I'm trying to use a Make substitution reference to alter a string. The problem being that the string happens to contain an equals = symbol.
For example:
INPUT = -switch1 -switch2=potato -switch3
OUTPUT = $(INPUT:-switch2=%=-switch2=turnip)
all:
#echo TEST : $(OUTPUT)
so in the form $(var:a=b), INPUT is var, -switch2=% is a and -switch2=turnip is b.
Obviously that doesn't work because = is a special character in this context, but I've no idea how to make it realize that this is part of string a.
I've tried quoting, backslashes, alternative escape characters and putting -switch2=% in a variable and using that instead. All to no avail.
I know I can use patsubst, but that'd be adding the first non-POSIX extension to the file and i'd prefer to not be that guy.
Any suggestions appreciated!
You'll have to use a full patsubst function. Substitution references are just a shortcut for patsubst:
OUTPUT = $(patsubst -switch2=%,-switch2=turnip,$(INPUT))
Related
It's a weird problem
to_be_stripped="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120"
And two strings below:
s1="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120\\[Content_Types].xml"
s2="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120\\_rels\.rels"
When I use the command below:
s1.strip(to_be_stripped)
s2.strip(to_be_stripped)
I get these outputs:
'[Content_Types].x'
'_rels\\.'
If I use lstrip(), they will be:
'[Content_Types].xml'
'_rels\\.rels'
Which is the right outputs.
However, if we replace all Project Known with zeus_pipeline:
to_be_stripped="D:\\Users\\UserKnown\\PycharmProjects\\zeus_pipeline\\PT\\collections\\120"
And:
s2="D:\\Users\\UserKnown\\PycharmProjects\\zeus_pipeline\\PT\\collections\\120\\_rels\.rels"
s2.lstrip(to_be_stripped)will be '.rels'
If I use / instead of \\, nothing goes wrong. I am wondering why this problem happens.
strip isn't meant to remove full strings exactly. Rather, you give it a string, and every character in that string is removed from the start and of the string to be stripped.
In your case, the variable to_be_stripped contains the characters m and l, so those are stripped from the end of s1. However, it doesn't contain the character x, so the stripping stops there and no characters beyond that are removed.
Check out this question. The accepted answer is probably more extensive than you need - I like another user's suggestion of using replace instead of strip. This would look like:
s1.replace(to_be_stripped, "")
I have been trying to remove in batch script "=" character from string by using this:
set Path_var=%Path_var:^==%
Unfortunately this does not work... I tried also some other common solutions like:
set Path_var=%Path_var:"="=%
set Path_var=%Path_var:'='=%
But without success. Maybe it would be worthy also to explain for what I need it as I am aware you may be able to provide better solution. I extract one line from xml configuration file. The line is following:
<burning addDicomViewer="true" finalizeMedium="true" dicomViewer="C:\user\App_folder\App-name_subfolder_1.1.1_Setup" burnVerification="true" numberOfCopies="0" cleanupProjectData="false" volumeName="Patient Medium"/>
I need to extract from this line this path: "C:\user\App_folder\App-name_subfolder_1.1.1_Setup" (The path will not always be the same)
My strategy was to simply remove definite number of characters before the path as I know this setting will always be the same and therefore the length of the string won`t change.
set /p Path_var= < temp_file01.txt
set Path_var=%Path_var:~81,100%
Then I wanted to use simply substitution to remove the rest. For example:
set Path_var=%Path_var:burnVerification=%
But I ran into problem that my string contains characters like "=" which I can not remove by this method. (Because obviously there are handled as operators) What I was also wondering is what I should do if there will be a space character in my path. Then when I attempt to remove the empty characters at the end I also invalidate my path.
I know batch scripts are not the best for manipulation with strings, but I do not have other choice as my boss want me to use scripting language which does not need compiling.
I asked my work colleague for help and he came with following PowerShell solution:
$path_temp_file01 = "C:\user\temp\tmpFile_backup_script01.txt"
$path_temp_file02 = "C:\user\temp\tmpFile_backup_script02.txt"
$string = [IO.File]::ReadAllText($path_temp_file01)
$Start = $string.IndexOf("C:")
$string = $string.substring($Start)
$End =$string.IndexOf("""")
$string = $string.substring(0,$End)
$string > $path_temp_file02
It works for me. I post it here in the case someone needs similar solution or has better idea how to do that.
I have inherited a cookbook that sets some attributes in the ./attributes/default.rb file as per normal.
However, we have a problem with one of the lines is, which is:
default["obscured"]["mysql"] = "#{node['jboss']['jboss_home']}/modules/com/mysql/jdbc/main"
When run, it write this into the node as:
{}/com/mysql/jdbc/main
I can confirm that the node['jboss']['jboss_home'] attribute exists and has correct values.
So, I cannot see any problem with the above, except that every other declaration of this type in our cookbooks has single quotes on the attribute to be set (i.e. left side), not double quotes. I haven't heard this of as being an issue before, but I am pretty new to chef.
Is there any rule that says they must be single quotes?
The answer is that there is no rule.
Using double-quotes in something like this is completely fine:
default["obscured"]["mysql"] = blah blah
The reason I know that is that I just found one being set, with double quotes, that actually works. :-)
What you have there is fine, how are you confirming the value of node['jboss']['jboss_home'] and how are you using it in the template?
In Ruby single and double quoted literals both become Strings but single quotes are relatively literal while double quotes allow backslash escapes and #{} interpolation.
You are most likely hitting the derived attributes problem:
https://coderanger.net/derived-attributes/
The attribute code in your cookbook is getting parsed before the jboss_home attribute is being set. One way or another the solution is to move the interpolation into recipe code.
You could just use a plain old ruby variable instead of the attribute you are trying to construct -- particularly if nothing else in your system ever sets that attribute.
You also should be able to delete the declaration from your attributes file and use this in recipe code as well:
node.default_unless["obscured"]["mysql"] =
"#{node['jboss']['jboss_home']}/modules/com/mysql/jdbc/main"
Although you need to place that statement early in your run_list, before you ever use node["obscured"]["mysql"] as an argument to any resource.
I am totally confused about the following macro modifier in a Makefile I have come onto,
TOOLS = $(TOOL_ROOTS:%=$(OBJDIR)%$(TOOL_SUFFIX))
Here
TOOL_ROOTS=some filename prefixes
OBJDIR=$HOME/obj/
TOOL_SUFFIX=.so
Can someone tell me what this line actually means?
TOOL_ROOTS must be getting assigned some value other than the empty string at some point or that does nothing (which I'll show in a moment).
So first things first just expanding the variables takes us from:
TOOLS = $(TOOL_ROOTS:%=$(OBJDIR)%$(TOOL_SUFFIX))
to:
TOOLS = $(:%=~/obj%.so)
(which we can immediately see doesn't look right, and as I'll explain in a moment doesn't do anything)
So lets pretend it has a value instead.
TOOL_ROOTS = shovel axe hammer
And try the expansion again:
TOOLS = $(shovel axe hammer:%=~/obj%.so)
(That OBJDIR definition looks odd also. I would expect it to be ~/obj/ or something... and that's ignoring that ~ is a bad choice here and that $HOME would be much better.)
The next thing we need to know is what that syntax is all about. Well it is a Substitution Reference.
A substitution reference substitutes the value of a variable with alterations that you specify. It has the form ‘$(var:a=b)’ (or ‘${var:a=b}’) and its meaning is to take the value of the variable var, replace every a at the end of a word with b in that value, and substitute the resulting string.
When we say “at the end of a word”, we mean that a must appear either followed by whitespace or at the end of the value in order to be replaced; other occurrences of a in the value are unaltered. For example:
foo := a.o b.o c.o
bar := $(foo:.o=.c)
sets ‘bar’ to ‘a.c b.c c.c’. See Setting Variables.
A substitution reference is actually an abbreviation for use of the patsubst expansion function (see Functions for String Substitution and Analysis). We provide substitution references as well as patsubst for compatibility with other implementations of make.
Another type of substitution reference lets you use the full power of the patsubst function. It has the same form ‘$(var:a=b)’ described above, except that now a must contain a single ‘%’ character. This case is equivalent to ‘$(patsubst a,b,$(var))’. See Functions for String Substitution and Analysis, for a description of the patsubst function.
For example:
foo := a.o b.o c.o
bar := $(foo:%.o=%.c)
sets ‘bar’ to ‘a.c b.c c.c’.
So, the first % there is matching the entirety of every word in the value of the variable (here shovel axe hammer) and then replacing each value with the expansion of the second part.
So shovel becomes ~/objshovel.so, etc. and we end up with:
TOOLS = ~/objshovel.so ~/objaxe.so ~/objhammer.so
See what I meant about OBJDIR being odd before? OBJDIR=~/obj/ would have left us with this instead:
TOOLS = ~/obj/shovel.so ~/obj/axe.so ~/obj/hammer.so
which makes a lot more sense to me.
I'm currently teaching myself Lua for iOS game development, since I've heard lots of very good things about it. I'm really impressed by the level of documentation there is for the language, which makes learning it that much easier.
My problem is that I've found a Lua concept that nobody seems to have a "beginner's" explanation for: nested brackets for quotes. For example, I was taught that long strings with escaped single and double quotes like the following:
string_1 = "This is an \"escaped\" word and \"here\'s\" another."
could also be written without the overall surrounding quotes. Instead one would simply replace them with double brackets, like the following:
string_2 = [[This is an "escaped" word and "here's" another.]]
Those both make complete sense to me. But I can also write the string_2 line with "nested brackets," which include equal signs between both sets of the double brackets, as follows:
string_3 = [===[This is an "escaped" word and "here's" another.]===]
My question is simple. What is the point of the syntax used in string_3? It gives the same result as string_1 and string_2 when given as an an input for print(), so I don't understand why nested brackets even exist. Can somebody please help a noob (me) gain some perspective?
It would be used if your string contains a substring that is equal to the delimiter. For example, the following would be invalid:
string_2 = [[This is an "escaped" word, the characters ]].]]
Therefore, in order for it to work as expected, you would need to use a different string delimiter, like in the following:
string_3 = [===[This is an "escaped" word, the characters ]].]===]
I think it's safe to say that not a lot of string literals contain the substring ]], in which case there may never be a reason to use the above syntax.
It helps to, well, nest them:
print [==[malucart[[bbbb]]]bbbb]==]
Will print:
malucart[[bbbb]]]bbbb
But if that's not useful enough, you can use them to put whole programs in a string:
loadstring([===[print "o m g"]===])()
Will print:
o m g
I personally use them for my static/dynamic library implementation. In the case you don't know if the program has a closing bracket with the same amount of =s, you should determine it with something like this:
local c = 0
while contains(prog, "]" .. string.rep("=", c) .. "]") do
c = c + 1
end
-- do stuff