Wireshark packet capture basics in Cybersecurity - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start capturing packets on the default network interface in Wireshark.

Cybersecurity

wireshark -i [1]

Drag options to blanks, or click blank then click option'

Anone

Beth0

Call

Ddefault

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'default', which is not a valid interface name.

Using 'all', which is not a valid interface name in this context.

The -i eth0 option tells Wireshark to capture packets on the eth0 network interface, which is commonly used as the default.

2fill in blank

medium

Complete the command to capture only TCP packets using a capture filter in Wireshark.

Cybersecurity

wireshark -f "[1]"

Drag options to blanks, or click blank then click option'

Atcp

Budp

Cicmp

Darp

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'udp' which captures UDP packets instead of TCP.

Using 'icmp' which captures ping packets.

The capture filter tcp limits the capture to TCP packets only.

3fill in blank

hard

Fix the error in the command to save captured packets to a file named capture.pcap.

Cybersecurity

wireshark -w [1]

Drag options to blanks, or click blank then click option'

Acapture.log

Bcapture.txt

Ccapture.csv

Dcapture.pcap

Attempts:

3 left

💡 Hint

Common Mistakes

Using .txt or .csv extensions which are not valid for packet capture files.

Using .log which is a generic log file format.

The -w option saves the capture to a file, which should have the .pcap extension for packet capture files.

4fill in blank

hard

Fill both blanks to create a display filter that shows only HTTP traffic from source IP 192.168.1.10.

Cybersecurity

http && ip.src == [1] && [2]

Drag options to blanks, or click blank then click option'

A192.168.1.10

Btcp.port == 80

Cudp.port == 53

Dip.dst == 192.168.1.10

Attempts:

3 left

💡 Hint

Common Mistakes

Using destination IP instead of source IP.

Using UDP port 53 which is for DNS, not HTTP.

The filter ip.src == 192.168.1.10 selects packets from that source IP, and tcp.port == 80 filters HTTP traffic which uses TCP port 80.

5fill in blank

hard

Fill all three blanks to create a capture filter that captures only ICMP packets from network 10.0.0.0/24 excluding broadcast address.

Cybersecurity

icmp and src net [1] and not dst host [2] and not dst host [3]

Drag options to blanks, or click blank then click option'

A10.0.0.0/24

B10.0.0.255

C255.255.255.255

D10.0.0.1

Attempts:

3 left

💡 Hint

Common Mistakes

Not excluding broadcast addresses causing extra packets to be captured.

Using wrong subnet mask or IP addresses.

The filter captures ICMP packets from the 10.0.0.0/24 network, excluding the broadcast addresses 10.0.0.255 and 255.255.255.255.

Practice

1. What is the primary purpose of Wireshark in cybersecurity?

easy

A. To capture and analyze network packets in real time

B. To encrypt network traffic for security

C. To block unauthorized network access

D. To create virtual private networks (VPNs)

Solution

1. Step 1: Understand Wireshark's function

   Wireshark is a tool designed to capture and display network packets as they travel through a network.

2. Step 2: Identify the correct purpose

   Among the options, only capturing and analyzing packets matches Wireshark's main use.

3. Final Answer:

   To capture and analyze network packets in real time -> Option A

4. Quick Check:

   Wireshark captures packets = To capture and analyze network packets in real time [OK]

Hint: Wireshark shows network data live, not encrypt or block [OK]

Common Mistakes:

• Confusing Wireshark with firewall or VPN tools
• Thinking Wireshark encrypts data
• Assuming Wireshark blocks traffic

2. Which of the following is the correct way to start a packet capture in Wireshark?

easy

A. Click on 'File' then 'Open Capture'

B. Click on 'Capture' then 'Start'

C. Click on 'Analyze' then 'Filter'

D. Click on 'Edit' then 'Preferences'

Solution

1. Step 1: Identify the menu for starting capture

   In Wireshark, the 'Capture' menu contains options to start or stop capturing packets.

2. Step 2: Match the correct action

   Clicking 'Capture' then 'Start' begins the live packet capture process.

3. Final Answer:

   Click on 'Capture' then 'Start' -> Option B

4. Quick Check:

   Start capture via Capture menu = Click on 'Capture' then 'Start' [OK]

Hint: Start capture under 'Capture' menu, not 'File' or 'Edit' [OK]

Common Mistakes:

• Choosing 'File' to start capture instead of 'Capture'
• Confusing 'Analyze' with starting capture
• Looking in 'Edit' menu for capture options

3. Consider the following Wireshark filter: ip.src == 192.168.1.10. What does this filter do?

medium

A. Shows packets where the destination IP is 192.168.1.10

B. Shows packets with any IP address except 192.168.1.10

C. Shows packets where the source IP is 192.168.1.10

D. Shows packets where either source or destination IP is 192.168.1.10

Solution

1. Step 1: Understand the filter syntax

   The filter ip.src == 192.168.1.10 means packets where the source IP address equals 192.168.1.10.

2. Step 2: Match filter meaning to options

   Only Shows packets where the source IP is 192.168.1.10 correctly describes packets with source IP 192.168.1.10.

3. Final Answer:

   Shows packets where the source IP is 192.168.1.10 -> Option C

4. Quick Check:

   ip.src filter = source IP = Shows packets where the source IP is 192.168.1.10 [OK]

Hint: ip.src means source IP, ip.dst means destination IP [OK]

Common Mistakes:

• Confusing source IP with destination IP
• Assuming filter matches both source and destination
• Thinking filter excludes the IP address

4. You tried to filter packets with tcp.port == 80 but no packets appear. What could be a likely reason?

medium

A. Port 80 is not used for TCP traffic

B. The filter syntax is incorrect

C. Wireshark does not support filtering by port

D. You captured packets on the wrong network interface

Solution

1. Step 1: Check filter syntax correctness

   The filter tcp.port == 80 is valid syntax to filter TCP packets on port 80.

2. Step 2: Consider capture context

   If no packets appear, a common cause is capturing on the wrong network interface where no HTTP traffic (port 80) passes.

3. Final Answer:

   You captured packets on the wrong network interface -> Option D

4. Quick Check:

   Wrong interface capture = no matching packets = You captured packets on the wrong network interface [OK]

Hint: No packets? Check if capturing on correct network interface [OK]

Common Mistakes:

• Assuming filter syntax is wrong without checking
• Believing Wireshark can't filter by port
• Thinking port 80 is not TCP by default

5. You want to capture only HTTP traffic from a specific device with IP 10.0.0.5 using Wireshark. Which filter should you apply?

hard

A. ip.addr == 10.0.0.5 and tcp.port == 80

B. ip.src == 10.0.0.5 or ip.dst == 10.0.0.5 and tcp.port == 80

C. ip.addr == 10.0.0.5 or tcp.port == 80

D. ip.src == 10.0.0.5 and tcp.port == 80

Solution

1. Step 1: Define the filter requirements

   You want packets where the device IP is either source or destination and the traffic is HTTP (TCP port 80).

2. Step 2: Analyze each filter option

   ip.addr == 10.0.0.5 and tcp.port == 80 uses ip.addr == 10.0.0.5 which matches source or destination IP, combined with tcp.port == 80 to filter HTTP traffic. This matches the requirement exactly.

3. Step 3: Identify issues in other options

   ip.src == 10.0.0.5 or ip.dst == 10.0.0.5 and tcp.port == 80 lacks parentheses, so 'or' and 'and' precedence causes incorrect filtering. ip.addr == 10.0.0.5 or tcp.port == 80 matches any packet with IP 10.0.0.5 or any TCP port 80 packet, which is too broad. ip.src == 10.0.0.5 and tcp.port == 80 only matches packets where 10.0.0.5 is source, missing destination packets.

4. Final Answer:

   ip.addr == 10.0.0.5 and tcp.port == 80 -> Option A

5. Quick Check:

   ip.addr covers both ends + tcp.port 80 = ip.addr == 10.0.0.5 and tcp.port == 80 [OK]

Hint: Use ip.addr for both source/destination IPs in filters [OK]

Common Mistakes:

• Not using parentheses causing wrong logic in filters
• Using only ip.src or ip.dst missing half the traffic
• Using 'or' instead of 'and' causing too many packets