i have oid name and when i pass that oid name to an api it should give data type of that oid_name using pysnmp.
suppose i pass case1 : ssCpuRawSystem it should give me Integer32 . case2 : ssCpuRawUser it should give me Counter32 case3 : ssErrorName it should give me DisplayString
Depending on your preferred granularity, you can use PySNMP's MIB API,
from pysnmp.proto.rfc1902 import ObjectIdentifier
from pysnmp.smi import builder, view, compiler
# Create MIB builder
mibBuilder = builder.MibBuilder()
# Optionally compile MIBs
compiler.addMibCompiler(mibBuilder, sources=["/usr/share/snmp/mibs"])
mibBuilder.loadTexts = True
# Load MIB modules
mibBuilder.loadModules("SNMPv2-MIB")
# mibBuilder.addMibSources(builder.DirMibSource('/Users/lextm/pysnmp.com/pysnmp/mibs'))
# mibBuilder.loadModule('LEXTUDIO-MIB')
# Create MIB view controller
mibViewController = view.MibViewController(mibBuilder)
# Create an OID object
oid = ObjectIdentifier("1.3.6.1.2.1.1.3.0")
# Get the MIB name and symbol name for the OID
modName, symName, suffix = mibViewController.getNodeLocation(oid)
# Get the MIB node for the OID
(mibNode,) = mibBuilder.importSymbols(modName, symName)
# Get the description of the MIB node
description = mibNode.getDescription()
# Print the results
print("OID: %s" % oid)
print("MIB name: %s" % modName)
print("Symbol name: %s" % symName)
print("Description: %s" % description)
print("Syntax: %s" % mibNode.getSyntax().__class__.__name__) # <- The syntax object contains everything you want to know
Taken from official example