Windows OID 'for CPU, memory, disk Utilization2007-10-31 11: 03 windows CPU utilization:
Oid:. 1.3.6.1.2.1.25.3.3.1.2 (.iso.org. DOD. Internet. Mgmt. mib-2.host.hrDevice.hrProcessorTable.hrProcessorEntry.hrProcessorLoad) in HOST-RESOURCES-MIB
1.3.6.1.2.1.25.3.3.1.2.2 and 1.3.6.1.2.1.25.3.3.1.2.3
Windows Memory utilization:
Total: oid: 1.3.6.1.2.1.25.2.2.0 unit (Kb) -- Windows 2003
1.3.6.1.2.1.25.2.2.0 = 2096304 [integer]
Memory usage processing is consistent with HDD total, HDD usage
Windows disk utilization:
Total OID:
Usage OID:
1. Disk Type OID (hrstrogetype ):
Hrstoragetypes
1.3.6.1.2.1.25.2.1
Object Identifier
Hrstorageother
1.3.6.1.2.1.25.2.1.1
Object Identifier
Hrstorageram
1.3.6.1.2.1.25.2.1.2
Object Identifier
Hrstoragevirtualmemory
1.3.6.1.2.1.25.2.1.3
Object Identifier
Hrstoragefixeddisk
1.3.6.1.2.1.25.2.1.4
Object Identifier
Hrstorageremovabledisk
1.3.6.1.2.1.25.2.1.5
Object Identifier
Hrstoragefloppydisk
1.3.6.1.2.1.25.2.1.6
Object Identifier
Hrstoragecompactdisc
1.3.6.1.2.1.25.2.1.7
Object Identifier
Hrstorageramdisk
1.3.6.1.2.1.25.2.1.8
Object Identifier
Hrstorageram indicates memory, hrstoragefixeddisk indicates hard disk, hrstoragevirtualmemory indicates virtual memory, and hrstoragecompactdisc indicates Disc
2. differentiate Disk/partition types
1.3.6.1.2.1.25.2.3.1.2.1 = 1.3.6.1.2.1.25.2.1.4 [objectidentifier] // Hard Disk
1.3.6.1.2.1.25.2.3.1.2.2 = 1.3.6.1.2.1.25.2.1.4 [objectidentifier] // Hard Disk
1.3.6.1.2.1.25.2.3.1.2.3 = 1.3.6.1.2.1.25.2.1.4 [objectidentifier] // Hard Disk
1.3.6.1.2.1.25.2.3.1.2.4 = 1.3.6.1.2.1.25.2.1.4 [objectidentifier] // Hard Disk
1.3.6.1.2.1.25.2.3.1.2.5 = 1.3.6.1.2.1.25.2.1.7 [objectidentifier] // CD
1.3.6.1.2.1.25.2.3.1.2.6 = 1.3.6.1.2.1.25.2.1.7 [objectidentifier] // CD
1.3.6.1.2.1.25.2.3.1.2.7 = 1.3.6.1.2.1.25.2.1.3 [objectidentifier] // virtual memory
1.3.6.1.2.1.25.2.3.1.2.8 = 1.3.6.1.2.1.25.2.1.2 [objectidentifier] // physical memory
3. Disk/partition information OID:
1.3.6.1.2.1.25.2.3.1.3.1 = "C: \ label: Serial Number dc26ea33" [ETS]
1.3.6.1.2.1.25.2.3.1.3.2 = "D: \ label: Test serial number 9629f314" [ETS]
1.3.6.1.2.1.25.2.3.1.3.3 = "E: \ label: test_backup serial number 94389268" [octets]
1.3.6.1.2.1.25.2.3.1.3.4 = "F: \ label: Others serial number 5c4002e8" [octets]
1.3.6.1.2.1.25.2.3.1.3.5 = "G: \" [octets]
1.3.6.1.2.1.25.2.3.1.3.6 = "H: \ label: Oracle10g serial number b1c04dbe" [octets]
1.3.6.1.2.1.25.2.3.1.3.7 = "Virtual Memory" [octets]
1.3.6.1.2.1.25.2.3.1.3.8 = "physical memory" [octets]
4. storageallocationunits OID: the size of each shard/block.
1.3.6.1.2.1.25.2.3.1.4.1 = 512 [integer]
1.3.6.1.2.1.25.2.3.1.4.2 = 4096 [integer]
1.3.6.1.2.1.25.2.3.1.4.3 = 4096 [integer]
1.3.6.1.2.1.25.2.3.1.4.4 = 4096 [integer]
1.3.6.1.2.1.25.2.3.1.4.5 = 0 [integer]
1.3.6.1.2.1.25.2.3.1.4.6 = 2048 [integer]
1.3.6.1.2.1.25.2.3.1.4.7 = 65536 [integer]
1.3.6.1.2.1.25.2.3.1.4.8 = 65536 [integer]
5. storagesize OID: the number of disks/shards and the total size of a disk.
1.3.6.1.2.1.25.2.3.1.5.1 = 25157789 [integer]
1.3.6.1.2.1.25.2.3.1.5.2 = 2815383 [integer]
1.3.6.1.2.1.25.2.3.1.5.3 = 4096567 [integer]
1.3.6.1.2.1.25.2.3.1.5.4 = 11006525 [integer]
1.3.6.1.2.1.25.2.3.1.5.5 = 0 [integer]
1.3.6.1.2.1.25.2.3.1.5.6 = 317298 [integer]
1.3.6.1.2.1.25.2.3.1.5.7 = 63101 [integer]
1.3.6.1.2.1.25.2.3.1.5.8 = 32754 [integer]
6. storageused OID: // used block/shard
1.3.6.1.2.1.25.2.3.1.6.1 = 7427924 [integer]
1.3.6.1.2.1.25.2.3.1.6.2 = 861039 [integer]
1.3.6.1.2.1.25.2.3.1.6.3 = 16820 [integer]
1.3.6.1.2.1.25.2.3.1.6.4 = 5788717 [integer]
1.3.6.1.2.1.25.2.3.1.6.5 = 0 [integer]
1.3.6.1.2.1.25.2.3.1.6.6 = 317298 [integer]
1.3.6.1.2.1.25.2.3.1.6.7 = 12172 [integer]
1.3.6.1.2.1.25.2.3.1.6.8 = 13270 [integer]
7. storageallocationfailures OID: // blocks that fail to be allocated/blocks
1.3.6.1.2.1.25.2.3.1.7.1 = 0 [counter32]
1.3.6.1.2.1.25.2.3.1.7.2 = 0 [counter32]
1.3.6.1.2.1.25.2.3.1.7.3 = 0 [counter32]
1.3.6.1.2.1.25.2.3.1.7.4 = 0 [counter32]
1.3.6.1.2.1.25.2.3.1.7.5 = 0 [counter32]
1.3.6.1.2.1.25.2.3.1.7.6 = 0 [counter32]
1.3.6.1.2.1.25.2.3.1.7.7 = 0 [counter32]
1.3.6.1.2.1.25.2.3.1.7.8 = 0 [counter32]
8. In this example
We can see that the drive letter is C: D: e: F:. At the same time, we can get information about virtual memory and physical memory.
C: The disk size is 512x25157789 = 12880787968 bytes or 12 GB.
C: disk space: 512x7427924 = 3803097088 bytes or 3.54 GB
C: The disk usage is 3.54/12 * 100% = 29.5%
C: disk space allocation error 512 × 0 = 0 bytes
The total disk size is 512*25157789 + 4096*2815383 + 4096*4096567 + 4096*11006525 = about 80g
Disk size: 512*7427924 + 4096*861039 + 4096*16820 + 4096*5788717 = about 29 GB
The usage of the entire hard disk is 29/80 * 100% = 36.2%
Physical memory
Total Ram 65536*32754 = 2096256 kb, 2047.125 m about 2g. The result is similar to 1.3.6.1.2.1.25.2.2.0.
Used Ram 65536*13270 = 829 m
Memory usage 829/2048 * 100% = 40.4%
From: http://hi.baidu.com/wangj998/blog/item/35bbadfc13876f84b801a051.html
