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Python: Iterate through directory to find specific text file, open it, write some of its contents to a new file, close it, and move on to the next dir

I have a script that takes an input text file then finds data in it, puts that data as a variable, then later I call that variable to write to a new file. This snippet of code is just for reading the txt file and storing the data from it as variables.
searchfile = open('C://Users//Me//DynamicFolder//report//summary.txt','r', encoding='utf-8')
slab_count=0
slab_number=[]
slab_total=0
for line in searchfile:
if "Slab" in line:
slab_num = ([float(s) for s in re.findall(r'[-+]?(?:\d*\.\d+|\d+)', line)])
slab_percent = slab_num[-1]
slab_number.append(slab_percent)
slab_count=slab_count+1
slab_total=0
for slab_percent in slab_number:
slab_total+=slab_percent
searchfile.close()
I am using xlsxwriter to write the variables to an excel doc.
My question is, how do I iterate this to search through a given directories sub-directories for summary.txt when there is a dynamic folder.
So C://Users//Me//DynamicFolder//report//summary.txt is a path to one of the files. There are several folders I named DynamicFolder that are there because another process puts them there, they change their names all the time. I need have this script go into each of those dynamic folders to a subdir called report, this is a static name and is always the same. So each of those dynamicfolders has another subdir called report, and in the report folder is a file called summary.txt. I am trying to go through each of those dynamicfolders into the subdir report > summary.txt and then opening and writing data from those txt files.
How do I iterate or loop this? Right now I have 18 folders with those DynamicFolder names that will change when they are over written. How can I put this snip of code to iterate through?
for path in Path('C://Users//Me//DynamicFolder//report//summary.txt').rglob('summary.txt'):
report folder is not the only folder with a summary.txt file, but its the only folder with the file I want. So this code above pulls ALL summary.txt files from all subdir's under the DynamicFolder (not just report folder). I am wondering if I can make this JUST do the 'report' subdir folders under DynamicFolders, and somehow use this to iterate the rest of my code?

zipping files in python without folder structure

I want to create a zip file of some files somewhere on the disk in python. I successfully got the path to the folder and each file name so I did:
with zp(os.path.join(self.savePath, self.selectedIndex + ".zip"), "w") as zip:
for file in filesToZip:
zip.write(self.folderPath + file)
Everything works fine, but the zipfile that is output contains the entire folder structure leading up to the files. Is there a way to only zip the files and not the folders with it?

From the documentation:
ZipFile.write(filename, arcname=None, compress_type=None,
compresslevel=None)
Write the file named filename to the archive, giving it the archive
name arcname (by default, this will be the same as filename, but
without a drive letter and with leading path separators removed).
So, just specify an explicit arcname:
with zp(os.path.join(self.savePath, self.selectedIndex + ".zip"), "w") as zip:
for file in filesToZip:
zip.write(self.folderPath + file, arcname=file)

Maybe I' misunderstand the question but could someone explain to me why the answer is not:
zip.write(file, arcname=os.path.basename(file))
It works for me but, again, I might be missing something in the question...

Determine Filename of Unzipped File

Say you unzip a file called file123.zip with zipfile.ZipFile, which yields an unzipped file saved to a known path. However, this unzipped file has a completely random name. How do you determine this completely random filename? Or is there some way to control what the name of the unzipped file is?
I am trying to implement this in python.

By "random" I assume that you mean that the files are named arbitrarily.
You can use ZipFile.read() which unzips the file and returns its contents as a string of bytes. You can then write that string to a named file of your choice.
from zipfile import ZipFile
with ZipFile('file123.zip') as zf:
for i, name in enumerate(zf.namelist()):
with open('outfile_{}'.format(i), 'wb') as f:
f.write(zf.read(name))
This will write each file from the archive to a file named output_n in the current directory. The names of the files contained in the archive are obtained with ZipFile.namelist(). I've used enumerate() as a simple method of generating the file names, however, you could substitute that with whatever naming scheme you require.

If the filename is completely random you can first check for all filenames in a particular directory using os.listdir(). Now you know the filename and can do whatever you want with it :)
See this topic for more information.

extract zip file without folder python

I am currently using extratall function in python to unzip, after unziping it also creates a folder like: myfile.zip -> myfile/myfile.zip , how do i get rid of myfile flder and just unzip it to the current folder without the folder, is it possible ?

I use the standard module zipfile. There is the method extract which provides what I think you want. This method has the optional argument path to either extract the content to the current working directory or the the given path
import os, zipfile
os.chdir('path/of/my.zip')
with zipfile.ZipFile('my.zip') as Z :
for elem in Z.namelist() :
Z.extract(elem, 'path/where/extract/to')
If you omit the 'path/where/extract/to' the files from the ZIP-File will be extracted to the directory of the ZIP-File.

import shutil
# loop over everything in the zip
for name in myzip.namelist():
# open the entry so we can copy it
member = myzip.open(name)
with open(os.path.basename(name), 'wb') as outfile:
# copy it directly to the output directory,
# without creating the intermediate directory
shutil.copyfileobj(member, outfile)

Adding file to existing zipfile

I'm using python's zipfile module.
Having a zip file located in a path of:
/home/user/a/b/c/test.zip
And having another file created under /home/user/a/b/c/1.txt
I want to add this file to existing zip, I did:
zip = zipfile.ZipFile('/home/user/a/b/c/test.zip','a')
zip.write('/home/user/a/b/c/1.txt')
zip.close()`
And got all the subfolders appears in path when unzipping the file, how do I just enter the zip file without path's subfolders?
I tried also :
zip.write(os.path.basename('/home/user/a/b/c/1.txt'))
And got an error that file doesn't exist, although it does.

You got very close:
zip.write(path_to_file, os.path.basename(path_to_file))
should do the trick for you.
Explanation: The zip.write function accepts a second argument (the arcname) which is the filename to be stored in the zip archive, see the documentation for zipfile more details.
os.path.basename() strips off the directories in the path for you, so that the file will be stored in the archive under just it's name.
Note that if you only zip.write(os.path.basename(path_to_file)) it will look for the file in the current directory where it (as the error says) does not exist.

import zipfile
# Open a zip file at the given filepath. If it doesn't exist, create one.
# If the directory does not exist, it fails with FileNotFoundError
filepath = '/home/user/a/b/c/test.zip'
with zipfile.ZipFile(filepath, 'a') as zipf:
# Add a file located at the source_path to the destination within the zip
# file. It will overwrite existing files if the names collide, but it
# will give a warning
source_path = '/home/user/a/b/c/1.txt'
destination = 'foobar.txt'
zipf.write(source_path, destination)