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tcl array question - key with quotes
I have the following code:
set my_list1 {"a" "b"}
set my_list2 {"#1" "#2"}
array set my_array {}
foreach li1 $my_list1 li2 $my_list2 {
set my_array($li1) $li2
}
puts $my_array("a")
On the list line I get ERROR "can't read my_array("a"): no such element in array"
Why?
I have it, because when I write
set newVar "a"
puts $my_array($newVar)
it returns the value!
This is just one of those things in Tcl. The array element is not my_array("a") -- it's my_array(a). Don't include the quotes when referencing the array. They're actually not necessary, although in that case note harmful, when you're installing the data into the array in the first place -- i.e.,
set my_list1 {a b}
would be just fine.
Tcl looks enough like a "normal" programming language that it's easy to forget how primitive its parser really is. Remember that everything is broken down into "words" by whitespace. If a double-quote character isn't preceded by whitespace, it isn't at the start of a word, and it no longer has any special significance. A reference to an array element is a single word, and after variable interpolation, it has to have exactly the right text. You can't put quote marks around the element name because simply those quote marks are not part of the correct text of that word.
Related
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Command not found error in Bash variable assignment
(5 answers)
Closed 2 years ago.
I need to setup an array and use that to loop through and list the files. I started with this setup and it works.
file = (x1 x2 x3 x4 x5)
for var in ${file[#]}; do
echo $var
done
I have a requirement to add the timestamp on these variables, I tried a few ways and none of the seem to work. Here is one way I tried to define the array:
today=$(date +"%Y%m%d")
file = (x1_"$today" x2_"$today" x3_"$today" x4_"$today" x5_"$today")
What would be the best way to define the array, where the array element has a variable?
The way you're using "$today" is correct, no problem there.
Don't put spaces around the equal sign. It needs to be file=(...) in both cases. Spaces are not allowed in assignments.
You can use Shell Check to catch these kinds of errors. Here's what it prints:
file = (x1_"$today" x2_"$today" x3_"$today" x4_"$today" x5_"$today")
^-- SC2283: Remove spaces around = to assign (or use [ ] to compare, or quote '=' if literal).
^-- SC1036: '(' is invalid here. Did you forget to escape it?
^-- SC1088: Parsing stopped here. Invalid use of parentheses?
I'm trying to create a for-loop to go through all the items from an array, and add the items to a string. The tags are given as a single string with format "tag1 tag2 tag3", and the tagging parameter can be given as many times as I want with the single command with syntax "-tag tag1 -tag -tag2 -tag tag3". I'm unable to create a for loop for the job, and I'm a little confused what is wrong with my code.
TAGS="asd fgh jkl zxc bnm" # Amount of tags varies, but there is always at least one
ARRAY=($TAGS)
TAGSTOBEADDED=""
for i in "$ARRAY[#]"
do
STRINGTOBEADDED="-tag ${ARRAY[$i]}"
$TAGSTOBEADDED=$TAGSTOBEADDED+$STRINGTOBEADDED
done
command $TAGSTOBEADDED
First, your array sintax is wrong as #oguz ismail said. To iter through array items you shold use this:
for i in "${ARRAY[#]}"; { echo $i;}
Second $TAGSTOBEADDED=$TAGSTOBEADDED+$STRINGTOBEADDED this is also fail.
Variables are set like so var="$var 123" you don't need $ in front of var name if you want to change it. Back to code. In this example you dont even need an array, just use TAGS var(without ""):
for i in $TAGS; { TAGSTOBEADDED+="-tag $i"; }
First: avoid storing lists of things in space-delimited strings (as you're currently doing with TAGS and TAGSTOBEADDED) -- there are a bunch of things that can go wrong if they have any "funny" characters (or if IFS gets changed). Use an array instead. Storing them as a string and then converting doesn't help; all of the same potential problems apply during the conversion.
I also recommend using lower- or mixed-case variable names in scripts, since there are a bunch of all-caps names with special meanings, and accidentally using one of those for something else can have weird effects. So, to define the array of tags, I'd just use this:
tags=(asd fgh jkl zxc bnm)
You also have a number of syntax errors in the script. In this line:
for i in "$ARRAY[#]"
... the shell will try to expand $ARRAY as a plain variable (not an array), and then treat "[#]" as just some unrelated characters that go after it. You need braces around the variable refence (like "${ARRAY[#]}") any time you're doing anything nontrivial with a variable reference. BTW, this idiom -- including double-quotes, braces, square-brackets and at-sign -- is what you almost always want when getting the contents of an array.
In this line:
STRINGTOBEADDED="-tag ${ARRAY[$i]}"
$i will expand to one of the array elements, not its index. That is, it'll expand to something like:
STRINGTOBEADDED="-tag ${ARRAY[asd]}"
...which doesn't make any sense. You just want
STRINGTOBEADDED="-tag $i"
...except you don't want that either, because (as I said before) storing lists of things space-delimited in a string is a bad idea. But I'll get to that because fixing it will involve the next line:
$TAGSTOBEADDED=$TAGSTOBEADDED+$STRINGTOBEADDED
There are two problems here: you don't want a dollar sign on the variable being assigned to ($varname gets the value of a variable; anytime you're setting it, don't use the $). Also the + isn't needed to add strings, you just stick them end to end. Well, you'd need to add a space in between, something like one of these:
TAGSTOBEADDED=$TAGSTOBEADDED" "$STRINGTOBEADDED
TAGSTOBEADDED="$TAGSTOBEADDED $STRINGTOBEADDED"
(Generally, you should have double-quotes around all variable references; on the right side of a plain assignment is one of the few places it's safe to leave them unquoted, but I tend to prefer to just double-quote always rather than try to remember all of the exceptions about where it's safe and where it isn't. Plus, quoting just the space looks weird.)
But you don't want to do that either, because (again) space-delimited strings are a bad way to do things. Use an array. So before the loop, create an empty array instead of an empty string:
tagstobeadded=()
...and then inside the loop, append to it with +=( ):
tagstobeadded+=(-tag "$i")
...and then at the end, use it with all the appropriate quotes, braces, etc:
command "${tagstobeadded[#]}"
So, with all of these changes, here's what I'd recommend:
tags=(asd fgh jkl zxc bnm)
tagstobeadded=()
for i in "${tags[#]}"
do
tagstobeadded+=(-tag "$i")
done
command "${tagstobeadded[#]}"
I am trying to read below CSV file content line by line in Perl.
CSV File Content:
A7777777.A777777777.XXX3604,XXX,3604,YES,9
B9694396.B216905785.YYY0018,YYY,0018,YES,13
C9694396.C216905785.ZZZ0028,ZZZ,0028,YES,16
I am able to split line content using below code and able to verify the content too:
#column_fields1 = split(',', $_);
print $column_fields1[0],"\n";
I am also trying to find the second part on the first column of CSV file (i.e., A777777777 or B216905785 or C216905785) – the first column delimited with . using the below code and I am unable to get it.
Instead, just a new line printed.
my ($v1, $v2, $v3) = split(".", $column_fields1[0]);
print $v2,"\n";
Can someone suggest me how to split the array element and get the above value?
On my functionality, I need the first column value altogether at someplace and just only the second part at someplace.
Below is my code:
use strict;
use warnings;
my $dailybillable_tab_section1_file = "./sql/demanding_01_T.csv";
open(FILE, $dailybillable_tab_section1_file) or die "Could not read from $dailybillable_tab_section1_file, program halting.";
my #column_fields1;
my #column_fields2;
while (<FILE>)
{
chomp;
#column_fields1 = split(',', $_);
print $column_fields1[0],"\n";
my ($v1, $v2, $v3) = split(".",$column_fields1[0]);
print $v2,"\n";
if($v2 ne 'A777777777')
{
…
…
…
}
else
{
…
…
…
}
}
close FILE;
split takes a regex as its first argument. You can pass it a string (as in your code), but the contents of the string will simply be interpreted as a regex at runtime.
That's not a problem for , (which has no special meaning in a regex), but it breaks with . (which matches any (non-newline) character in a regex).
Your attempt to fix the problem with split "\." fails because "\." is identical to ".": The backslash has its normal string escape meaning, but since . isn't special in strings, escaping it has no effect. You can see this by just printing the resulting string:
print "\.\n"; # outputs '.', same as print ".\n";
That . is then interpreted as a regex, causing the problems you have observed.
The normal fix is to just pass a regex to split:
split /\./, $string
Now the backslash is interpreted as part of the regex, forcing . to match itself literally.
If you really wanted to pass a string to split (I'm not sure why you'd want to do that), you could also do it like this:
split "\\.", $string
The first backslash escapes the second backslash, giving a two character string (\.), which when interpreted as a regex means the same thing as /\./.
If you look at the documentation for split(), you'll see it gives the following ways to call the function:
split /PATTERN/,EXPR,LIMIT
split /PATTERN/,EXPR
split /PATTERN/
split
In three of those examples, the first argument to the function is /PATTERN/. That is, split() expects to be given a regular expression which defines how the input string is split apart.
It's very important to realise that this argument is a regex, not a string. Unfortunately, Perl's parser doesn't insist on that. It allows you to use a first argument which looks like a string (as you have done). But no matter how it looks, it's not a string. It's a regex.
So you have confused yourself by using code like this:
split(".",$COLUMN_FIELDS1[0])
If you had made the first argument look like a regex, then you would be more likely to realise that the first argument is a regex and that, therefore, a dot needs to be escaped to prevent it being interpreted as a metacharacter.
split(/\./, $COLUMN_FIELDS1[0])
Update: It's generally accepted among Perl programmers, that variable with upper case names are constants and don't change their values. By using upper case names for standard variables, you are likely to confuse the next person who edits your code (who could well be you in six months time).
The bash shell script can split a given string by space into a 1D array.
str="a b c d e"
arr=($str)
# arr[0] is a, arr[1] is b, etc. arr is now an array, but what is the magic behind?
But, what exactly happened when we can arr=($str)? My understanding is the parenthesis here creates a subshell, but what happen after that?
In an assignment, the parentheses simply indicate that an array is being created; this is independent of the use of parentheses as a compound command.
This isn't the recommended way to split a string, though. Suppose you have the string
str="a * b"
arr=($str)
When $str is expanded, the value undergoes both word-splitting (which is what allows the array to have multiple elements) and pathname expansion. Your array will now have a as its first element, b as its last element, but one or more elements in between, depending on how many files in the current working directly * matches. A better solution is to use the read command.
read -ra arr <<< "$str"
Now the read command itself splits the value of $str without also applying pathname expansion to the result.
It seems you've confused
arr=($str) # An array is created with word-splitted str
with
(some command) # executing some command in a subshell
Note that
arr=($str) is different from arr=("$str")in that in the latter, the double quotes prevents word splitting ie the array will contain only one value -> a b c d e.
You can check the difference between the two by the below
echo "${#arr[#]}"
I want to make an array with string values that have square brackets. but every time I keep getting output unexpected.
selections=()
for i in $choices
do
selections+=("role[${filenames[$i]}]")
done
echo ${selections[#]}
If choices were 1 and 2, and the array filenames[1] and filenames[2] held the values 'A', 'B' I want the selections array to hold the strings role[A], and role[B]
instead the output I get is just roles.
I can make the code you presented produce the output you wanted, or not, depending on the values I assign to variables filenames and choices.
First, I observe that bash indexed arrays are indexed starting at 0, not 1. If you are using the values 1 and 2 as indices into array filenames, and if that is an indexed array with only two elements, then it may be that ${filenames[2]} expands to nothing. This would be the result if you initialize filenames like so:
# NOT WHAT YOU WANT:
filenames=(A B)
Instead, either assign array elements individually, or add a dummy value at index 0:
# Could work:
filenames=('' A B)
Next, I'm suspicious of choices. Since you're playing with arrays, I speculate that you may have initialized choices as an array, like so:
# NOT CONSISTENT WITH YOUR LATER USAGE:
choices=(1 2)
If you expand an array-valued variable without specifying an index, it is as if you specified index 0. With the above initialization, then, $choices would expand to just 1, not 1 2 as you intend. There are two possibilities: either initialize choices as a flat string:
# Could work:
choices='1 2'
or expand it differently:
# or expand it this way:
for i in "${choices[#]}"
. Do not overlook the quotes, by the way: that particular form will expand to one word per array element, but without the quotes the array elements would be subject to word splitting and other expansions (though that's moot for the particular values you're using in this case).
The quoting applies also, in general, to your echo command: if you do not quote the expansion then you have to analyze the code much more carefully to be confident that it will do what you intend in all cases. It will be subject not only to word splitting, but pathname expansion and a few others. In your case, there is a potential for pathname expansion to be performed, depending on the names of the files in the working directory (thanks #CharlesDuffy). It is far safer to just quote.
Anyway, here is a complete demonstration incorporating your code verbatim and producing the output you want:
#!/bin/bash
filenames=('' 'A' 'B')
choices="1 2"
selections=()
for i in $choices
do
selections+=("role[${filenames[$i]}]")
done
echo ${selections[#]}
# better:
# echo "${selections[#]}"
Output:
role[A] role[B]
Finally, as I observed in comments, there is no way that your code could output "roles", as you claim it does, given the inputs (variable values) you claim it has. If that's in fact what you see, then either it is not related to the code you presented at all, or your inputs are different than you claim.