I am trying to replace single quote in a string with double quote using replace function with data factory expressions.
For example, replace single quote in the following string
hello'world ---> hello''world
#replace(pipeline().parameters.tst,''','''')
The above code is not working. Need help in fixing the code
You can declare a new parameter with the value ' (single quote). You can look at the following demonstration for reference.
I have taken 2 parameters, text with the value hello'world and replace_char with the value '.
I used a set variable activity to store the output of the replace() function (for demonstration) into variable named output (String). Now, I modified the value as:
#replace(pipeline().parameters.text,pipeline().parameters.replace_char,'"')
This successfully helps in replacing a single quote with double quote character.
NOTE: The \ in the output variable value indicates that the " is to be considered as a character inside the string value.
Use two single quotes to escape a ' character in string functions.
For example, expression #concat('Baba', '''s ', 'book store') will return below result.
Baba's book store
https://learn.microsoft.com/en-us/azure/data-factory/control-flow-expression-language-functions#escaping-single-quote-character
Related
I have varibale which contains raw data as shown below.
I want to replace the comma inside double quotes to nothing/blank.
I used replaceAll(',',''), but all other commas are also getting replaced.
So need regex function to identify pattern like "123,456,775" and then replace here comma into blank.
var = '
2/5/2023,25,"717,990","18,132,406"
2/4/2023,27,"725,674","19,403,116"
2/3/2023,35,"728,501","25,578,008"
1/31/2023,37,"716,580","26,358,186"
2/1/2023,37,"720,466","26,494,010"
1/30/2023,37,"715,685","26,517,878"
2/2/2023,37,"723,545","26,603,765" '
Tried replaceAll, but did not work
If you just want to replace "," with "", you have to escape the quotes this will do:
var.replaceAll(/\",\"/, /\"\"/)
If you want to replace commas inside the number strings, "725,674" with "725674" you will have to use a regex and capture groups, like this:
var.replaceAll(/(\"\d+),(\d+\")/, /$1$2/)
It will change for three groupings, like "18,132,406", you will have to use three capture groups.
I want to update a string in quotes in yaml file in Jenkins Job. While updating the file, single quotes around the string are replaced by triple quotes. Following is the method that I have written:
{
def fileName = 'config.yml'
datas = readYaml file: fileName
var = "'" + params.ReleaseBranchName + "'"
println var // this shows output as expected, string in single quotes -> 'rel-21.9'
datas.branchName = var
println datas // this prints the yaml with single quotes -> productiveBranch='rel-21.9',
writeYaml file: fileName, data: datas, overwrite: true //this show value in triple quotes -> productiveBranch: '''rel-21.9'''
}
Could someone suggest how can I save string with single quotes in yaml file? Thanks!
The value of var, as written, is 'rel-21.9', i.e. the single quotes are part of the value.
In YAML input, when 'rel-21.9' is encountered, the single quotes are not part of the value; they are part of the syntax and enclose the value, so the value is rel-21.9.
Therefore, if you want your value to be rel-21.9, which is most probably what you want, do not put single quotes in the value; just do var = params.ReleaseBranchName.
Your code does not do anything with var; I assume what you're trying is to put it into datas. This would result in YAML writing out "'rel-21.9'" (not triple single quotes, that can't happen since it would be invalid YAML). By surrounding the value with double quotes, the single quotes become part of the value just like your code requested.
When you do not put single quotes into the data, YAML will probably serialise it without any quotes. This is expected since rel-21.9 does not contain any special characters that would require quoting. There are ways to force a YAML processor to quote a value, but they are complex and I am unsure whether the API is exposed to Groovy. For references, this is how you would do it in Java.
Since you are editing a YAML file, you might want to read this question which details how and why updating YAML files in code can lead to style changes.
Trying to return a text (or string) value that contains double quotes.
I have a basic validation in one of my cells:
=IF(C8=C9,"test "new" test")
where test "new" test is returned when C8 is equal to C9.
Receiving the error 'There's a problem with this formula.'
I believe the double quotes around the word new are causing the problem. If I delete them, the error is no longer coming. However, I need these double quotes as part of the return value.
=IF(C8=C9,"test ""new"" test")
Double-quotes in strings need to be escaped by doubling them up.
I have a string to modify as per the requirements.
For example:
The given string is:
str1 varchar = '123,456,789';
I want to show the string as:
'456,789'
Note: The first part (delimited) with comma, I want to remove from string and show the rest of string.
In SQL Server I used STUFF() function.
SELECT STUFF('123,456,789',1,4,'');
Result:
456,789
Question: Is there any string function in PostgreSQL 9.3 version to do the same job?
you can use regular expressions:
select substring('123,456,789' from ',(.*)$');
The comma matches the first comma found in the string. The part inside the brackets (.*) is returned from the function. The symbol $ means the end of the string.
A alternative solution without regular expressions:
select str, substring(str from position(',' in str)+1 for length(str)) from
(select '123,456,789'::text as str) as foo;
You could first turn the string to array and return second and third cell:
select array_to_string((regexp_split_to_array('123,456,789', ','))[2:3], ',')
Or you could use substring-function with regular expressions (pattern matching):
SELECT substring('123,456,789' from '[0-9]+,([0-9]+,[0-9]+)')
[0-9]+ means one or more digits
parentheses tell to return that part from the string
Both solutions work on your specific string.
Your The SQL Server example indicates you just want to remove the first 4 characters, which makes the rest of your question seem misleading because it completely ignores what's in the string. Only the positions matters.
Be that as it may, the simple and cheap way to cut off leading characters is with right():
SELECT right('123,456,789', -4);
SQL Fiddle.
I wrote the following line:
string QuoteTest2 = "Benjamin Netnayahu,\"BB\", said that: \"Israel will not fall\"";
This example went well, but what can I do in case I want to write a multi paragraph string including quotes?
The following example shows that puting '#' before the doesn't cut it..
string QuoteTest2 = #"Benjamin Netnayahu,\"BB\", said that: \"Israel will not fall\"";
The string ends and the second quote and the over just gives me errors, what should I do?
Use double quotes to escape ""
e.g.
string QuoteTest2 = #"Benjamin Netnayahu,""BB"", said that: ""Israel will not fall""";