CODEWITHSUNDEEP
bashif-statement
Maria Ines Parnisari
Maria Ines Parnisari

Reputation: 17496

Bash "if" statement throws "too many arguments"

This is the code that produces the error: 

#!/bin/bash
types=([i]=Info [w]=Warning [e]=Error [se]=Severe)

function isValidType
{
    for type in "${!types[@]}"
    do
        if [ $1 == $type ]; then
            return 0
        fi
    done
    return 1
}

if [ isValidType "$msgType" -eq 1 ]; then # <---- error here
    echo "Invalid type."
    exit 0
fi

Upvotes: 1

Views: 3932

Answers (3)

rici
rici

Reputation: 241721

The syntax of an if statement is:

if <list>; then <list>; [elif <list>; then <list>;]* [else <list>;]? fi

where <list> is any sequence of "pipelines" separated by ;, &&, or ||. (A pipeline is one or more simple commands separated by | symbols.)

The if statement is evaluated by first executing the <list> following the if, and checking the return code, which will be the return code of the last simple command executed. Based on that, a decision is made as to whether to execute the <list> following then (if the first list succeeeded) or to proceed with the elif tests and/or else clause.

Nowhere in that syntax does a [ appear, and for a good reason. [ is actually a command. In fact, it is almost a synonym for test; the difference is that [ insists that its last argument be ].

It's sometimes convenient to use [ (although it is almost always more convenient to use [[, but that's an essay for another day), but it is in no way obligatory. If you just want to test whether a command succeeded or not, then do so:

if isValidType "$msgType"; then
  # it's valid
else
  # it's not valid
fi

If you only need to do something if it didn't work, use the ! special form:

if ! isValidType "$msgType"; then
  # it's not valid
fi

Upvotes: 3

Saulius
Saulius

Reputation: 21

To check if a function returns true or not the proper way to do this is:

if [ !isValidType ]; then
       // would throw a flag for any options passed that are invalid
       // error output
fi

This works if isValidType will be either one or zero. I think the problem is in the you are checking the array.

Hope this helps.

Upvotes: -1

P.P
P.P

Reputation: 121387

Change this

if [ isValidType "$msgType" -eq 1 ]; then

to

isValidType "$msgType"
if [ $? -eq 1 ]; then

[ is the test command which accepts an expression and doesn't work as you designed (Comparing the return value of the function).

Upvotes: 2

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