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#1 toyotaki   User is offline

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shell script: while loop / equal check statement

Posted 25 March 2011 - 10:28 AM


I am having this code where it seems that there is a mistake at the while loop. Can someone help me find out what's going on?

while [ $accept -eq 0 ]
        echo "Give a name:"
        read name
        accept=`java CheckName $name`

What I am trying to do is to have a variable "accept" which is initialized with the value 0 (number). While the value of accept is 0 I want to go into the loop and give the user the order "Give a name:" and let the user give a name. After reading the name I call a .java program that checks the name and prints on screen 0 or 1. I want this printed message to be stored in variable "accept". I want the while loop to me executed until the .java program returns 1 which means that the name is accepted.

The shell script seems to have a mistake on to line with the while check statement
while [ $accept -eq 0 ]

The mistake shown on screen is:" [: too many arguments"

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#2 brds   User is offline

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Re: shell script: while loop / equal check statement

Posted 25 March 2011 - 12:34 PM

If you checked the man page for your shell it would provide you with the correct syntax for while. It looks like your using sh/bash/ksh or some flavour thereof so Id have to say...

You are missing a ; on your while statement.
while [ $accept -eq 0 ];

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