I have this expression to print the first N chars of the flow file content. The problem is when the content length is less than N. What's a good way to say "up to first N chars". That is I will not get an index out of bounds error when the size is less than N.
${incoming_content.0:substring(0,100)}
Here my N=100
Thank you.
If the content of your flow files are not too big, you can use ExtractText to convert contents of flowfile as attribute and then use UpdateAttribute processor to run following logic
${incoming.0:length():le(100):ifElse(${incoming.0},${incoming.0:substring(0,100)})}
which is essentially checks if length of flowfile contents (in attribute) is less than equals to 100, if true, return entire string as is, else return substring up to 100 characters
I was also getting the index out of bound error, and hence created the below expression:
${firstName:length():gt(0):ifElse(${firstName:substring(0,5)},'')}
:length() => fetching the length of string ,
:gt(0) => return boolean true if length is greater than 0 ,
:ifElse => If true, then get first 5 characters, Else return empty string
I have tested the above expression having firstName attribute value as null, "" and "very long name".
Here is the link to Nifi Expression guide
Related
In painless I would like to create a script which reads a keyword field called 'objldn' and extracts only five consecutive characters sometimes present in a precise position. Infact, in the keyword field 'objldn' there are a large variety of long strings among which there are some of them with a third underscore. After the third underscore, if it is present, I can fetch the consecutive 5 chars.
Whith the following lines of code I implement what I want:
def LU = doc['objldn'].value.splitOnToken('_');
return LU[3].substring(0, 5);
But the system returnes an error message "out of bounds":
Request error: array_index_out_of_bounds_exception, Index 3 out of
bounds for length 3 in "def LU =
doc['objldn'].value.splitOnToken('_'); ..." (Painless script)
error executing runtime field or scripted field on index pattern
return LU[3].substring(0, 5);
^---- HERE
may be it is due to the fact that many strings do not have the third underscore or do not even have one and therefore I need to implement firstly a IF statement which evaluates if a third underscore is in the string and only if it is present it proceeds to execute splitOnToken()... but I am not able to do it correctly. Can you help me to add the IF statement in the script please?
Why not simply checking the length of the LU array?
def LU = doc['objldn'].value.splitOnToken('_');
return LU.length >= 4 ? LU[3].substring(0, 5) : null;
I have enabled XPath 2.0 configuration synapse.xpath.dom.failover.enabled=true in synapse.properties but still unable to get string padding done. Is there any expression to achieve it?
Edit :
The length of a particular string needs to be 10 chars, if it is lesser than it, we have to pad it with the special character '%'.
Eg., Input = 'WSO2', after padding it should be 'WSO2%%%%%%'
Thanks in advance
This can be achieved using XPath 1.0 like so, assuming that "WSO2" will be replaced by dynamic input string in the actual implementation :
substring(concat('WSO2', '%%%%%%%%%%'), 1, 10)
The above XPath basically works by concatenating string of 10 specific for-padding characters to the original input string, and then substring the result to get only the first 10 characters. Found this trick in the following XSL question : XSL left-right justification with Padding
To put this in a more generic formula :
substring(concat('input_string', '%%%%....'), 1, n)
input_string : string to which padding operation will applied
% : character used for padding, repeated n times
n : fixed number of characters expected in the output string
The solution from #har07 is fine if you have a reasonable upper bound on the value of n, but if you don't, you can create a string containing '%' repeated $n times using
XPath 3.0: string-join((1 to $n)!"%")
XPath 2.0: string-join(for $x in 1 to $n return "%", "")
I have a string called indicators, that the original developer of this application used to store single characters to indicate certain components of a model. I need to change the 7th character in the string, which I tried to do with the following code:
indicators[6] = "R"
The problem, I discovered quickly, was that the string is not always 7 characters long. For example, I have one set of values with U 2, that I need to convert to U 2 R (adding an additional space after the 2). Is there an easy way to force character count with Ruby?
use String.ljust(integer, padstr=' ')
If integer is greater than the length of [the receiver], returns a new String of
length integer with [the return value] left justified and padded with padstr;
otherwise, returns [an unmodified version of the receiver].
indicators = indicators.ljust(7)
indicators[6] = "R"
I am working on a watermarking project that embeds binary values (i.e 1s and 0s) in the image, for which I have to take the input from the user, and check certain conditions such as
1) no empty string
2) no other character or special character
3) no number other than 0 and 1
is entered.
The following code just checks the first condition. Is there any default function in Matlab to check whether entered string is binary
int_state = get(handles.edit1,'String'); %edit1 is the Tag of edit box
if isempty(int_state)`
fprintf('Error: Enter Text first\n');
else
%computation code
end
There is no such standard function, but the check can be easily implemented.
Use this error condition:
isempty(int_state) || any(~ismember(int_state, '01'))
It returns false (no error) if the string is non-empty and composed of '0's and '1's only.
The function ismember returns a boolean array that indicates for every character in int_state whether it is contained in the second argument, '01'. The advantage is that this can be generalized to arbitrary sets of allowed characters.
I think the 2nd and 3rd can be combined together as 1 condition: your input string can only be a combination of 0 and 1? If it is so, then a small trick with findstr can do that:
if length(findstr(input_str, '1')) + length(findstr(input_str, '0')) == length(input_str)
condition_satisfied;
end
tf = isnumeric(A) returns true if A is a numeric array and false otherwise.
A numeric array is any of the numeric types and any subclasses of those types.
isnumeric(A)
ans =
1 (when A is numeric).
Dim str as String
str = "30 40 50 60"
I want to count the number of substrings.
Expected Output: 4
(because there are 4 total values: 30, 40, 50, 60)
How can I accomplish this in VB6?
You could try this:
arrStr = Split(str, " ")
strCnt = UBound(arrStr) + 1
msgBox strCnt
Of course, if you've got Option Explicit set (which you should..) then declare the variables above first..
Your request doesn't make any sense. A string is a sequence of text. The fact that that sequence of text contains numbers separated by spaces is quite irrelevant. Your string looks like this:
30 40 50 60
There are not 4 separate values, there is only one value, shown above—a single string.
You could also view the string as containing 11 individual characters, so it could be argued that the "count" of the string would be 11, but this doesn't get you any further towards your goal.
In order to get the result that you expect, you need to split the string into multiple strings at each space, producing 4 separate strings, each containing a 2-digit numeric value.
Of course, the real question is why you're storing this value in a string in the first place. If they're numeric values, you should store them in an array (for example, an array of Integers). Then you can easily obtain the number of elements in the array using the LBound() and UBound() functions.
I agree with everything Cody stated.
If you really wanted to you could loop through the string character by character and count the number of times you find your delimiter. In your example, it is space delimited, so you would simply count the number of spaces and add 1, but as Cody stated, those are not separate values..
Are you trying to parse text here or what? Regardless, I think what you really need to do is store your data into an array. Make your life easier, not more difficult.