create array in loop with variable in array's name
Hi all, I'm just a beginer and I got stucked I'm trying to create multiple array with variable in array's name. I would expect to have 3 arrays and each of them would contain 3 elements like array1=(a b c) array2=(a b c ) array3=(a b c) ; what means that first element of array1[0]=a and for example third element of array3[2]=c. This is the way how I do it : for (( n=1; n4; n++ )) do array$n=(a b c) done ...but there is syntax error and I can not fix it. I know I can declare these 3 arrays as : array1=(a b c) array2=(a b c) but this is just a simple example; in fact I want to read from several files to array$n and I can not go through ... can I declare array's name with some variable in it's name pls help; thanks in advance for any hint -- View this message in context: http://old.nabble.com/create-array-in-loop-with-variable-in-array%27s-name-tp28712785p28712785.html Sent from the Gnu - Bash mailing list archive at Nabble.com.
Re: create array in loop with variable in array's name
Am 29.05.2010 01:04, schrieb pikta: Hi all, I'm just a beginer and I got stucked I'm trying to create multiple array with variable in array's name. I would expect to have 3 arrays and each of them would contain 3 elements like array1=(a b c) array2=(a b c ) array3=(a b c) ; what means that first element of array1[0]=a and for example third element of array3[2]=c. This is the way how I do it : for (( n=1; n4; n++ )) do array$n=(a b c) done ...but there is syntax error That's because array$n isn't a valid name. Use eval: eval array$n=(a b c) Bernd -- Bernd Eggink http://sudrala.de
