp_path
=
f
'{os.path.splitext(word_path)[0]}'
# 拷贝源文件后重命名再解压
splitext
=
os.path.splitext(word_path)
zip_path
=
shutil.copy(word_path, f
'{splitext[0]}_new{splitext[1]}'
)
with zipfile.ZipFile(zip_path,
'r'
) as f:
for
file
in
f.namelist():
f.extract(
file
, tmp_path)
os.remove(zip_path)
# 注：word图片在zip文件内的word/media目录下
pic_path
=
os.path.join(tmp_path,
'word/media'
)
if
not
os.path.exists(pic_path):
shutil.rmtree(tmp_path)
return
'no pictures found'
pictures
=
os.listdir(pic_path)
if
not
os.path.exists(result_path):
os.makedirs(result_path)
for
picture
in
pictures:
# 根据word的文件名生成图片的名称
word_name
=
os.path.splitext(word_path)[
0
]
if
os.sep
in
word_name:
new_name
=
word_name.split(
'\\'
)[
-
1
]
else
:
new_name
=
word_name.split(
'/'
)[
-
1
]
picture_name
=
f
'{new_name}_{picture}'
shutil.copy(os.path.join(pic_path, picture), os.path.join(result_path, picture_name))
shutil.rmtree(tmp_path)
return
(os.path.join(result_path, pic)
for
pic
in
os.listdir(result_path))
|
word_path可以支持所有类型路径，如
p
=
r
"C:\Users\Desktop\test\小说.docx"
p1
=
"C:/Users/Desktop/test/小说.docx"
p2
=
"C:\\Users\\Desktop\\test\\小说.docx"
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利用三方库docx实现图片提取(推荐)
import
docx
import
os, re
def
get_pictures(word_path, result_path):
"""
图片提取
:param word_path: word路径
:param result_path: 结果路径
:return:
"""
doc
=
docx.Document(word_path)
dict_rel
=
doc.part._rels
for
rel
in
dict_rel:
rel
=
dict_rel[rel]
if
"image"
in
rel.target_ref:
if
not
os.path.exists(result_path):
os.makedirs(result_path)
img_name
=
re.findall(
"/(.*)"
, rel.target_ref)[
0
]
word_name
=
os.path.splitext(word_path)[
0
]
if
os.sep
in
word_name:
new_name
=
word_name.split(
'\\'
)[
-
1
]
else
:
new_name
=
word_name.split(
'/'
)[
-
1
]
img_name
=
f
'{new_name}_{img_name}'
with
open
(f
'{result_path}/{img_name}'
,
"wb"
) as f:
f.write(rel.target_part.blob)
|