Match multiple occurences of string A in string B in business objects - sap

Is there a way to match multiple occurences of string A in string B? Match only return a boolean and Pos only returns the index of the first match it finds.
Maybe split the string into an iterable array based on a common delimiter and return a count somehow?

Well this is one way to solve it. It's not pretty but it works.
=(Length([stringB]) - Length(Replace([stringB];"stringA";""))) / Length("stringA")

Related

string operation in ruby

I have a below string array object which has 4 elements. I want to compare the elements of this list to an string and want to check if string is part of this list.
list = ["starter_v2", "professional_q", "custom_v", "basic_t9"]
str = "starter"
if list.include?str #should return true but returning false as it is checking the full string not the substring
Above if condition should return true, however it is returning false.
Can someone suggest me how to fix this, as I am new to ruby and want to compare strings.
For my usecase, in list object I will always have entries with "_" followed by an alphabetic character and I will compare this by string without "_"
Enumerable#include? checks if a given value matches any value in the given enumerable. A substring is not equivalent to a string that contains it, so this check fails.
Instead, you want to check if any string in the array matches your substring. Ruby has handy facilities for this: Enumerable#any? lets you iterate an enumerable, yielding each element to a block, and then will return true if any invocation of the block returns true.
So, you can use:
list.any? {|element| element.include?(str) }
What this will do is check each entry in list to see if str is included in it; once a match is found, it'll stop iterating and return true. If it goes through the entire list without finding a match, it'll return false.
You could also use use element.start_with? if you know that your search string should always match the first part of the string, or you could use a more complex condition which splits each element on underscore and compares the first part, or you could use a regex. The important part is that the block returns true when you want to indicate a match.

regular expression, how to replace a part of a text preserving its length

I have, in a database, records that are serialized PHP strings that I must obfuscate emails if there are any. The simplest record is like {s:20:"pika.chu#pokemon.com"}. It is basically saying: this is a string of length 20 which is pika.chu#pokemon.com. This field can be kilobytes long with lot of emails (or none) and sometimes it is empty.
I wish I could use a SQL regular expression function to obfuscate the user part of the email while preserving the length of the string in order not to break the PHP serialization. The example email above shall be turned into {s:20:"xxxxxxxx#pokemon.com"} where the number of x matches the length of pika.chu.
Any thoughts?
Here is a more complete example of what can be found as serialized PHP:
a:4:{s:7:"locales";a:3:{i:0;s:5:"fr_FR";i:1;s:5:"de_DE";i:2;s:5:"en_US";}s:9:"publisher";s:18:"john#something.com";s:7:"authors";a:2:{i:0;s:21:"william#something.com";i:1;s:19:"debbie#software.org";}s:12:"published_at";O:8:"DateTime":3:{s:4:"date";s:26:"2022-01-26 13:05:26.531289";s:13:"timezone_type";i:3;s:8:"timezone";s:3:"UTC";}}
I tried to do it using native functions but it not worked because functions like REGEXP_REPLACE don't let you manipulate the match to get the size of it, for example.
Instead, I've created a UDF to do that:
CREATE TEMP FUNCTION hideEmail(str STRING)
RETURNS STRING
LANGUAGE js AS """
return str
.replace(/([a-zA-Z.0-9_\\+-:]*)#/g, function(txt){return '*'.repeat(txt.length-1)+"#";})
""";
select hideEmail('a:4:{s:7:"locales";a:3:{i:0;s:5:"fr_FR";i:1;s:5:"de_DE";i:2;s:5:"en_US";}s:9:"publisher";s:18:"john#something.com";s:7:"authors";a:2:{i:0;s:21:"william#something.com";i:1;s:19:"debbie#software.org";}s:12:"published_at";O:8:"DateTime":3:{s:4:"date";s:26:"2022-01-26 13:05:26.531289";s:13:"timezone_type";i:3;s:8:"timezone";s:3:"UTC";}}')
Result:
a:4:{s:7:"locales";a:3:{i:0;s:5:"fr_FR";i:1;s:5:"de_DE";i:2;s:5:"en_US";}s:9:"publisher";s:18:"****#something.com";s:7:"authors";a:2:{i:0;s:21:"*******#something.com";i:1;s:19:"******#software.org";}s:12:"published_at";O:8:"DateTime":3:{s:4:"date";s:26:"2022-01-26 13:05:26.531289";s:13:"timezone_type";i:3;s:8:"timezone";s:3:"UTC";}}

Extract all characters before a period with HiveQL regex?

I have a table that looks like:
bl.ah
foo.bar
bar.fight
And I'd like to use HiveQL's regexp_extract to return
bl
foo
bar
Given the docs data about regexp_extract:
regexp_extract(string subject, string pattern, int index)
Returns the string extracted using the pattern. For example, regexp_extract('foothebar', 'foo(.*?)(bar)', 2) returns 'bar.' Note that some care is necessary in using predefined character classes: using '\s' as the second argument will match the letter s; '\s' is necessary to match whitespace, etc. The 'index' parameter is the Java regex Matcher group() method index. See docs/api/java/util/regex/Matcher.html for more information on the 'index' or Java regex group() method.
So, if you have a table with a single column (let's call it description for our example) you should be able to use regexp_extract as follows to get the data before a period, if one exists, or the entire string in the absence of a period:
regexp_extract(description,'^([^\.]+)\.?',1)
The components of the regex are as follows:
^ start of string
([^\.]+) any non-period character one or more times, in a capture group
\.? a period either once or no times
Because the part of the string we're interested in will be in the first (and only) capture group, we refer to it by passing the index parameter a value of 1.

Convert a StringBuilder to a Jagged Array

I have built a VB.Net class that will be used in VBA for reading text files. I've set it up so the user can specify what tables in the file he wants to return. What I have done is build a StringBuilder of the tables, then return it as a jagged array, but I can't quite get the conversion of the builder to array part right. I'd like the the first level to be split on "NewLine" and the second level to be split on ",".
Is this possible without having to use multiple arrays and\or loops?
This will create the jagged array:
Dim myArray = (From row In myStringBuilder.ToString().Split({vbCrLf}, StringSplitOptions.None)
Select (From col In row.Split(","c)
Select col
).ToArray()
).ToArray()
Explanation:
First, we convert the StringBuilder to a String: myStringBuilder.ToString()
Then we split on line breaks: Split({vbCrLf}, StringSplitOptions.None). Since a line break consists of two characters in Windows, we use the Split overload that accepts a string array (hence the braces).
Within the row we split the line on commas: Split(","c). The c specifies that this is a character instead of a string.
Finally, we convert this enumerable of enumerables into an array of arrays by applying ToArray to the outer as well as the inner LINQ expression.
You could represent your jagged array using nested lists and generics. The outer (row) would be a generic list and the inner (col) could be a list of strings.
Other approaches could leverage XML or LINQ but would be less efficient.

Weird results when splitting strings in VB.NET

I was getting weird results when doing multiple splits on a string, so I decided to make a simple test to figure out what was going on
testString "1234567891011121314151617181920"
If I wanted to get whats between 10 to 20 in Javascript I would do this:
var results = testString.split("10")[1].split("20")[0]
Which would return 111213141516171819
However when I do this in VB I get 111
Split(testString,"10")(1).Split("20")(0)
It seems the 2nd split is only recognizing the first character no matter what I put.
So it's stopping when it finds the next "2" in the string, even "2abc" would have the same outcome even though that string doesn't even exist.
String.Split does not have an overload that takes only a String. The argument is a Char array or String array. Your string is probably being converted to a char array. Explicitly pass a string array like so:
testString.Split(New String() { "10" }, StringSplitOptions.None)
Try wrapping the second split so it's fashioned like the first one, i.e.:
Split( Split(testString,"10")(1), "20" )(0)"
Vb treats the delimiter argument only as a single character.
This is a tricky scenario that I have seen trip people up before, so I think it is worth a little more explanation than the other answers give. In your original format Split(testString,"10")(1).Split("20")(0), you are unknowingly using two DIFFERENT Split functions.
The first Split(testString,"10") is using the Microsoft.VisualBasic.Strings.Split function, which takes String type parameters. http://msdn.microsoft.com/en-us/library/microsoft.visualbasic.strings.split(v=vs.110).aspx
The second .Split("20")(0) is using System.String.Split method, which does not have an overload that takes a String parameter. http://msdn.microsoft.com/en-us/library/System.String.Split(v=vs.110).aspx
So what was happening is:
Split(testString,"10") uses Microsoft.VisualBasic.Strings.Split, which
returns new String() {"123456789", "11121314151617181920"}
(1) means get 1st position of the returned array, which is "11121314151617181920"
"11121314151617181920".Split("20")(0) uses System.String.Split, and attempts to split on string separator "20"
NOTE: The string "20" param gets implicitly converted to a char "2" because the only single parameter overload of String.Split has a signature of Public Function Split (ParamArray separator As Char()) As String(). The ParamArray parameter option allows you to pass a comma delimited list of values into the function, similar to how String.Format works with a dynamic # of replacement values. http://msdn.microsoft.com/en-us/library/538f81ec.aspx
Step 3 code becomes "11121314151617181920".Split(new Char() {CChar("20")})(0), which using literal values is "11121314151617181920".Split(new Char() {"2"c})(0). The result is {"111", "13141516171819", "0"}. Get the 0th position, returns "111".
So to avoid confusion, you should convert your code to use the same version of Split on both sides.
Either of the 2 examples below should work:
Example 1: Using Microsoft.VisualBasic.Strings.Split:
Split( Split(testString,"10")(1), "20" )(0)
Example 2: Using System.String.Split:
testString _
.Split(New String() {"10"}, StringSplitOptions.None)(1) _
.Split(New String() {"20"}, StringSplitOptions.None)(0)