pastebin - splunk_showsource_to_xml.py

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Paste Description for splunk_showsource_to_xml.py

This is a simple utility to convert Splunk's simple dashboard/form XML format used in Splunk 4.0 and help you convert it into the advanced XML view format using the provided "showsource" feature.

splunk_showsource_to_xml.py
Wednesday, November 18th, 2009 at 9:12:48am MST

#!/usr/bin/env python
"""
A simple utility to convert Splunk's simple dashboard/form XML format used in
Splunk 4.0 and help you convert it into the advanced XML view format using the
provided "showsource" feature. In the future, an automated tool will most likely
be available. However, in the mean time it is quite tedious to convert all of
that by hand, so I wrote this to quickly convert the python dictionaries dumped
on the web page into the advance XML form structure one module at a time.
 
 
Basic instructions:
 
 * Point your browser to your view and add "?showsource=true" to the end of the
   URL, for example: http://localhost:8000/app/myapp/myview?showsource=true
 * View the source (or download the HTML). This is because of embedded "<" and ">"
   characters.
 * Run a single expression (Found between the <pre></pre> tags) or the entire
   file though this script. (Standard in / standard out. I found this works very
   nicely in Komodo Edit)
 
Notes/Limitations:
 
 * This utility only converts a single dictionary ("module") into the
   corresponding XML syntax at a time. This script has no knowledge of any
   nested structuring of this modules: you will have to do that yourself. (Or
   update this script to figure the nesting out for you. This should be possible
   using the <div> tags in the HTML.  But I didn't want to put the effort into it.
 * Minimal XML escaping is handled here. If "<>" are found then a CDATA element
   is used. And "&" is escaped in regular strings.  This was good enough for my
   needs, but may need to be improved upon.
 
This was written/tested with Splunk 4.0.6.
 
*** Feel free to use this script however you would like. It is provided as-is.
There is no warranty of any kind. ***
 
"""
 
__author__ = "Lowell Alleman"
__version__ = "0.2"
 
import sys
import re
 
INDENT = "  "
 
def write_dict(d, stream, indent=0):
    " Write out a nested dictionary structure in the XML <param> format."
    prefix = INDENT * indent
    for key, value in d.items():
        if value is None:
            continue
        stream.write("%s<param name=\"%s\">" % (prefix,key))
        multiline = False
        if type(value) == dict:
            stream.write("\n")
            write_dict(value, stream, indent+1)
            multiline = True
        elif type(value) in (list, tuple):
            write_list(value, stream, indent+1)
            multiline = True
        elif type(value) == str and ("<" in value or ">" in value):
            stream.write("<![CDATA[%s]]>" % value)
        else:
            stream.write(str(value).replace("&", "&amp;"))
        if multiline:
            stream.write(prefix)
        stream.write("</param>\n")
 
def write_list(lst, stream, indent=0):
    prefix = INDENT * indent
    for item in lst:
        stream.write(prefix + "<list>\n")
        if type(item) == dict:
            write_dict(item, stream, indent+1)
        else:
            raise TypeError("Always expecting a list of dictionariy. "
                            "Found type: %s" % type(item))
        stream.write(prefix + "</list>\n")
 
def convert_to_xml(d, stream):
    " Write out a <module>/<param> XML structure for the given dictionary. "
    d = dict(d)
    className = d["className"]
    if "params" in d:
        params = d["params"]
        del d["params"]
    else:
        params = {}
    del d["className"]
    
    stream.write('<module name="%s" ' % className)
    for key,value in d.items():
        stream.write('%s="%s" ' % (key, value))
    stream.write(">\n")
    write_dict(params, stream)
    stream.write("\n")
    stream.write("</module>\n")
    stream.flush()
 
 
def convert_py_to_xml(py_dict_string, xml_out):
    """
    Take a single python dictionary string expression and convert it to XML. Any
    errors are simply written to standard error and the process continues.
    """
    try:
        # Convert python dictionary string into a real python dictionary
        d = eval(py_dict_string)
        convert_to_xml(d, xml_out)
    except Exception, e:
        sys.stderr.write("Failed to convert to Splunk Advanced XML format"
                         "(%s)\nInput expression:\n%s\n\n" % (e, py_dict_string))
 
 
if __name__ == '__main__':
    # Read entire input stream at once
    chunk = sys.stdin.read()
    
    # Lazy approach to handle an entire HTML file based on the fact that the
    # each module has it's own dictionary within a "<pre>" block.
    if "<pre>" in chunk:
        chunks = re.findall(r"<pre>(.*?)</pre>", chunk, re.IGNORECASE | re.MULTILINE | re.DOTALL)
    else:
        chunks = [ chunk ]
    
    for chunk in chunks:
        convert_py_to_xml(chunk, sys.stdout)

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Tags: python xml views splunk showsource

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#!/usr/bin/env python
"""
A simple utility to convert Splunk's simple dashboard/form XML format used in
Splunk 4.0 and help you convert it into the advanced XML view format using the
provided "showsource" feature. In the future, an automated tool will most likely
be available. However, in the mean time it is quite tedious to convert all of
that by hand, so I wrote this to quickly convert the python dictionaries dumped
on the web page into the advance XML form structure one module at a time.

Basic instructions:

* Point your browser to your view and add "?showsource=true" to the end of the
   URL, for example: http://localhost:8000/app/myapp/myview?showsource=true
 * View the source (or download the HTML). This is because of embedded "<" and ">"
   characters.
 * Run a single expression (Found between the <pre></pre> tags) or the entire
   file though this script. (Standard in / standard out. I found this works very
   nicely in Komodo Edit)

Notes/Limitations:

* This utility only converts a single dictionary ("module") into the
   corresponding XML syntax at a time. This script has no knowledge of any
   nested structuring of this modules: you will have to do that yourself. (Or
   update this script to figure the nesting out for you. This should be possible
   using the <div> tags in the HTML.  But I didn't want to put the effort into it.
 * Minimal XML escaping is handled here. If "<>" are found then a CDATA element
   is used. And "&" is escaped in regular strings.  This was good enough for my
   needs, but may need to be improved upon.

This was written/tested with Splunk 4.0.6.

*** Feel free to use this script however you would like. It is provided as-is.
There is no warranty of any kind. ***

"""

__author__ = "Lowell Alleman"
__version__ = "0.2"

import sys
import re

INDENT = "  "

def write_dict(d, stream, indent=0):
    " Write out a nested dictionary structure in the XML <param> format."
    prefix = INDENT * indent
    for key, value in d.items():
        if value is None:
            continue
        stream.write("%s<param name=\"%s\">" % (prefix,key))
        multiline = False
        if type(value) == dict:
            stream.write("\n")
            write_dict(value, stream, indent+1)
            multiline = True
        elif type(value) in (list, tuple):
            write_list(value, stream, indent+1)
            multiline = True
        elif type(value) == str and ("<" in value or ">" in value):
            stream.write("<![CDATA[%s]]>" % value)
        else:
            stream.write(str(value).replace("&", "&amp;"))
        if multiline:
            stream.write(prefix)
        stream.write("</param>\n")

def write_list(lst, stream, indent=0):
    prefix = INDENT * indent
    for item in lst:
        stream.write(prefix + "<list>\n")
        if type(item) == dict:
            write_dict(item, stream, indent+1)
        else:
            raise TypeError("Always expecting a list of dictionariy. "
                            "Found type: %s" % type(item))
        stream.write(prefix + "</list>\n")

def convert_to_xml(d, stream):
    " Write out a <module>/<param> XML structure for the given dictionary. "
    d = dict(d)
    className = d["className"]
    if "params" in d:
        params = d["params"]
        del d["params"]
    else:
        params = {}
    del d["className"]
    
    stream.write('<module name="%s" ' % className)
    for key,value in d.items():
        stream.write('%s="%s" ' % (key, value))
    stream.write(">\n")
    write_dict(params, stream)
    stream.write("\n")
    stream.write("</module>\n")
    stream.flush()

def convert_py_to_xml(py_dict_string, xml_out):
    """
    Take a single python dictionary string expression and convert it to XML. Any
    errors are simply written to standard error and the process continues.
    """
    try:
        # Convert python dictionary string into a real python dictionary
        d = eval(py_dict_string)
        convert_to_xml(d, xml_out)
    except Exception, e:
        sys.stderr.write("Failed to convert to Splunk Advanced XML format"
                         "(%s)\nInput expression:\n%s\n\n" % (e, py_dict_string))

if __name__ == '__main__':
    # Read entire input stream at once
    chunk = sys.stdin.read()
    
    # Lazy approach to handle an entire HTML file based on the fact that the
    # each module has it's own dictionary within a "<pre>" block.
    if "<pre>" in chunk:
        chunks = re.findall(r"<pre>(.*?)</pre>", chunk, re.IGNORECASE | re.MULTILINE | re.DOTALL)
    else:
        chunks = [ chunk ]
    
    for chunk in chunks:
        convert_py_to_xml(chunk, sys.stdout)

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pastebin - splunk_showsource_to_xml.py - post number 1676548

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splunk_showsource_to_xml.py
Wednesday, November 18th, 2009 at 9:12:48am MST

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Paste Description for splunk_showsource_to_xml.py

splunk_showsource_to_xml.pyWednesday, November 18th, 2009 at 9:12:48am MST

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splunk_showsource_to_xml.py
Wednesday, November 18th, 2009 at 9:12:48am MST