SSH - How to delete all .htaccess files inside the wp-content folders? - ssh

I have thousands of .htaccess files added by malware, I need to login via SSH and delete them, but only when they are inside wp-content folder, outside this folder they need to stay.
What's the cmd please?

Note this is for Python
Try This:
import os
file_type = input("Enter file type: ")
folder_path = input("Enter folder path: ")
filelist = []
for root, dirs, files in os.walk(folder_path):
for file in files:
#append the file name to the list
filelist.append(os.path.join(root,file))
#print all the file names
for name in filelist:
delete_file = str(name)
if delete_file.endswith(file_type):
os.remove(delete_file)
print("File deleted successfully.")
break
else:
print("No files found.")

Related

Save Pandas or Pyspark dataframe from Databricks to Azure Blob Storage

Is there a way I can save a Pyspark or Pandas dataframe from Databricks to a blob storage without mounting or installing libraries?
I was able to achieve this after mounting the storage container into Databricks and using the library com.crealytics.spark.excel, but I was wondering if I can do the same without the library or without mounting because I will be working on clusters without these 2 permissions.
Here the code for saving the dataframe locally to dbfs.
# export
from os import path
folder = "export"
name = "export"
file_path_name_on_dbfs = path.join("/tmp", folder, name)
# Writing to DBFS
# .coalesce(1) used to generate only 1 file, if the dataframe is too big this won't work so you'll have multiple files qnd you need to copy them later one by one
sampleDF \
.coalesce(1) \
.write \
.mode("overwrite") \
.option("header", "true") \
.option("delimiter", ";") \
.option("encoding", "UTF-8") \
.csv(file_path_name_on_dbfs)
# path of destination, which will be sent to az storage
dest = file_path_name_on_dbfs + ".csv"
# Renaming part-000...csv to our file name
target_file = list(filter(lambda file: file.name.startswith("part-00000"), dbutils.fs.ls(file_path_name_on_dbfs)))
if len(target_file) > 0:
dbutils.fs.mv(target_file[0].path, dest)
dbutils.fs.cp(dest, f"file://{dest}") # this line is added for community edition only cause /dbfs is not recognized, so we copy the file locally
dbutils.fs.rm(file_path_name_on_dbfs,True)
The code that will send the file into az storage :
import requests
sas="YOUR_SAS_TOKEN_PREVIOUSLY_CREATED" # follow the link below to create SAS token (using sas is slightly more secure than raw key storage)
blob_account_name = "YOUR_BLOB_ACCOUNT_NAME"
container = "YOUR_CONTAINER_NAME"
destination_path_w_name = "export/export.csv"
url = f"https://{blob_account_name}.blob.core.windows.net/{container}/{destination_path_w_name}?{sas}"
# here we read the content of our previously exported df -> csv
# if you are not on community edition you might want to use /dbfs + dest
payload=open(dest).read()
headers = {
'x-ms-blob-type': 'BlockBlob',
'Content-Type': 'text/csv' # you can change the content type according to your needs
}
response = requests.request("PUT", url, headers=headers, data=payload)
# if response.status_code is 201 it means your file was created successfully
print(response.status_code)
Follow this link to setup a SAS token
Remember that anyone who got the sas token can access your storage depending on permissions you set while creating the sas token
Code for Excel export version (using com.crealytics:spark-excel_2.12:0.14.0)
Saving the dataframe :
data = [
('a',25,'ast'),
('b',15,'phone'),
('c',32,'dlp'),
('d',45,'rare'),
('e',60,'phq' )
]
colums = ["column1" ,"column2" ,"column3"]
sampleDF = spark.createDataFrame(data=data, schema = colums)
sampleDF.show()
# export
from os import path
folder = "export"
name = "export"
file_path_name_on_dbfs = path.join("/tmp", folder, name)
# Writing to DBFS
sampleDF.write.format("com.crealytics.spark.excel")\
.option("header", "true")\
.mode("overwrite")\
.save(file_path_name_on_dbfs + ".xlsx")
# excel
dest = file_path_name_on_dbfs + ".xlsx"
dbutils.fs.cp(dest, f"file://{dest}") # this line is added for community edition only cause /dbfs is not recognized, so we copy the file locally
Uploading the file to azure storage :
import requests
sas="YOUR_SAS_TOKEN_PREVIOUSLY_CREATED" # follow the link below to create SAS token (using sas is slightly more secure than raw key storage)
blob_account_name = "YOUR_BLOB_ACCOUNT_NAME"
container = "YOUR_CONTAINER_NAME"
destination_path_w_name = "export/export.xlsx"
# destination_path_w_name = "export/export.csv"
url = f"https://{blob_account_name}.blob.core.windows.net/{container}/{destination_path_w_name}?{sas}"
# here we read the content of our previously exported df -> csv
# if you are not on community edition you might want to use /dbfs + dest
# payload=open(dest).read()
payload=open(dest, 'rb').read()
headers = {
'x-ms-blob-type': 'BlockBlob',
# 'Content-Type': 'text/csv'
'Content-Type': 'application/vnd.openxmlformats-officedocument.spreadsheetml.sheet'
}
response = requests.request("PUT", url, headers=headers, data=payload)
# if response.status_code is 201 it means your file was created successfully
print(response.status_code)

How to write a python-for-android recipe for a package in a local directory and not a zip file url?

I have
buildozer.spec
recipes/
myrecipe/
__init__.py
mypackage/
setup.py
code.py
But when I try to write a recipe with a file:// URL as seen when googling this issue, I get an error Exception: Given path is neither a file nor a directory: /home/user/project/.buildozer/android/platform/build-armeabi-v7a/packages/mypackage/mypackage (not the mypackage twice).
How can I achieve this?
There is an IncludedFilesBehaviour mixin just for this, just give it a relative path with src_filename:
from pythonforandroid.recipe import IncludedFilesBehaviour, CppCompiledComponentsPythonRecipe
import os
import sys
class MyRecipe(IncludedFilesBehaviour, CppCompiledComponentsPythonRecipe):
version = 'stable'
src_filename = "../../../phase-engine"
name = 'phase-engine'
depends = ['setuptools']
call_hostpython_via_targetpython = False
install_in_hostpython = True
def get_recipe_env(self, arch):
env = super().get_recipe_env(arch)
env['LDFLAGS'] += ' -lc++_shared'
return env
recipe = MyRecipe()

Uploading Multiple files in AWS S3 from terraform

I want to upload multiple files to AWS S3 from a specific folder in my local device. I am running into the following error.
Here is my terraform code.
resource "aws_s3_bucket" "testbucket" {
bucket = "test-terraform-pawan-1"
acl = "private"
tags = {
Name = "test-terraform"
Environment = "test"
}
}
resource "aws_s3_bucket_object" "uploadfile" {
bucket = "test-terraform-pawan-1"
key = "index.html"
source = "/home/pawan/Documents/Projects/"
}
How can I solve this problem?
As of Terraform 0.12.8, you can use the fileset function to get a list of files for a given path and pattern. Combined with for_each, you should be able to upload every file as its own aws_s3_bucket_object:
resource "aws_s3_bucket_object" "dist" {
for_each = fileset("/home/pawan/Documents/Projects/", "*")
bucket = "test-terraform-pawan-1"
key = each.value
source = "/home/pawan/Documents/Projects/${each.value}"
# etag makes the file update when it changes; see https://stackoverflow.com/questions/56107258/terraform-upload-file-to-s3-on-every-apply
etag = filemd5("/home/pawan/Documents/Projects/${each.value}")
}
See terraform-providers/terraform-provider-aws : aws_s3_bucket_object: support for directory uploads #3020 on GitHub.
Note: This does not set metadata like content_type, and as far as I can tell there is no built-in way for Terraform to infer the content type of a file. This metadata is important for things like HTTP access from the browser working correctly. If that's important to you, you should look into specifying each file manually instead of trying to automatically grab everything out of a folder.
You are trying to upload a directory, whereas Terraform expects a single file in the source field. It is not yet supported to upload a folder to an S3 bucket.
However, you can invoke awscli commands using null_resource provisioner, as suggested here.
resource "null_resource" "remove_and_upload_to_s3" {
provisioner "local-exec" {
command = "aws s3 sync ${path.module}/s3Contents s3://${aws_s3_bucket.site.id}"
}
}
Since June 9, 2020, terraform has a built-in way to infer the content type (and a few other attributes) of a file which you may need as you upload to a S3 bucket
HCL format:
module "template_files" {
source = "hashicorp/dir/template"
base_dir = "${path.module}/src"
template_vars = {
# Pass in any values that you wish to use in your templates.
vpc_id = "vpc-abc123"
}
}
resource "aws_s3_bucket_object" "static_files" {
for_each = module.template_files.files
bucket = "example"
key = each.key
content_type = each.value.content_type
# The template_files module guarantees that only one of these two attributes
# will be set for each file, depending on whether it is an in-memory template
# rendering result or a static file on disk.
source = each.value.source_path
content = each.value.content
# Unless the bucket has encryption enabled, the ETag of each object is an
# MD5 hash of that object.
etag = each.value.digests.md5
}
JSON format:
{
"resource": {
"aws_s3_bucket_object": {
"static_files": {
"for_each": "${module.template_files.files}"
#...
}}}}
#...
}
Source: https://registry.terraform.io/modules/hashicorp/dir/template/latest
My objective was to make this dynamic, so whenever i create a folder in a directory, terraform automatically uploads that new folder and its contents into S3 bucket with the same key structure.
Heres how i did it.
First you have to get a local variable with a list of each Folder and the files under it. Then we can loop through that list to upload the source to S3 bucket.
Example: I have a folder called "Directories" with 2 sub folders called "Folder1" and "Folder2" each with their own files.
- Directories
- Folder1
* test_file_1.txt
* test_file_2.txt
- Folder2
* test_file_3.txt
Step 1: Get the local var.
locals{
folder_files = flatten([for d in flatten(fileset("${path.module}/Directories/*", "*")) : trim( d, "../") ])
}
Output looks like this:
folder_files = [
"Folder1/test_file_1.txt",
"Folder1/test_file_2.txt",
"Folder2/test_file_3.txt",
]
Step 2: dynamically upload s3 objects
resource "aws_s3_object" "this" {
for_each = { for idx, file in local.folder_files : idx => file }
bucket = aws_s3_bucket.this.bucket
key = "/Directories/${each.value}"
source = "${path.module}/Directories/${each.value}"
etag = "${path.module}/Directories/${each.value}"
}
This loops over the local var,
So in your S3 bucket, you will have uploaded in the same structure, the local Directory and its sub directories and files:
Directory
- Folder1
- test_file_1.txt
- test_file_2.txt
- Folder2
- test_file_3.txt

Cant read web2py uploaded .txt from the shell

I have a simple table:
db.define_table('myfiles',
Field('title','string'),
Field('myfile','upload))
Then i run my app from shell:
python web2py.py -S myapp -M
Choose my file_path:
file_path = os.path.join(request.folder,'upload',db.myfiles[1].myfile)
but then i try to read my uploaded file, i get "File not open for reading"
with open(file_path, 'wb') as f: data = f.readlines()
I even tried the same process with copy-paste my file to private folder but still get the same error.
First, the default folder for uploaded files is "uploads", not "upload":
file_path = os.path.join(request.folder, 'uploads', db.myfiles[1].myfile)
Second, you should open the file for reading rather than writing:
with open(file_path, 'rb') as f:
data = f.readlines()

how to remove the file after download in specified path

filepath = self.class.instance_variable_get(:#filename)
# puts" #{:#filename}"
qget = params['clientquery']
if !qget.nil? then
begin
systemCmd = "bash /home/abc/t.sh \"#{qget}\" \"#{filepath}\""
puts systemCmd
output = system("#{systemCmd} 2>&1")
data = File.read(filepath)
send_data data, filename: File.basename(filepath),
type: 'application/csv',
disposition: 'attachment'
ensure
# delfile = File.basename("/tmp/download.csv")
FileUtils.remove_entry_secure File.basename("/tmp/download.csv")
# File.delete(delfile)
# redirect_to '/report'
end
FileUtils.remove_entry_secure File.basename("/tmp/download.csv") using this code i try to remove file after downloading but it not working
if i comment the line FileUtils.remove_entry_secure File.basename("/tmp/download.csv")
The file downloaded but i want remove that file after download the file
I think permission problem.could you please verify permission for /tmp folder.
because FileUtils.remove_entry_secure method will check all permission,user and group and it will remove.
Please refer click here