Not sure if I understand the point of writing .json config in a single line like this with both forward and backwards slashes?:
{"name":"zendframework\/zendservice-twitter","description":"OOP wrapper for the Twitter web service","type":"library","keywords":["zf2","twitter"],"homepage":"http:\/\/packages.zendframework.com\/","license":"BSD-3-Clause","autoload":{"psr-0":{"ZendService":"library\/"}},"repositories":[{"type":"composer","url":"http:\/\/packages.zendframework.com\/"}],"require":{"php":">=5.3.3","zendframework\/zend-http":">=2.0.0","zendframework\/zend-uri":">=2.0.0","zendframework\/zend-version":">=2.0.0"},"extra":{"branch-alias":{"dev-master":"2.0.x-dev"}},"version":"2.0.1","dist":{"url":"http:\/\/packages.zendframework.com\/composer\/ZendService_Twitter-2.0.1.zip","type":"zip"}}
The backslash \ is an escape character, so every time you see \/, the / is being escaped.
The fact that the entire object is written on one line is most likely an artifact of whatever generated the JSON (I'm assuming it was not handwritten).
Related
For a gradle script, I am composing strings that will be used as command line for a subsequent gradle Test-task. One of the strings is the user's password, which eventually will be passed to the called (exec'ed) "java ..." call using the JVM's -D option, e.g. -Dpassword=foobar.
What complicates things here is, that this password can/should of course contain special characters, that may interfere with the use of the string as command line. In other words: I need to escape special characters (which is OS-specific). :-(
Now to my actual question:
I want to use the String.replaceAll method, i.e. replaceAll(list_of_special characters, EscapeCharacter + Ref_to_matched_character),
e.g. simplified something like replaceAll("[#$%^&]", "^$1")
'^' meaning the escape character and '$1' meaning the matched character here.
Is that possible, i.e. can one refer to the matched pattern in the second argument of replaceAll?
Is that possible, i.e. can one refer to the matched pattern in the second argument of replaceAll?
yes, it's possible
'a#b$c'.replaceAll('([#$%^&])', '^$1')
returns
a^#b^$c
Thanks for the responses and the reviews improving readability. Meanwhile I got my expression working. For those interested:
// handles gthe following: `~!##$%^&*()_+-={}|[]\:;"'<>?,./
escaped = original.replaceAll('[~!##\\$\\%\\^\\&\\*\\(\\)_\\+-={}\\|\\[\\]\\\\:;\"\\\'<>\\?,\\./]', '^$0') // for Windows - cmd.exe
I have to read a yaml file, modify it and write back using pyYAML. Every thing works fine except when there is multi-line string values in single quotes e.g. if input yaml file looks like
FOO:
- Bar: '{"HELLO":
"WORLD"}'
then reading it as data=yaml.load(open("foo.yaml")) and writing it yaml.dump(data, fref, default_flow_style=False) generates something like
FOO:
- Bar: '{"HELLO": "WORLD"}'
i.e. without the extra line for Bar value. Strange thing is that if input file has something like
FOO:
- Bar: '{"HELLO":
"WORLD"}'
i.e. one extra new line for Bar value then writing it back generates the correct number of new lines. Any idea what I am doing wrong?
You are not doing anything wrong, but you probably should have read more of the YAML specification.
According to the (outdated) 1.1 spec that PyYAML implements, within
single quoted scalars:
In a multi-line single-quoted scalar, line breaks are subject to (flow) line folding, and any trailing white space is excluded from the content.
And line-folding:
Line folding allows long lines to be broken for readability, while retaining the original semantics of a single long line. When folding is done, any line break ending an empty line is preserved. In addition, any specific line breaks are also preserved, even when ending a non-empty line.
This means that your first two examples are the same, as the
line-break is read as if there is a space.
The third example is different, because it actually contains a newline after loading, because "any line break ending an empty line is preserved".
In order to understand why that dumps back as it was loaded, you have to know that PyYAML doesn't
maintain any information about the quoting (nor about the single newline in the first example), it
just loads that scalar into a Python string. During dumping PyYAML evaluates how that string
can best be written and the options it considers (unless you try to force things using the default_style argument to dump()): plain style, single quoted style, double quoted style.
PyYAML will use plain style (without quotes) when possible, but since
the string starts with {, this leads to confusion (collision) with
that character's use as the start of a flow style mapping. So quoting
is necessary. Since there are also double quotes in the string, and
there are no characters that need backslash escaping the "cleanest"
representation that PyYAML can choose is single quoted style, and in
that style it needs to represent a line-break by including an emtpy
line withing the single quoted scalar.
I would personally prefer using a block style literal scalar to represent your last example:
FOO:
- Bar: |
{"HELLO":
"WORLD"}
but if you load, then dump that using PyYAML its readability would be lost.
Although worded differently in the YAML 1.2 specification (released almost 10 years ago) the line-folding works the same, so this would "work" in a similar way with a more up-to-date YAML loader/dumper. My package ruamel.yaml, for loading/dumping YAML 1.2 will properly maintain the block style if you set the attribute preserve_quotes = True on the YAML() instance, but it will still get rid of the newline in your first example. This could be implemented (as is shown by ruamel.yaml preserving appropriate newline positions in folded style block scalars), but nobody ever asked for that, probably because if people want that kind of control over wrapping they use a block style to start with.
I have many text files, and I need to add some text (e.g. MNP) to the beginning of the first line in each file.
How can I do this in Notepad++?
(I'm using v6.6.9)
Make sure to backup your work beforehand, and set proper extension of files to affect and folder to search through before you do this.
You can use regular expressions. Several places around the internet claim that the regex \A works, but it wasn't working for me, it was cycling byte by byte through. I found that \A^ sticks to 0 position of the file.
Oddly, I additionally found that I couldn't replace \A or \A^ and have it take effect. This is what worked for me.
Find: \A^(.*?)
Replace MNP\1
Truthfully, the \1 in Replace isn't even necessary since I'm cheating and basically telling notepad to look for 0 characters.
This should work just as well.
Find: \A^.*?
Replace MNP
Please backup your work beforehand though.
Alternatively, this also seems to work.
Find: .{0}(.*)
Replace: MNP\1
It effectively looks for 0 characters followed by the whole document/line (depending on whether . matches newline is checked, this choice won't matter for the outcome however).
I'm redirecting output of an API call to file
however I always get the following characters surrounding the value I need
domainid='^[[39;49;00m^[[33;01m75307d12-e3f4-4a96-ac23-e2a9439f8299^[[39;49;00m'
Desired output
domainid='75307d12-e3f4-4a96-ac23-e2a9439f8299'
I really have no idea how to clean the output and make it look like the above.
Any suggestions will be highly appreciated.
Thank you
Those are ANSI control characters, or escape sequences, and they typically are used to add colors, underline, and so forth to your output.
First order of business is to check if your API command line tool supports a no-color mode. That would solve your problem at the source.
Barring that, try this Server Fault answer, which has a command to clear ANSI sequences out of a text file using sed.
You could remove the undesired characters by replacing the line with just the submatches you want to keep:
... | sed -r "s/(domainid=).*([0-9a-f]{8}(-[0-9a-f]{4}){3}-[0-9a-f]{12}).*/\1'\2'/i"
I have been looking at regular expressions to try and do this, but the most I can do is find the start of a line with ^, but not replace it.
I can then find the first characters on a line to replace, but can not do it in such a way with keeping it intact.
Unfortunately I don´t have access to a tool like cut since I am on a windows machine...so is there any way to do what I want with just regexp?
Use notepad++. It offers a way to record an sequence of actions which then can be repeated for all lines in the file.
Did you try replacing the regular expression ^ with the text you want to put at the start of each line? Also you should use the multiline option (also called m in some regex dialects) if you want ^ to match the start of every line in your input rather than just the first.
string s = "test test\ntest2 test2";
s = Regex.Replace(s, "^", "foo", RegexOptions.Multiline);
Console.WriteLine(s);
Result:
footest test
footest2 test2
I used to program on the mainframe and got used to SPF panels. I was thrilled to find a Windows version of the same editor at Command Technology. Makes problems like this drop-dead simple. You can use expressions to exclude or include lines, then apply transforms on just the excluded or included lines and do so inside of column boundaries. You can even take the contents of one set of lines and overlay the contents of another set of lines entirely or within column boundaries which makes it very easy to generate mass assignments of values to variables and similar tasks. I use Notepad++ for most stuff but keep a copy of SPFSE around for special-purpose editing like this. It's not cheap but once you figure out how to use it, it pays for itself in time saved.