Simple question for someone who know little C shell programming
I am posting some code lines of my program
........
set gramma = $<
else if ($gramma == 's') then
echo " ta defterolepta pou menoun ine " secs_remain
else if ($gramma == 'm') then
# min = $secs_remain % 60
# secs = $secs_remain - ($min*60)
echo " o xronos pou meni ine " min " lepta ke " secs " defterolepta "
else if ($gramma == 'h') then
.........
My question is about the character. As you can see I am reading a character from the keyboard.
On the if command, does my character need the ' and '?
Or can i simply write the character?
For example
if ($gramma == 's') then
or
if ($gramma == s) then
I know that if it was a string it will be
if ($gramma == "sexy") then
and if a number
if ($gramma == 4) then
What about a single character?
I think you are asking should the right hand side of a comparison always be wrapped in single quote chars? == 'm' (right hand side)
AND OF course, the answer is it depends
Single quotes mean that any variable values inside will NOT be expanded to the value that was assigned. If you you have variable values, quote them with double-quote chars.
You can leave out single OR double quotes if you're really,really,really sure you're never going to have a value on the right-hand-side that evaluates to a multi-'word' value with white space chars. You really need quoting (single or double) to avoid problems with white space chars.
I think the general consensus among people that write books about shell coding is that you always want to surround the left-and-right-hand sides of a string comparison operations (==, !=, ~=) with appropriate quotes, double if variable expansion required, single if not.
Related
In the string
"#{x ? (x.to_s + ' is ') : ''}ok", Rubocop's Style/StringConcatenation suggests avoiding the +.
But that requires a nested string interpolation
"#{x ? '#{x.to_s} is ' : ''}ok)",
which at least in Ruby 2.7 is not expanded: #{x.to_s} is treated like any other literal.
Is the + version alright because it's on the fringes of what a style guide could cover, or must one introduce a temporary variable?
tmp = x ? '#{x.to_s} is ' : ''
"#{tmp}ok"
Context: the string is sent to a logfile. ok is actually a long list of details. x is worth logging, but only when it exists.
Yes, a variable will make this more readable (imo):
prefix = "#{x} is " if x
"#{prefix}ok"
(this relies on the fact that nil#to_s == '')
Given that "ok" is actually:(according to the comments)
"...a long string that has even more interpolations. Duplicating that string isn't DRY".
I would go with
ok = generate_some_long_string()
ok.prepend("#{x} is ") if x
ok
This does mutate ok but based on my understanding of the question this may actually be desirable.
Nesting Interpolation
As an aside and I would not recommend it (because it is difficult to read) but nesting interpolation is completely valid ruby e.g.
x="val"
"#{x ? "#{x} is " : ""}ok"
#=> "val is ok"
This works because what is inside the interpolation closure is treated like any other ruby code. The inner double quotes open and close a new String rather than closing the first and opening another because the interpolation closure is waiting for a closing curly brace. You could technically do this at any depth.
"#{x ? "#{"the #{y = x}ue of"} #{x} is " : ""}#{y.inspect}"
#=> "the value of val is \"val\""
I have a script in bash that calls a TCL script for each element on my network which performs some actions based on the type of the element. This is part of the code that checks whether or not the hostname contains a specific pattern(e.g. *CGN01) and then gives the appropriate command to that machine.
if {[string match "{*CGN01}" $hostname] || $hostname == "AthMet1BG01"} {
expect {
"*#" {send "admin show inventory\r"; send "exit\r"; exp_continue}
eof
}
}
With the code i quoted above i get no error BUT when the hostname is "PhiMSC1CGN01" then the code inside the if is not executed which means that the expression is not correct.
I have tried everything (use of "()" or "{}" or"[]" inside the if) but when i dont put "" on the pattern i get an error like:
invalid bareword "string"
in expression "(string match {*DR0* *1TS0* *...";
should be "$string" or "{string}" or "string(...)" or ...
(parsing expression "(string match {*DR0* *...")
invoked from within
"if {$hostname == "AthMar1BG03" || [string match *CGN01 $hostname]...
or this:
expected boolean value but got "[string match -nocase "*CGN01" $hostname]==0"
while executing
"if {$hostname == "AthMar1BG03" || {[string match -nocase "*CGN01" $hostname]==0}...
when i tried to use ==0 or ==1 on the expression.
My TCL-Version is 8.3 and i cant update it because the machine has no internet connecticity :(
Please help me i am trying to fix this for over a month...
If you want to match a string that is either exactly AthMet1BG01 or any string that ends with CGN01, you should use
if {[string match *CGN01 $hostname] || $hostname == "AthMet1BG01"} {
(For Tcl 8.5 or later, use eq instead of ==.)
Some comments on your attempts:
(The notes about the expression language used by if go for expr and while as well. It is fully described in the documentation for expr.)
To invoke a command inside the condition and substitute its result, it needs to be enclosed in brackets ([ ]). Parentheses (( )) can be used to set the priority of subexpressions within the condition, but don't indicate a command substitution.
Normally, inside the condition strings need to be enclosed in double quotes or braces ({ }). This is because the expression language that is used to express the condition needs to distinguish between e.g. numbers and strings, which Tcl in general doesn't. Inside a command substitution within a condition, you don't need to use quotes or braces, as long as there are no characters in the string that you need to quote.
The string {abc} contains the characters abc. The string "{abc}" contains the characters {abc}, because the double quotes make the braces normal characters (the reverse also holds). [string match "{*bar}" $str] matches the string {foobar} (with the braces as part of the text), but not foobar.
If you put braces around a command substitution, {[incr foo]}, it becomes just the string [incr foo], i.e. the command isn't invoked and no substitution is made. If you use {[incr foo]==1} you get the string [incr foo]==1. The correct way to write this within an expression is [incr foo]==1, with optional whitespace around the ==.
All this is kind of hard to grok, but when you have it is really easy to use. Tcl is stubborn as a mule about interpreting strings, but carries heavy loads if you treat her right.
ETA an alternate matcher (see comments)
You can write your own alternate string matcher:
proc altmatch {patterns string} {
foreach pattern $patterns {
if {[string match $pattern $string]} {
return 1
}
}
return 0
}
If any of the patterns match, you get 1; if none of the patterns match, you get 0.
% altmatch {*bar f?o} foobar
1
% altmatch {*bar f?o} fao
1
% altmatch {*bar f?o} foa
0
For those who have a modern Tcl version, you can actually add it to the string ensemble so it works like other string commands. Put it in the right namespace:
proc ::tcl::string::altmatch {patterns string} {
... as before ...
and install it like this:
% set map [namespace ensemble configure string -map]
% dict set map altmatch ::tcl::string::altmatch
% namespace ensemble configure string -map $map
Documentation:
expr,
string,
Summary of Tcl language syntax
This command:
if {[string match "{*CGN01}" $hostname] || $hostname == "AthMet1BG01"} {
is syntactically valid but I really don't think that you want to use that pattern with string match. I'd guess that you really want:
if {[string match "*CGN01" $hostname] || $hostname == "AthMet1BG01"} {
The {braces} inside that pattern are not actually meaningful (string match only does a subset of the full capabilities of a glob match) so with your erroneous pattern you're actually trying to match a { at the start of $hostname, any number of characters, and then CGN01} at the end of $hostname. With the literal braces. Simply removing the braces lets PhiMSC1CGN01 match.
I would like to find a regex that will pick out all commas that fall outside quote sets.
For example:
'foo' => 'bar',
'foofoo' => 'bar,bar'
This would pick out the single comma on line 1, after 'bar',
I don't really care about single vs double quotes.
Has anyone got any thoughts? I feel like this should be possible with readaheads, but my regex fu is too weak.
This will match any string up to and including the first non-quoted ",". Is that what you are wanting?
/^([^"]|"[^"]*")*?(,)/
If you want all of them (and as a counter-example to the guy who said it wasn't possible) you could write:
/(,)(?=(?:[^"]|"[^"]*")*$)/
which will match all of them. Thus
'test, a "comma,", bob, ",sam,",here'.gsub(/(,)(?=(?:[^"]|"[^"]*")*$)/,';')
replaces all the commas not inside quotes with semicolons, and produces:
'test; a "comma,"; bob; ",sam,";here'
If you need it to work across line breaks just add the m (multiline) flag.
The below regexes would match all the comma's which are present outside the double quotes,
,(?=(?:[^"]*"[^"]*")*[^"]*$)
DEMO
OR(PCRE only)
"[^"]*"(*SKIP)(*F)|,
"[^"]*" matches all the double quoted block. That is, in this buz,"bar,foo" input, this regex would match "bar,foo" only. Now the following (*SKIP)(*F) makes the match to fail. Then it moves on to the pattern which was next to | symbol and tries to match characters from the remaining string. That is, in our output , next to pattern | will match only the comma which was just after to buz . Note that this won't match the comma which was present inside double quotes, because we already make the double quoted part to skip.
DEMO
The below regex would match all the comma's which are present inside the double quotes,
,(?!(?:[^"]*"[^"]*")*[^"]*$)
DEMO
While it's possible to hack it with a regex (and I enjoy abusing regexes as much as the next guy), you'll get in trouble sooner or later trying to handle substrings without a more advanced parser. Possible ways to get in trouble include mixed quotes, and escaped quotes.
This function will split a string on commas, but not those commas that are within a single- or double-quoted string. It can be easily extended with additional characters to use as quotes (though character pairs like « » would need a few more lines of code) and will even tell you if you forgot to close a quote in your data:
function splitNotStrings(str){
var parse=[], inString=false, escape=0, end=0
for(var i=0, c; c=str[i]; i++){ // looping over the characters in str
if(c==='\\'){ escape^=1; continue} // 1 when odd number of consecutive \
if(c===','){
if(!inString){
parse.push(str.slice(end, i))
end=i+1
}
}
else if(splitNotStrings.quotes.indexOf(c)>-1 && !escape){
if(c===inString) inString=false
else if(!inString) inString=c
}
escape=0
}
// now we finished parsing, strings should be closed
if(inString) throw SyntaxError('expected matching '+inString)
if(end<i) parse.push(str.slice(end, i))
return parse
}
splitNotStrings.quotes="'\"" // add other (symmetrical) quotes here
Try this regular expression:
(?:"(?:[^\\"]+|\\(?:\\\\)*[\\"])*"|'(?:[^\\']+|\\(?:\\\\)*[\\'])*')\s*=>\s*(?:"(?:[^\\"]+|\\(?:\\\\)*[\\"])*"|'(?:[^\\']+|\\(?:\\\\)*[\\'])*')\s*,
This does also allow strings like “'foo\'bar' => 'bar\\',”.
MarkusQ's answer worked great for me for about a year, until it didn't. I just got a stack overflow error on a line with about 120 commas and 3682 characters total. In Java, like this:
String[] cells = line.split("[\t,](?=(?:[^\"]|\"[^\"]*\")*$)", -1);
Here's my extremely inelegant replacement that doesn't stack overflow:
private String[] extractCellsFromLine(String line) {
List<String> cellList = new ArrayList<String>();
while (true) {
String[] firstCellAndRest;
if (line.startsWith("\"")) {
firstCellAndRest = line.split("([\t,])(?=(?:[^\"]|\"[^\"]*\")*$)", 2);
}
else {
firstCellAndRest = line.split("[\t,]", 2);
}
cellList.add(firstCellAndRest[0]);
if (firstCellAndRest.length == 1) {
break;
}
line = firstCellAndRest[1];
}
return cellList.toArray(new String[cellList.size()]);
}
#SocialCensus, The example you gave in the comment to MarkusQ, where you throw in ' alongside the ", doesn't work with the example MarkusQ gave right above that if we change sam to sam's: (test, a "comma,", bob, ",sam's,",here) has no match against (,)(?=(?:[^"']|["|'][^"']")$). In fact, the problem itself, "I don't really care about single vs double quotes", is ambiguous. You have to be clear what you mean by quoting either with " or with '. For example, is nesting allowed or not? If so, to how many levels? If only 1 nested level, what happens to a comma outside the inner nested quotation but inside the outer nesting quotation? You should also consider that single quotes happen by themselves as apostrophes (ie, like the counter-example I gave earlier with sam's). Finally, the regex you made doesn't really treat single quotes on par with double quotes since it assumes the last type of quotation mark is necessarily a double quote -- and replacing that last double quote with ['|"] also has a problem if the text doesn't come with correct quoting (or if apostrophes are used), though, I suppose we probably could assume all quotes are correctly delineated.
MarkusQ's regexp answers the question: find all commas that have an even number of double quotes after it (ie, are outside double quotes) and disregard all commas that have an odd number of double quotes after it (ie, are inside double quotes). This is generally the same solution as what you probably want, but let's look at a few anomalies. First, if someone leaves off a quotation mark at the end, then this regexp finds all the wrong commas rather than finding the desired ones or failing to match any. Of course, if a double quote is missing, all bets are off since it might not be clear if the missing one belongs at the end or instead belongs at the beginning; however, there is a case that is legitimate and where the regex could conceivably fail (this is the second "anomaly"). If you adjust the regexp to go across text lines, then you should be aware that quoting multiple consecutive paragraphs requires that you place a single double quote at the beginning of each paragraph and leave out the quote at the end of each paragraph except for at the end of the very last paragraph. This means that over the space of those paragraphs, the regex will fail in some places and succeed in others.
Examples and brief discussions of paragraph quoting and of nested quoting can be found here http://en.wikipedia.org/wiki/Quotation_mark .
What is the opposite of Regexp.escape ?
> Regexp.escape('A & B')
=> "A\\ &\\ B"
> # do something, to get the next result: (something like Regexp.unescape(A\\ &\\ B))
=> "A & B"
How can I get the original value?
replaces = Hash.new { |hash,key| key } # simple trick to return key if there is no value in hash
replaces['t'] = "\t"
replaces['n'] = "\n"
replaces['r'] = "\r"
replaces['f'] = "\f"
replaces['v'] = "\v"
rx = Regexp.escape('A & B')
str = rx.gsub(/\\(.)/){ replaces[$1] }
Also make sure to #puts output in irb, because #inspect escapes characters by default.
Basically escaping/quoting looks for meta-characters, and prepends \ character (which has to be escaped for string interpretation in source code). But if we find any control character from list: \t, \n, \r, \f, \v, then quoting outputs \ character followed by this special character translated to ascii.
UPDATE:
My solution had problems with special characters (\n, \t ans so on), I updated it after investigating source code for rb_reg_quote method.
UPDATE 2:
replaces is hash, which converts escaped characters (thats why it is used in block attached to gsub) to unescaped ones. It is indexed by character without escape character (second character in sequence) and searches for unescaped value. The only defined values are control-characters, but there is also default_proc attached (block attached to Hash.new), which returns key if there is no value found in hash. So it works like this:
for "n" it returns "\n", the same for all other escaped control characters, because it is value associated with key
for "(" it returns "(", because there is no value associated with "(" key, hash calls #default_proc, which returns key itself
The only characters escaped by Regexp.escape are meta characters and control characters, so we don't have to worry about alphanumerics.
Take a look at http://ruby-doc.org/core-2.0.0/Hash.html#method-i-default_proc for documentation on #defoult_proc
You can perhaps use something like this?
def unescape(s)
eval %Q{"#{s}"}
end
puts unescape('A\\ &\\ B')
Credits to this question.
codepad demo
If you are okay with a regex solution, you can use this:
res = s.gsub(/\\(?!\\)|(\\)\\/, "\\1")
codepad demo
Try this
>> r = Regexp.escape("A & B (and * c [ e] + )")
# => "A\\ &\\ B\\ \\(and\\ \\*\\ c\\ \\[\\ e\\]\\ \\+\\ \\)"
>> r.gsub("\\(","(").gsub("\\)",")").gsub("\\[","[").gsub("\\]","]").gsub("\\{","{").gsub("\\}","}").gsub("\\.",".").gsub("\\?","?").gsub("\\+","+").gsub("\\*","*").gsub("\\ "," ")
# => "A & B (and * c [ e] + )"
Basically, these (, ), [, ], {, }, ., ?, +, * are the meta characters in regex. And also \ which is used as an escape character.
The chain of gsub() calls replace the escaped patterns with corresponding actual value.
I am sure there is a way to DRY this up.
Update: DRY version as suggested by user2503775
>> r.gsub("\\","")
Update:
following are the special characters in regex
[,],{,},(,),|,-,*,.,\\,?,+,^,$,<space>,#,\t,\f,\v,\n,\r
using a regex replace using \\(?=([\\\*\+\?\|\{\[\(\)\^\$\.\#\ ]))\
should give you the string unescaped, you would only have to replace \r\n sequences with there CrLf counterparts.
"There\ is\ a\ \?\ after\ the\ \(white\)\ car\.\ \r\n\ it\ should\ be\ http://car\.com\?\r\n"
is unescaped to :
"There is a ? after the (white) car. \r\n it should be http://car.com?\r\n"
and removing the \r\n gives you :
There is a ? after the (white) car.
it should be http://car.com?
I am trying to replace a two letter state abbreviation with text then the abbreviation.
Eventually I want to find and replace the rest. How do I capture the value found? .... I tried \1 and {1}
AL 32.2679134368897 -86.5251510620117
AR 35.2315113544464 -92.2926173210144
AZ 33.3440766538127 -111.955985217148
CO 39.7098631425337 -104.899092934348
if( usState == "AZ") dpos= "33.4736704187888" + " " + "-112.043138087587";
if( usState == "CA") dpos= "36.0783581515733" + " " + " -119.868895584259";
if( usState == "CO") dpos= "39.8950788035537" + " " + " -104.831521872318";
if( usState == "CT") dpos= "41.6001570945562" + " " + " -72.6606015937273";
Update
$1 does not work.
I am finding: [A-Z][A-Z]
replacing with: if( usState == "$1
Oddly enough, Visual Studio Regular Expressions are different than normal .Net regular expressions. They have a slightly different syntax for tags and replaces. In order to tag a piece of text for later matching you must wrap it in braces {}. Then you can use \n in the replacement strings where n is the nth tagged expression. For your scenario here are the strings you should use
Find: {[A-Z][A-Z]}
Replace: if( usState == "\1")
My regex matcher matches $1. Try that.
I might not have understood your problem, but why don't you record a temporary macro to do the transformation?
Since this questions seems to be a duplicate of https://stackoverflow.com/a/3147177/154480 but I found this one first: since Visual Studio 2012, you can use (pattern) and $1. As an example for this specific question, find ([A-Z]{2}) by if( usState == "$1")
Enclose the [A-Z][A-Z] within parentheses, which captures it; then, use \1 in your replacement string to refer to the capture.