I am testing string comparison with BRO, and got some runtime errors. Hope you guys could take a look and give me some hints.
For example i have two strings, let's say str_A and str_B, str_A is sort of a pattern, like: str_A = "\x13\x02\xf0\x80";
And str_B is a payload(contents) string from the function:
event tcp_packet(c: connection, is_orig: bool, flags: string, seq: count, ack: count, len: count, contents: string)
I compared the two of the strings with: if(str_A in str_B), which reduced the runtime errors like:
1467860547.182543 error: string with embedded NUL: "\x13\x00\xf0\x13"
1467860547.182543 error: string without NUL terminator: "\x13\x00\xf0\x13\x02\xf0\x80\x02\x00\x00\xc0\x01\x00\x00\x00\x00\x87\x02"
It looks like the 'x00' in the middle of the pattern string was considered as a terminator, and for the latter there wasn't a NUL at the end of the str_B.
So the (silly) question is how i could append a NUL at the end of str_B within BRO? and how to make BRO ignore the embeded NUL in the middle of a string when comparing? Many Thanks.
This was figured all right by translating(calling the function string_to_ascii_hex()) the hex-string into an ASCII-hex-string.
Related
I want to show a message of type E for which I have to first create a string. The string has mixed string and integer variables to be joined.
Since only strings can be concatenated, I copy integer variable into string variable, make a whole string and concatenate.
Is there a conversion function such as to_string(integer_variable) that can convert integers to string?
PROGRAM abc.
DATA: im_acc_no TYPE i VALUE 100,
lv_acc_no TYPE string,
lv_msg TYPE string.
START-OF-SELECTION.
lv_acc_no = im_acc_no.
CONCATENATE 'Acnt# ' lv_acc_no ' does not exist' INTO lv_msg.
MESSAGE lv_msg TYPE 'E'.
There is the CONV operator (SAP help) which can do something similar to to_string but it is not allowed in the CONCATENATE, so won't help you in your scenario.
You could use the && operator (SAP help) to create the message in-place in the MESSAGE command like:
MESSAGE |Acnt# | && lv_acc_no && | does not exist| type 'E'.
Side note: do not use this variant of the MESSAGE command, it might be easy to program but it makes it hard to investigate where a message is being generated. For this reason it is better to actually create a message in SE91 and use that. Variable replacements (&) in the message also handle integers just fine.
I tried many ways to get a single backslash from an executed (I don't mean an input from html).
I can get special characters as tab, new line and many others then escape them to \\t or \\n or \\(someother character) but I cannot get a single backslash when a non-special character is next to it.
I don't want something like:
str = "\apple"; // I want this, to return:
console.log(str); // \apple
and if I try to get character at 0 then I get a instead of \.
(See ES2015 update at the end of the answer.)
You've tagged your question both string and regex.
In JavaScript, the backslash has special meaning both in string literals and in regular expressions. If you want an actual backslash in the string or regex, you have to write two: \\.
The following string starts with one backslash, the first one you see in the literal is an escape character starting an escape sequence. The \\ escape sequence tells the parser to put a single backslash in the string:
var str = "\\I have one backslash";
The following regular expression will match a single backslash (not two); again, the first one you see in the literal is an escape character starting an escape sequence. The \\ escape sequence tells the parser to put a single backslash character in the regular expression pattern:
var rex = /\\/;
If you're using a string to create a regular expression (rather than using a regular expression literal as I did above), note that you're dealing with two levels: The string level, and the regular expression level. So to create a regular expression using a string that matches a single backslash, you end up using four:
// Matches *one* backslash
var rex = new RegExp("\\\\");
That's because first, you're writing a string literal, but you want to actually put backslashes in the resulting string, so you do that with \\ for each one backslash you want. But your regex also requires two \\ for every one real backslash you want, and so it needs to see two backslashes in the string. Hence, a total of four. This is one of the reasons I avoid using new RegExp(string) whenver I can; I get confused easily. :-)
ES2015 and ES2018 update
Fast-forward to 2015, and as Dolphin_Wood points out the new ES2015 standard gives us template literals, tag functions, and the String.raw function:
// Yes, this unlikely-looking syntax is actually valid ES2015
let str = String.raw`\apple`;
str ends up having the characters \, a, p, p, l, and e in it. Just be careful there are no ${ in your template literal, since ${ starts a substitution in a template literal. E.g.:
let foo = "bar";
let str = String.raw`\apple${foo}`;
...ends up being \applebar.
Try String.raw method:
str = String.raw`\apple` // "\apple"
Reference here: String.raw()
\ is an escape character, when followed by a non-special character it doesn't become a literal \. Instead, you have to double it \\.
console.log("\apple"); //-> "apple"
console.log("\\apple"); //-> "\apple"
There is no way to get the original, raw string definition or create a literal string without escape characters.
please try the below one it works for me and I'm getting the output with backslash
String sss="dfsdf\\dfds";
System.out.println(sss);
I am trying to take a string like "Hello my name is Nick" and transform it to "Hello+my+name+is+Nick" to be passed through a URL. This would be easily done by replacing all the spaces with a + char however I also need to replace all special characters (. , ! &) with their ASCII values. I have searched the net but cannot find anything. I wonder if anyone knows of existing code to do this as its a fairly common task?
I think you're looking for this: HttpUtility.UrlEncode Method (String)
Handles non-URL compliant characters and spaces.
I need my VB.net to write a file containing the following line
objWriter.WriteLine ("TEXTA " (FILEA) " TEXTB")
Unfortunatly the variable (FILEA) is causing problems i now get the error
Comma, ')', or valid expression continuation expected.
Could someone explain this please?
You're not concatenating (joining) the strings proerly...
objWriter.WriteLine ("TEXTA " & FILEA & " TEXTB")
A better style to get into the habit of using is:
objWriter.WriteLine (string.format("TEXTA {0} TEXTB", FILEA))
The FILEA variable replaces the {0} placeholder in the format string. Depending on what the writer you're using is, you may have a formatted overload so you could just do:
objWriter.WriteLine ("TEXTA {0} TEXTB", FILEA)
And since you asked for an explanation;
The compiler is asking you what exactly you want it to do - you've given it 3 variables (String, variable, String) and haven't told it that you want to join them together - It's saying that after the first string "TEXTA", there should either be the closing bracket (to end the method call), a comma (to pass another parameter to the method) OR a "valid continuation expression" - ie something that tells it what to do with the next bit. in this case, you want a continuation expression, specifically an ampersand to signify "concatenate with the next 'thing'".
Presumably you're looking for string concatenation? Try this:
objWriter.WriteLine("TEXTA" & FILEA & "TEXTB");
Note that FILEA isn't exactly a conventional variable name... which leads me to suspect there may be something else you're trying to achieve. Could you give more details?
I am trying to use a copy command for Windows and we have directories such as c:\oracle.
While trying to execute one such, we get the following error:
source_file=folder+"\"
^
SyntaxError: Lexical error at line 17, column 23. Encountered: "\r" (13), after : ""
Here folder is my path of c:\oracle and while trying to add file to it like:
source=folder+"\"+src_file
I am not able to do so. Any suggestion on how to solve this issue?
I tried with / but my copy windows calling source in os.command is getting "the syntax is incorrect" and the only way to solve it is to use \ but I am getting the above error in doing so.
Please suggest. Thanks for your help
Thanks.
Short answer:
You need:
source_file = folder + "\\" + src_file
Long answer:
The problem with
source_file = folder + "\" + src_file
is that \ is the escape character. What it's doing in this particular case is escaping the " so that it's treated as a character of the string rather than the string terminator, similar to:
source_file = folder + "X + src_file
which would have the same problem.
In other words, you're trying to construct a string consisting of ", some other text and the end of line (\r, the carriage return character). That's where your error is coming from:
Encountered: "\r" (13)
Paxdiablo is absolutely correct about why \ isn't working for you. However, you could also solve your problem by using os.path.normpath instead of trying to construct the proper platform-specific path characters yourself.
In all programming languages I know of, you can't put a quote inside a string like this: "this is a quote: "." The reason for this is that the first quote opens the string, the second then closes it (!), and then the third one opens another string - with the following two problems:
whatever is between the quotes #2 and #3 is probably not valid code;
the quote #3 is probably not being closed.
There are two common mechanisms of solving this: doubling and escaping. Escaping is far more common, and what it means is you put a special character (usually \) in front of characters that you don't want to be interpreted in their usual value. Thus, "no, *this* is a quote: \"." is a proper string, where the quote #2 is not closing the string - and the character \ does not appear.
However, now you have another problem - how do you actually make the escape character appear in a string? Simple: escape it! "This is an escape: \\!" is how you do it: the backslash #1 is the escape character, and the backslash #2 is the escapee: it will not be interpreted with its usual escape semantics, but as a simple backslash character.
Thus, your line should say this:
source=folder+"\\"+src_file
BTW: upvote for both #paxdiablo (who got in before my diatribe) and #Nick (who has a proper Pythonic way to do what you want to do)