ISTQB Day 4
Static Testing:
Verification : Checking Documents i.e BRS, SRS, Test Cases, Test Data, Test Scenario
Test Plan
R BRS & SRS
D HLD & LLD
C Source Code :
T Test Cases, Test Data, Test Scenario, Test Plan, Defect Report
Dep Mix Documents
Mantainance
Static Analysis: Checking Source Code with help of Tool/Compiler
i.e Multiple markers at this line
1. Syntax error, insert “;” to complete BlockStatements
2. g cannot be resolved to a variable
3. Unreachable catch block for NullPointerException. It is already handled by the catch block for Exception
Whole Team Approach:
5-7 Team Size
DOR(Definition of Ready): Entry Criteria : When to Start
Entry Criteria: 1.TC Ready & Test Data
2.Environment
3.Build Should be Available
DOD(Definition of Done) : Exit Criteria : When to Stop
When to Stop:
1.90 % Quality should be there.
2.No High Priority Defect
3.Time & Budget is Over
STLC:
Test Planning
Monitoring & Controlling
Test Design
Test Implementation
Test Execution (Actual Testing starts here since Build/AUT is available here)
Test Clouser
R BRS : 1 Req
SRS : 2-3 Req in SRS
Design : 5-6 Design
Source : 7-8 Source
Testing: No Defect 10- 15 Test Cases Wrong
Dep :
Maintenance
Defect found Later in Phases, Cost more to Fix
SDLC : Developer Follows
STLC : Tester Follows
Review Process Activities :
Life Cycle for Review
https://drive.google.com/drive/folders/14Uhv9n2qBcBIuY7QO3m7EBw-maa3r9oc
Test Design Tech:
Technique : A Proven & efficient way of doing something
Black Test Design Tech:
White Box Test Design Tech :
ECP: Equivalence Class Partitioning:
Driving Lic.
Test Data according to ECP.
1.Group inputs according to their Behavior so that they can give same output.
2.Take 1 value from Each Partition
3. i.e 15, 25, 65 or 17, 20, 61
18-60
1 17|18 60|61
invalid Valid Invalid
(Number of Partition Tested/Number of Partitions available)*100
3/3*100 =100 %
if(age>=17 & age<=61)
allow
else
not allow
End if
if(age>18 & age<60)
allow
else
not allow
End if
Dis-Advantage of ECP : It ignores boundary Values
Test Data according to BVA.
1.Group inputs according to their Behavior so that they can give same output.
2. Take Valid Partition & Do +1 & -1 to Edges/Boundary of Valid Partition
3. 3 Value BVA : 17, 18, 19, 59, 60, 61
2 Value BVA : 17, 18, 60, 61
Valid Boundary: 18 & 60
Invalid Boundary: 17 & 61
1 17|18 60|61
invalid Valid Invalid
17 18 19 59, 60, 61
if(x=10)
allow
End if
2 Value BVA
9, 10
3 Value BVA
9, 10, 11
1 9 |10 100
invalid Valid



