I'm getting JCL error when executing this job.
// IF &O = O AND
// (&GEG = SD.SD.SYNC.DB2.LBER OR
// &GEG = SD.LA.SYNC.DB2.LBER OR
// &GEG = SD.TW.SYNC.DB2.LBER)
//STEP015 EXEC COZ,PNAME=IN0603ZC
Expected continuation not found. When I type '+' after every line for the IF, the total length is greater than 80 which also gives problems. Please help me :)
Per the z/OS MVS JCL syntax reference:
//[name] IF [(]relational-expression[)] THEN [comments]
.
. action when relational-expression is true
.
//[name] ELSE [comments]
.
. action when relational-expression is false
.
//[name] ENDIF [comments]
The IF statement consists of the characters // in columns 1 and 2 and
the five fields: name, operation (IF), the relational-expression, the
characters THEN, and comments. The relational-expression can be
enclosed in parentheses.
The ELSE statement consists of the characters // in columns 1 and 2
and the three fields: name, operation (ELSE), and comments.
The ENDIF statement consists of the characters // in columns 1 and 2
and the three fields: name, operation (ENDIF), and comments.
Your JCL is missing THEN
We had to create another solution. Like Bill Woodger said before: You can test only on numerics in a JCL. Thankx all for your help.
Related
I wrote code (based on How to collect data from multiple cells (quantity of 2) and return an output with 2 comma-separated values in one cell? AppScripts) that returns an output of comma separated values. It works great.
For example,
Input:
A1 = 1
B1 = 2
Results in Output:
1, 2
Here is that code:
[formSheet.getRange("E25").setValue(row[2]) + ", " +
formSheet.getRange("F25").setValue(row[2])],
The context is a spreadsheet-based Form. (No, the generic Google Forms does not work for our purposes).
Now I got myself in a pickle because users are requesting the Feature of being able to Retrieve and Edit their Form.
Ok. I can do that.
Until....
I get to those Output cells that are comma separated. Whoops. Is there a way to reverse this process, so that each value goes back into its respective Input cell?
Output (source):
1, 2
Retrieved Results:
A1 = 1
B1 = 2
Thanks!
Just split and setValues:
formSheet.getRange('E25:F25').setValues([
outputSheet.getRange('A25').getValue().split(', ')
])
I have to write a lex program that has these rules:
Identifiers: String of alphanumeric (and _), starting with an alphabetic character
Literals: Integers and strings
Comments: Start with ! character, go to until the end of the line
Here is what I came up with
[a-zA-Z][a-zA-Z0-9]+ return(ID);
[+-]?[0-9]+ return(INTEGER);
[a-zA-Z]+ return ( STRING);
!.*\n return ( COMMENT );
However, I still get a lot of errors when I compile this lex file.
What do you think the error is?
It would have helped if you'd shown more clearly what the problem was with your code. For example, did you get an error message or did it not function as desired?
There are a couple of problems with your code, but it is mainly correct. The first issue I see is that you have not divided your lex program into the necessary parts with the %% divider. The first part of a lex program is the declarations section, where regular expression patterns are specified. The second part is where the action that match patterns are specified. The (optional) third section is where any code (for the compiler) is placed. Code for the compiler can also be placed in the declaration section when delineated by %{ and %} at the start of a line.
If we put your code through lex we would get this error:
"SoNov16.l", line 1: bad character: [
"SoNov16.l", line 1: unknown error processing section 1
"SoNov16.l", line 1: unknown error processing section 1
"SoNov16.l", line 1: bad character: ]
"SoNov16.l", line 1: bad character: +
"SoNov16.l", line 1: unknown error processing section 1
"SoNov16.l", line 1: bad character: (
"SoNov16.l", line 1: unknown error processing section 1
"SoNov16.l", line 1: bad character: )
"SoNov16.l", line 1: bad character: ;
Did you get something like that? In your example code you are specifying actions (the return(ID); is an example of an action) and thus your code is for the second section. You therefore need to put a %% line ahead of it. It will then be a valid lex program.
You code is dependant on (probably) a parser, which consumes (and declares) the tokens. For testing purposes it is often easier to just print the tokens first. I solved this problem by making a C macro which will do the print and can be redefined to do the return at a later stage. Something like this:
%{
#define TOKEN(t) printf("String: %s Matched: " #t "\n",yytext)
%}
%%
[a-zA-Z][a-zA-Z0-9]+ TOKEN(ID);
[+-]?[0-9]+ TOKEN(INTEGER);
[a-zA-Z]+ TOKEN (STRING);
!.*\n TOKEN (COMMENT);
If we build and test this, we get the following:
abc
String: abc Matched: ID
abc123
String: abc123 Matched: ID
! comment text
String: ! comment text
Matched: COMMENT
Not quite correct. We can see that the ID rule is matching what should be a string. This is due to the ordering of the rules. We have to put the String rule first to ensure it matches first - unless of course you were supposed to match strings inside some quotes? You also missed the underline from the ID pattern. Its also a good idea to match and discard any whitespace characters:
%{
#define TOKEN(t) printf("String: %s Matched: " #t "\n",yytext)
%}
%%
[a-zA-Z]+ TOKEN (STRING);
[a-zA-Z][a-zA-Z0-9_]+ TOKEN(ID);
[+-]?[0-9]+ TOKEN(INTEGER);
!.*\n TOKEN (COMMENT);
[ \t\r\n]+ ;
Which when tested shows:
abc
String: abc Matched: STRING
abc123_
String: abc123_ Matched: ID
-1234
String: -1234 Matched: INTEGER
abc abc123 ! comment text
String: abc Matched: STRING
String: abc123 Matched: ID
String: ! comment text
Matched: COMMENT
Just in case you wanted strings in quotes, that is easy too:
%{
#define TOKEN(t) printf("String: %s Matched: " #t "\n",yytext)
%}
%%
\"[^"]+\" TOKEN (STRING);
[a-zA-Z][a-zA-Z0-9_]+ TOKEN(ID);
[+-]?[0-9]+ TOKEN(INTEGER);
!.*\n TOKEN (COMMENT );
[ \t\r\n] ;
"abc"
String: "abc" Matched: STRING
I want to use fscanf for reading a text file containing 4 rows with an unknown number of columns. The newline is represented by two consecutive spaces.
It was suggested that I pass : as the sizeA parameter but it doesn't work.
How can I read in my data?
update: The file format is
String1 String2 String3
10 20 30
a b c
1 2 3
I have to fill 4 arrays, one for each row.
See if this will work for your application.
fid1=fopen('test.txt');
i=1;
check=0;
while check~=1
str=fscanf(fid1,'%s',1);
if strcmp(str,'')~=1;
string(i)={str};
end
i=i+1;
check=strcmp(str,'');
end
fclose(fid1);
X=reshape(string,[],4);
ar1=X(:,1)
ar2=X(:,2)
ar3=X(:,3)
ar4=X(:,4)
Once you have 'ar1','ar2','ar3','ar4' you can parse them however you want.
I have found a solution, i don't know if it is the only one but it works fine:
A=fscanf(fid,'%[^\n] *\n')
B=sscanf(A,'%c ')
Z=fscanf(fid,'%[^\n] *\n')
C=sscanf(Z,'%d')
....
You could use
rawText = getl(fid);
lines = regexp(thisLine,' ','split);
tokens = {};
for ix = 1:numel(lines)
tokens{end+1} = regexp(lines{ix},' ','split'};
end
This will give you a cell array of strings having the row and column shape or your original data.
To read an arbitrary line of text then break it up according the the formating information you have available. My example uses a single space character.
This uses regular expressions to define the separator. Regular expressions powerful but too complex to describe here. See the MATLAB help for regexp and regular expressions.
I have a any challenge. I must write brainfuck-code.
For a given number n appoint its last digit .
entrance
Input will consist of only one line in which there is only one integer n ( 1 < = n < = 2,000,000,000 ) , followed by a newline ' \ n' (ASCII 10).
exit
On the output has to find exactly one integer denoting the last digit of n .
example I
entrance: 32
exit: 2
example II:
entrance: 231231132
exit: 2
This is what I tried, but it didn't work:
+[>,]<.>++++++++++.
The last input is the newline. So you have to go two memory positions back to get the last digit of the number. And maybe you don't have to return a newline character, so the code is
,[>,]<<.
Nope sorry, real answer is
,[>,]<.
because your answer was getting one too far ;)
Depending on the interpreter, you might have to escape the return key by yourself. considering the return key is ASCII: 10, your code should look like this :
>,----- -----[+++++ +++++>,----- -----]<.
broken down :
> | //first operation (just in case your interpreter does not
support a negative pointer index)
,----- ----- | //first entry if it's a return; you don't even get in the loop
[
+++++ +++++ | //if the value was not ASCII 10; you want the original value back
>, | //every next entry
----- ----- | //check again for the the return,
you exit the loop only if the last entered value is 10
]
<. | //your current pointer is 0; you go back to the last valid entry
and you display it
Your issue is that a loop continues for forever until at the end of the loop the cell the pointer is currently on in equal to 0. Your code never prints in the loop, and never subtracts, so your loop will never end, and all that your code does is take an ASCII character as input, move one forward, take an ASCII character as input, and so on. All of your code after the end of the loop is useless, because that your loop will never end.
This may be a very simple task for many but I could not find anything appropriate for me.
I have a file name: filenm_A006.2011.269.10.47.G25_2010
I want to separate all its parts (separated by . and _) to use them separately. How can I do it with simple matlab commands?
Kind Regards,
Mushi
I recommend regexp:
fname = 'filenm_A006.2011.269.10.47.G25_2010';
parts = regexp(fname, '[^_.]+', 'match');
parts =
'filenm' 'A006' '2011' '269' '10' '47' 'G25' '2010'
You can now refer to parts{1} through parts{8} for the pieces. Explanation: the regexp pattern [^_.] means all characters not equal to _ or ., and the + means you want groups of at least 1 character. Then 'match' asks the regexp function to return a cell array of the strings of all the matches of that pattern. There are other regexp modes; for example, the indices of each piece of the file.
Use the command
strsplit.
cellArrayOfParts = strsplit(fileName,{'.' '_'});
You can use strsplit to split it:
strsplit('filenm_A006.2011.269.10.47.G25_2010',{'_','.'})
ans =
'filenm' 'A006' '2011' '269' '10' '47' 'G25' '2010'
Another option is to use regexp, like Peter suggested.