%global _empty_manifest_terminate_build 0
Name:		python-zipstream-new
Version:	1.1.8
Release:	1
Summary:	Zipfile generator that takes input files as well as streams
License:	GNU General Public License v3 (GPLv3)
URL:		https://github.com/arjan-s/python-zipstream
Source0:	https://mirrors.nju.edu.cn/pypi/web/packages/e5/f3/1b5228576f215b200c7e922a280a92e4494df33baae6e0280a6f45371f13/zipstream-new-1.1.8.tar.gz
BuildArch:	noarch


%description

# python-zipstream

zipstream.py is a zip archive generator based on python 3.3's zipfile.py. It was created to
generate a zip file generator for streaming (ie web apps). This is beneficial for when you
want to provide a downloadable archive of a large collection of regular files, which would be infeasible to
generate the archive prior to downloading or of a very large file that you do not want to store entirely on disk or on memory.

The archive is generated as an iterator of strings, which, when joined, form
the zip archive. For example, the following code snippet would write a zip
archive containing files from 'path' to a normal file:

```python
import zipstream

z = zipstream.ZipFile()
z.write('path/to/files')

with open('zipfile.zip', 'wb') as f:
    for data in z:
        f.write(data)
```

zipstream also allows to take as input a byte string iterable and to generate
the archive as an iterator.
This avoids storing large files on disk or in memory.
To do so you could use something like this snippet:

```python
def iterable():
    for _ in xrange(10):
        yield b'this is a byte string\x01\n'

z = zipstream.ZipFile()
z.write_iter('my_archive_iter', iterable())

with open('zipfile.zip', 'wb') as f:
    for data in z:
        f.write(data)
```

Of course both approach can be combined:

```python
def iterable():
    for _ in xrange(10):
        yield b'this is a byte string\x01\n'

z = zipstream.ZipFile()
z.write('path/to/files', 'my_archive_files')
z.write_iter('my_archive_iter', iterable())

with open('zipfile.zip', 'wb') as f:
    for data in z:
        f.write(data)
```

Since recent versions of web.py support returning iterators of strings to be
sent to the browser, to download a dynamically generated archive, you could
use something like this snippet:

```python
def GET(self):
    path = '/path/to/dir/of/files'
    zip_filename = 'files.zip'
    web.header('Content-type' , 'application/zip')
    web.header('Content-Disposition', 'attachment; filename="%s"' % (
        zip_filename,))
    return zipstream.ZipFile(path)
```

If the zlib module is available, zipstream.ZipFile can generate compressed zip
archives.

## Installation

```
pip install zipstream-new
```

## Requirements

  * Python 2.6+, 3.2+, pypy

## Examples

### flask

```python
from flask import Response

@app.route('/package.zip', methods=['GET'], endpoint='zipball')
def zipball():
    def generator():
        z = zipstream.ZipFile(mode='w', compression=zipstream.ZIP_DEFLATED)

        z.write('/path/to/file')

        for chunk in z:
            yield chunk

    response = Response(generator(), mimetype='application/zip')
    response.headers['Content-Disposition'] = 'attachment; filename={}'.format('files.zip')
    return response

# or

@app.route('/package.zip', methods=['GET'], endpoint='zipball')
def zipball():
    z = zipstream.ZipFile(mode='w', compression=zipstream.ZIP_DEFLATED)
    z.write('/path/to/file')

    response = Response(z, mimetype='application/zip')
    response.headers['Content-Disposition'] = 'attachment; filename={}'.format('files.zip')
    return response

# Partial flushing of the zip before closing

@app.route('/package.zip', methods=['GET'], endpoint='zipball')
def zipball():
    def generate_zip_with_manifest():
        z = zipstream.ZipFile(mode='w', compression=zipstream.ZIP_DEFLATED)

        manifest = []
        for filename in os.listdir('/path/to/files'):
            z.write(os.path.join('/path/to/files', filename), arcname=filename)
            yield from z.flush()
            manifest.append(filename)

        z.write_str('manifest.json', json.dumps(manifest).encode())

        yield from z

    response = Response(z, mimetype='application/zip')
    response.headers['Content-Disposition'] = 'attachment; filename={}'.format('files.zip')
    return response
```

### django 1.5+

```python
from django.http import StreamingHttpResponse

def zipball(request):
    z = zipstream.ZipFile(mode='w', compression=zipstream.ZIP_DEFLATED)
    z.write('/path/to/file')

    response = StreamingHttpResponse(z, content_type='application/zip')
    response['Content-Disposition'] = 'attachment; filename={}'.format('files.zip')
    return response
```

### webpy

```python
def GET(self):
    path = '/path/to/dir/of/files'
    zip_filename = 'files.zip'
    web.header('Content-type' , 'application/zip')
    web.header('Content-Disposition', 'attachment; filename="%s"' % (
        zip_filename,))
    return zipstream.ZipFile(path)
```

## Running tests

With python version > 2.6, just run the following command: `python -m unittest discover`

Alternatively, you can use `nose`.

If you want to run the tests on all supported Python versions, run `tox`.




%package -n python3-zipstream-new
Summary:	Zipfile generator that takes input files as well as streams
Provides:	python-zipstream-new
BuildRequires:	python3-devel
BuildRequires:	python3-setuptools
BuildRequires:	python3-pip
%description -n python3-zipstream-new

# python-zipstream

zipstream.py is a zip archive generator based on python 3.3's zipfile.py. It was created to
generate a zip file generator for streaming (ie web apps). This is beneficial for when you
want to provide a downloadable archive of a large collection of regular files, which would be infeasible to
generate the archive prior to downloading or of a very large file that you do not want to store entirely on disk or on memory.

The archive is generated as an iterator of strings, which, when joined, form
the zip archive. For example, the following code snippet would write a zip
archive containing files from 'path' to a normal file:

```python
import zipstream

z = zipstream.ZipFile()
z.write('path/to/files')

with open('zipfile.zip', 'wb') as f:
    for data in z:
        f.write(data)
```

zipstream also allows to take as input a byte string iterable and to generate
the archive as an iterator.
This avoids storing large files on disk or in memory.
To do so you could use something like this snippet:

```python
def iterable():
    for _ in xrange(10):
        yield b'this is a byte string\x01\n'

z = zipstream.ZipFile()
z.write_iter('my_archive_iter', iterable())

with open('zipfile.zip', 'wb') as f:
    for data in z:
        f.write(data)
```

Of course both approach can be combined:

```python
def iterable():
    for _ in xrange(10):
        yield b'this is a byte string\x01\n'

z = zipstream.ZipFile()
z.write('path/to/files', 'my_archive_files')
z.write_iter('my_archive_iter', iterable())

with open('zipfile.zip', 'wb') as f:
    for data in z:
        f.write(data)
```

Since recent versions of web.py support returning iterators of strings to be
sent to the browser, to download a dynamically generated archive, you could
use something like this snippet:

```python
def GET(self):
    path = '/path/to/dir/of/files'
    zip_filename = 'files.zip'
    web.header('Content-type' , 'application/zip')
    web.header('Content-Disposition', 'attachment; filename="%s"' % (
        zip_filename,))
    return zipstream.ZipFile(path)
```

If the zlib module is available, zipstream.ZipFile can generate compressed zip
archives.

## Installation

```
pip install zipstream-new
```

## Requirements

  * Python 2.6+, 3.2+, pypy

## Examples

### flask

```python
from flask import Response

@app.route('/package.zip', methods=['GET'], endpoint='zipball')
def zipball():
    def generator():
        z = zipstream.ZipFile(mode='w', compression=zipstream.ZIP_DEFLATED)

        z.write('/path/to/file')

        for chunk in z:
            yield chunk

    response = Response(generator(), mimetype='application/zip')
    response.headers['Content-Disposition'] = 'attachment; filename={}'.format('files.zip')
    return response

# or

@app.route('/package.zip', methods=['GET'], endpoint='zipball')
def zipball():
    z = zipstream.ZipFile(mode='w', compression=zipstream.ZIP_DEFLATED)
    z.write('/path/to/file')

    response = Response(z, mimetype='application/zip')
    response.headers['Content-Disposition'] = 'attachment; filename={}'.format('files.zip')
    return response

# Partial flushing of the zip before closing

@app.route('/package.zip', methods=['GET'], endpoint='zipball')
def zipball():
    def generate_zip_with_manifest():
        z = zipstream.ZipFile(mode='w', compression=zipstream.ZIP_DEFLATED)

        manifest = []
        for filename in os.listdir('/path/to/files'):
            z.write(os.path.join('/path/to/files', filename), arcname=filename)
            yield from z.flush()
            manifest.append(filename)

        z.write_str('manifest.json', json.dumps(manifest).encode())

        yield from z

    response = Response(z, mimetype='application/zip')
    response.headers['Content-Disposition'] = 'attachment; filename={}'.format('files.zip')
    return response
```

### django 1.5+

```python
from django.http import StreamingHttpResponse

def zipball(request):
    z = zipstream.ZipFile(mode='w', compression=zipstream.ZIP_DEFLATED)
    z.write('/path/to/file')

    response = StreamingHttpResponse(z, content_type='application/zip')
    response['Content-Disposition'] = 'attachment; filename={}'.format('files.zip')
    return response
```

### webpy

```python
def GET(self):
    path = '/path/to/dir/of/files'
    zip_filename = 'files.zip'
    web.header('Content-type' , 'application/zip')
    web.header('Content-Disposition', 'attachment; filename="%s"' % (
        zip_filename,))
    return zipstream.ZipFile(path)
```

## Running tests

With python version > 2.6, just run the following command: `python -m unittest discover`

Alternatively, you can use `nose`.

If you want to run the tests on all supported Python versions, run `tox`.




%package help
Summary:	Development documents and examples for zipstream-new
Provides:	python3-zipstream-new-doc
%description help

# python-zipstream

zipstream.py is a zip archive generator based on python 3.3's zipfile.py. It was created to
generate a zip file generator for streaming (ie web apps). This is beneficial for when you
want to provide a downloadable archive of a large collection of regular files, which would be infeasible to
generate the archive prior to downloading or of a very large file that you do not want to store entirely on disk or on memory.

The archive is generated as an iterator of strings, which, when joined, form
the zip archive. For example, the following code snippet would write a zip
archive containing files from 'path' to a normal file:

```python
import zipstream

z = zipstream.ZipFile()
z.write('path/to/files')

with open('zipfile.zip', 'wb') as f:
    for data in z:
        f.write(data)
```

zipstream also allows to take as input a byte string iterable and to generate
the archive as an iterator.
This avoids storing large files on disk or in memory.
To do so you could use something like this snippet:

```python
def iterable():
    for _ in xrange(10):
        yield b'this is a byte string\x01\n'

z = zipstream.ZipFile()
z.write_iter('my_archive_iter', iterable())

with open('zipfile.zip', 'wb') as f:
    for data in z:
        f.write(data)
```

Of course both approach can be combined:

```python
def iterable():
    for _ in xrange(10):
        yield b'this is a byte string\x01\n'

z = zipstream.ZipFile()
z.write('path/to/files', 'my_archive_files')
z.write_iter('my_archive_iter', iterable())

with open('zipfile.zip', 'wb') as f:
    for data in z:
        f.write(data)
```

Since recent versions of web.py support returning iterators of strings to be
sent to the browser, to download a dynamically generated archive, you could
use something like this snippet:

```python
def GET(self):
    path = '/path/to/dir/of/files'
    zip_filename = 'files.zip'
    web.header('Content-type' , 'application/zip')
    web.header('Content-Disposition', 'attachment; filename="%s"' % (
        zip_filename,))
    return zipstream.ZipFile(path)
```

If the zlib module is available, zipstream.ZipFile can generate compressed zip
archives.

## Installation

```
pip install zipstream-new
```

## Requirements

  * Python 2.6+, 3.2+, pypy

## Examples

### flask

```python
from flask import Response

@app.route('/package.zip', methods=['GET'], endpoint='zipball')
def zipball():
    def generator():
        z = zipstream.ZipFile(mode='w', compression=zipstream.ZIP_DEFLATED)

        z.write('/path/to/file')

        for chunk in z:
            yield chunk

    response = Response(generator(), mimetype='application/zip')
    response.headers['Content-Disposition'] = 'attachment; filename={}'.format('files.zip')
    return response

# or

@app.route('/package.zip', methods=['GET'], endpoint='zipball')
def zipball():
    z = zipstream.ZipFile(mode='w', compression=zipstream.ZIP_DEFLATED)
    z.write('/path/to/file')

    response = Response(z, mimetype='application/zip')
    response.headers['Content-Disposition'] = 'attachment; filename={}'.format('files.zip')
    return response

# Partial flushing of the zip before closing

@app.route('/package.zip', methods=['GET'], endpoint='zipball')
def zipball():
    def generate_zip_with_manifest():
        z = zipstream.ZipFile(mode='w', compression=zipstream.ZIP_DEFLATED)

        manifest = []
        for filename in os.listdir('/path/to/files'):
            z.write(os.path.join('/path/to/files', filename), arcname=filename)
            yield from z.flush()
            manifest.append(filename)

        z.write_str('manifest.json', json.dumps(manifest).encode())

        yield from z

    response = Response(z, mimetype='application/zip')
    response.headers['Content-Disposition'] = 'attachment; filename={}'.format('files.zip')
    return response
```

### django 1.5+

```python
from django.http import StreamingHttpResponse

def zipball(request):
    z = zipstream.ZipFile(mode='w', compression=zipstream.ZIP_DEFLATED)
    z.write('/path/to/file')

    response = StreamingHttpResponse(z, content_type='application/zip')
    response['Content-Disposition'] = 'attachment; filename={}'.format('files.zip')
    return response
```

### webpy

```python
def GET(self):
    path = '/path/to/dir/of/files'
    zip_filename = 'files.zip'
    web.header('Content-type' , 'application/zip')
    web.header('Content-Disposition', 'attachment; filename="%s"' % (
        zip_filename,))
    return zipstream.ZipFile(path)
```

## Running tests

With python version > 2.6, just run the following command: `python -m unittest discover`

Alternatively, you can use `nose`.

If you want to run the tests on all supported Python versions, run `tox`.




%prep
%autosetup -n zipstream-new-1.1.8

%build
%py3_build

%install
%py3_install
install -d -m755 %{buildroot}/%{_pkgdocdir}
if [ -d doc ]; then cp -arf doc %{buildroot}/%{_pkgdocdir}; fi
if [ -d docs ]; then cp -arf docs %{buildroot}/%{_pkgdocdir}; fi
if [ -d example ]; then cp -arf example %{buildroot}/%{_pkgdocdir}; fi
if [ -d examples ]; then cp -arf examples %{buildroot}/%{_pkgdocdir}; fi
pushd %{buildroot}
if [ -d usr/lib ]; then
	find usr/lib -type f -printf "/%h/%f\n" >> filelist.lst
fi
if [ -d usr/lib64 ]; then
	find usr/lib64 -type f -printf "/%h/%f\n" >> filelist.lst
fi
if [ -d usr/bin ]; then
	find usr/bin -type f -printf "/%h/%f\n" >> filelist.lst
fi
if [ -d usr/sbin ]; then
	find usr/sbin -type f -printf "/%h/%f\n" >> filelist.lst
fi
touch doclist.lst
if [ -d usr/share/man ]; then
	find usr/share/man -type f -printf "/%h/%f.gz\n" >> doclist.lst
fi
popd
mv %{buildroot}/filelist.lst .
mv %{buildroot}/doclist.lst .

%files -n python3-zipstream-new -f filelist.lst
%dir %{python3_sitelib}/*

%files help -f doclist.lst
%{_docdir}/*

%changelog
* Sun Apr 23 2023 Python_Bot <Python_Bot@openeuler.org> - 1.1.8-1
- Package Spec generated