Powershell giving wrong output in if else block

Question

Check the following code with simple if else blocks

cls
$a = Read-Host("Enter the marks ")

if (($a -le 100) -and ($a -ge 90)){

 Write-Host("The grade is A")

}
elseif (($a -le 89) -and ($a -ge 80)){

 Write-Host("The grade is B")

}

elseif (($a -le 79) -and ($a -ge 70)){

 Write-Host("The grade is C")

}

elseif (($a -le 69) -and ($a -ge 60)){

 Write-Host("The grade is D")

}
elseif($a -lt 60){
Write-Host("The grade is F")
}

And when I give input between 90 - 99 it returns without any output. And if I give 100 as input, I am getting "The grade is F", ie, the else block is executed, but it should not work according to the code. Can anyone explain why?

sunilvijendra · Accepted Answer · 2020-11-27 03:02:55Z

3

The reason for the issue is the data type. By default the data type is "String". The requirement here is to compare numbers hence the below solution to mark data type as 'int' is appropriate.

cls
[int]$a = Read-Host("Enter the marks ")
if (($a -le 100) -and ($a -ge 90)){
 Write-Host("The grade is A")
}
elseif (($a -le 89) -and ($a -ge 80)){
 Write-Host("The grade is B")
}
elseif (($a -le 79) -and ($a -ge 70)){
 Write-Host("The grade is C")
}
elseif (($a -le 69) -and ($a -ge 60)){
 Write-Host("The grade is D")
}
elseif($a -lt 60){
Write-Host("The grade is F")
}

I like it better

& {
    param(
        [ValidateRange(1, 100)]
        [int]$a
    )

    switch ($a){
        {$_ -in 90..100}{Write-Host("The grade is A")}
        {$_ -in 80..89}{Write-Host("The grade is B")}
        {$_ -in 70..79}{Write-Host("The grade is C")}
        {$_ -in 60..69}{Write-Host("The grade is D")}
        default {Write-Host("The grade is D")}
    }
    
} -a 95

$numberInString = Read-Host "imput number well be string"

    $numberInString -le 100
    [int]$numberInString -le 100

    $numberInString -ge 90
    [int]$numberInString -le 100

imput number well be sting: 95
False
True
True
True

If mark entered is say "99". Looking at the first comparison:

"99" -le "100"

The reason above result is "False" because "9" is greater than "1". String comparison occurs by comparing character by character until one or both strings reach end.

edited Nov 27, 2020 at 3:02

sunilvijendra

6785 silver badges4 bronze badges

answered Nov 24, 2020 at 8:58

Алексей Семенов

8565 silver badges11 bronze badges

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4 Comments

Roshin Raphel Over a year ago

Do you know why this occurs?

sunilvijendra Over a year ago

@RoshinRaphel, by default the type is String. If you add a line "$a.GetType()" after you read it, you can see the type. Same way when type is specified it is int

Roshin Raphel Over a year ago

@sunilvijendra So why is it working for all other cases outside the range 90-100

vonPryz Over a year ago

Please edit the answer and explain there why hand how comparing string "100" and int 100 provide the outcome. As of now, the answer tells how but not why - so it doesn't yet support understanding the actual issue.

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