How can I replace all GUID's in a Kusto query with no value.
e.g.
my data looks like
/page/1d58e797-a905-403f-ebd9-27ccf3f1d2cd/user/4d58e797-a905-403f-ebd9-27ccf3f1d2c3
and I want
/page//user/
You can use the replace function. Also, you can test regular expression here.
let input = '/page/1d58e797-a905-403f-ebd9-27ccf3f1d2cd/user/4d58e797-a905-403f-ebd9-27ccf3f1d2c3';
let rx = '[({]?[a-fA-F0-9]{8}[-]?([a-fA-F0-9]{4}[-]?){3}[a-fA-F0-9]{12}[})]?';
print replace(rx, '', input);
Try using the below regex It will remove guid (8-4-4-4-12) in url
let regex =/(\/[\w]{8}-[\w]{4}-[\w]{4}-[\w]{4}-[\w]{12})(\b|\/)/g
Related
Here is my original string:
"Chassis ID TLV\n\tMAC: 00:xx:xx:xx:xx:xx\nPort ID TLV\n\tIfname: Ethernet1/3\nTime to Live TLV\n\t120"
and i want the string to be formatted as :
"Chassis ID TLV;00:xx:xx:xx:xx:xx\nPort ID TLV;Ethernet1/3\nTime to Live TLV;120"
so i used following ruby string functions to do it:
y = x.gsub(/\t[a-zA-Z\d]+:/,"\t")
y = y.gsub(/\t /,"\t")
y = y.gsub("\n\t",";")
so i am looking for a one liner to do the above. since i am not used to regex, i tried doing it sequentially. i am messing it up when i try to do all of them together.
Replace the following construct
[\n\r]\t(?:\w+: )?
with ;, see a demo on regex101.com.
I'd tackle it as a few smaller steps:
input = "Chassis ID TLV\n\tMAC: 00:xx:xx:xx:xx:xx\nPort ID TLV\n\tIfname: Ethernet1/3\nTime to Live TLV\n\t120"
input.split(/\n\t?/).map { |s| s.sub(/\A[^:]+\:\s*/, '') }.join(';')
# => "Chassis ID TLV;00:xx:xx:xx:xx:xx;Port ID TLV;Ethernet1/3;Time to Live TLV;120"
That way you have control over each element instead of being entirely dependent on the regular expression to do it as one shot.
I have params like:
params[:id]= "\"ebfd11a9-3aa4-415a-ba72-1b6796ea1bf6\""
And i want to get expected result as below:
"ebfd11a9-3aa4-415a-ba72-1b6796ea1bf6"
How can I do this?
You can use gsub:
"\"ebfd11a9-3aa4-415a-ba72-1b6796ea1bf6\"".gsub("\"", "")
=> "ebfd11a9-3aa4-415a-ba72-1b6796ea1bf6"
Or, as #Stefan mentioned, delete:
"\"ebfd11a9-3aa4-415a-ba72-1b6796ea1bf6\"".delete("\"")
=> "ebfd11a9-3aa4-415a-ba72-1b6796ea1bf6"
If this is JSON data, which it could very well be in that format:
JSON.load(params[:id])
This handles things where there's somehow escaped strings in there, or the parameters are an array.
Just Use tr!
params[:id].tr!("\"","")
tr! will also change the main string
In case you do not want to change main string just use :
params[:id].tr("\"","")
Thanks Ilya
I want to remove some part of string from using ruby regex:
value = localhost:8393/foobar/1 test:foobartest
I want to remove "test" from my string [localhost:8393/foobar/1 test:foobartest] and rest of the value so that output should look like:
localhost:8393/foobar/1
How to do this in ruby? Can you share some sample code to achieve this?
Appreciated your help in advance!
Thanks!
I would do something like this:
value = 'localhost:8393/foobar/1 test:foobartest'
value.split.first
#=> "localhost:8393/foobar/1"
Or if you want to use an regexp:
value.sub(/ test.*/, '')
"localhost:8393/foobar/1"
I have a string containing a path:
/var/www/project/data/path/to/file.mp3
I need to get the substring starting with '/data' and delete all before it. So, I need to get only /data/path/to/file.mp3.
What would be the fastest solution?
'/var/www/project/data/path/to/file.mp3'.match(/\/data.*/)[0]
=> "/data/path/to/file.mp3"
could be as easy as:
string = '/var/www/project/data/path/to/file.mp3'
path = string[/\/data.*/]
puts path
=> /data/path/to/file.mp3
Using regular expression is a good way. Though I am not familiar with ruby, I think ruby should have some function like "substring()"(maybe another name in ruby).
Here is a demo by using javascript:
var str = "/var/www/project/data/path/to/file.mp3";
var startIndex = str.indexOf("/data");
var result = str.substring(startIndex );
And the link on jsfiddle demo
I think the code in ruby is similar, you can check the documentation. Hope it's helpful.
Please try this:
"/var/www/project/data/path/to/file.mp3".scan(/\/var\/www(\/.+)*/)
It should return you all occurrences.
When using the MongoDB shell, how do I use a guid datatype (which I have used as the _id in my collection).
The following format doesn't work:
>db.person.find({"_id","E3E45566-AFE4-A564-7876-AEFF6745FF"});
Thanks.
You can use easily:
.find({ "_id" : CSUUID("E3E45566-AFE4-A564-7876-AEFF6745FF")})
You have to compare the _id value against an instance of BinData (not against a string). Unfortunately the BinData constructor takes a Base64 string instead of a hex string.
Your GUID value is missing two hex digits at the end, so for the purposes of this example I will assume they are "00". The following values are equivalent:
hex: "E3E45566-AFE4-A564-7876-AEFF6745FF00" (ignoring dashes)
base64: "ZlXk4+SvZKV4dq7/Z0X/AA=="
So your query should be:
>db.person.find({_id : new BinData(3, "ZlXk4+SvZKV4dq7/Z0X/AA==")})
I am assuming that the binary subtype was correctly set to 3. If not, what driver was used to create the data?
You could use the following js function in front of your query like so:
function LUUID(uuid) {
var hex = uuid.replace(/[{}-]/g, ""); // removes extra characters
return new UUID(hex); //creates new UUID
}
db.person.find({"_id" : LUUID("E3E45566-AFE4-A564-7876-AEFF6745FF"});
You could save the function in .js file and load it or open it before you make your query and if you copy the value from your results you should rename the function with:
LUUID for Legacy UUID
JUUID for Java encoding
NUUID for .net encoding
CSUUID for c# encoding
PYUUID for python encoding
I know it's an old issue, but without any additional needs you can use this one:
find({_id:UUID('af64ab4f-1098-458a-a0a3-f0f6c93530b7')})
You can fix this issue by using split() and join() workaround:
for instance if I use "E3E45566-AFE4-A564-7876-AEFF6745FF" hex value with - inside UUID() function, it does not return BinData in mongo so please try removing all the - before passing to UUID function.
db.person.find({"_id":UUID("E3E45566-AFE4-A564-7876-AEFF6745FF".split("-").join(''))});
Or by defining a variable to do it in multiple line:
var uuid = UUID("E3E45566-AFE4-A564-7876-AEFF6745FF".split("-").join(''))
db.person.find({"_id":uuid});
or by creating a simple function:
function BUUID(uuid){
var str = uuid.split("-").join('');
return new UUID(str);
}
db.person.find({"_id": BUUID("E3E45566-AFE4-A564-7876-AEFF6745FF")}).pretty();