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1

I want to execute this condition 

while [ $(cat /path_of a file/) -eq 1 ]

The condition is only correct if

$(cat /path_of a file/)

is an integer and not a string !

How can I overcome this problem ?

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2 Answers 2

up vote 2 down vote

You can do this with = operator, like [ $(cat /path_of a file/) = 1 ]. It compares string representations of both arguments so it will work in all cases.

Using -eq instead of string comparison may be preferable in some cases ([ 01 -eq 1 ] is true while [ "01" = 1 ] is not) but in most cases it's just more dangerous. If, on the other hand, you really need integer comparison, you should use @Gnouc suggestion.

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better use quotes: [ "$(cat file)" = 1 ] -- if the file is empty, you'll get a syntax error otherwise – glenn jackman Apr 6 '14 at 23:27
up vote 1 down vote

Variable in bash is untyped. If you want the condition to always get evaluated with integers, you can use delcare -i to make variable always an integer. From bash manpage:

declare [-aAfFgilrtux] [-p] [name[=value] ...]
.....
-i     The variable is treated as an integer; arithmetic evaluation
       (see ARITHMETIC EVALUATION above) is performed  when the variable
       is assigned a value.

So your example becomes:

declare -i number
number="$(cat /path_of a file/)"

while [ "$number" -eq 1 ]

A note that if you have bash version 4.x, you should use new test [[...]] instead of [...].

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[[ is available in older versions of bash as well. – chepner Apr 6 '14 at 14:35

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