Question

In shell script what is the reason this statement is not working?

x=7

y=13

while [ $x -le $y ]

do

echo $x

x=x+2

done

why on while statement it has error?

2.x=7

y=13

while [$x \< $y ]

do

echo $x

let x=x+2

done

what does it mean \< ? and give me the reason how this statement works

Answer 1

Question 1:

Issue Description:

if you run the given script, you ll get an error "Integer expression expected"

Answer: It is because of bad arithmetic expansion. you have to use $ and the expression should be inside double brackets $((expression))

This program assigns x  with 7 and y with 13 initially. x is incremented by two for each iteration and prints x value until x exceeds y i.e, 13.

The script should be as follows:

x=7

y=13

while [ $x -le $y ]

do

echo $x

x=$((x+2))

done

Sample Output:

Question 2: what does it mean \< ? and give me the reason how this statement works

Answer: There is no such operator \< in shell scripting. If you run the script with '\<', you ll get an error. See the output below,

you have to use the below operators as appropriate.

• -le =>  Checks if the value of left operand is less than or equal to the value of right operand; if yes, then the condition becomes true.
• -ge  => Checks if the value of left operand is greater than or equal to the value of right operand; if yes, then the condition becomes true.
• -lt => Checks if the value of left operand is less than the value of right operand; if yes, then the condition becomes true.
• -gt => Checks if the value of left operand is greater than the value of right operand; if yes, then the condition becomes true.
• -ne => Checks if the value of two operands are equal or not; if values are not equal, then the condition becomes true.
• -eq => Checks if the value of two operands are equal or not; if yes, then the condition becomes true.

I have replaced the operator \< by -le, see the script and output below.

x=7

y=13

while [ $x -le $y ]

do

echo $x

let x=x+2

done

Sample output: