i want to parse a text file that represents a log. i want it to be powerful enough to handle all erros that might occur.. although i am clueless about the best practices and the errors i should account for . i will be using JAVA to implement this.
Sample log :
2012-07-16 10:23:40,558 - 127.0.0.1 - Paremter array[param1=1,param2=1,param3=0,] - 383
I already wrote a prasing code that works as follows :
public Parser(String log) {
this.log = log;
this.parse();
}
public void parse() {
String[] temp = new String[10];
String[] temp2 = new String[10];
temp = log.split(" - ");
key = temp[3];
id = Integer.parseInt(key);
String IP = temp[1];
String str;
String temp3 = temp[2].substring(temp[2].indexOf("g"), temp[2].indexOf("]"));
temp = temp3.split(",");
str = "param1";
boolean ordered = CheckOrder(temp);
if (ordered) {
for (int q = 0; q < temp.length; q++) {
temp[q] = temp[q].substring(temp[q].indexOf("=") + 1);
}
if (temp[0].equals("q")) {
param= 0;
} else if (temp[0].equals("k")) {
param= 1;
} else {
param= 2;
}
// Same way for all parameters
}
}
Check the javadoc of all the methods you use, and make sure to handle all the nominal and exceptional cases:
the file doesn't exist: an exception is being thrown. Handle this exception correctly
String.indexOf() doesn't find what it looks for. It returns -1. Handle this case correctly
String.split() doesn't return an array of the length I expect. Handle this case correctly
...
Split your big method into several sub-methods, each doing only one thing.
Write unit tests to check that your methods do what they're supposed to do, with all the possible inputs.
Note that "handling things correctly" might very well mean: throw an exception because the input is incorrect, if the contract is that the logs follow a well-defined format. In this case, it's the code generating the logs that is incorrect. But it's better to have an exception telling which format you expected and which format you got instead, rather than an obscure NullPointerException or ArrayIndexOutOfBoundsException.
The above applies to any kind of code you write, and not just to file parsing.
Side note:
String[] temp = new String[10];
temp = log.split(" - ");
What's the point in creating an array of 10 elements to discard it right after and replace it by another array (the one returned by log.split(" - ")).
Related
I'm trying to understand what is the proper way to achieve the following goal. Consider the following string:
/some/path/to/some/dir
I would like to split the path by / and get the last two string and connect them with _ so the output would be:
some_dir
I'm familiar with the split function but I'm not sure what is the proper way to write this code when speaking of code-styling.
I know that I have to check first if the string is valid. For example, the string dir is not valid.
What is the proper way to solve it?
You can play with the following. I omit error checks for the sake of simplicity.
class Test {
public static void main(String[] args) {
String s = "/some/path/to/some/dir";
String[] parts = s.split("/");
int len = parts.length;
String theLastTwoParts = parts[len - 2] + "_" + parts[len - 1];
System.out.println(theLastTwoParts);
}
}
You can use the below shown function for this purpose:
public String convertPath(String path) {
String[] str = path.split("/");
int length = str.length;
if(length < 2) {
//Customize the result here for this specific case
return "";
}
return str[length-2] + "_" + str[length-1];
}
If you're actually handling paths, you probably want to use the standard library's Path ecosystem. You can use it by
Path path = Paths.get(p);
int nameCount = path.getNameCount();
if (nameCount < 2) throw new RuntimeException();
String result = String.format("%s_%s", path.getName(nameCount-2), path.getName(nameCount-1));
See it here.
The advantage is that when you're working on Windows, it will also handle the different path separator, so it's more platform independent.
The question of "dir" being "invalid" raises the follow-up question of how you want it handled. Throwing a RuntimeException like I do is probably not going to hold up.
I have a long String with binary values. And i have a hash map that has the Binary digits as a key and char as a value. I have a function that supposed to read the binary string using 2 pointers and compare with hashmap and store the corresponding char in main.decodedTxt. However, im getting string out of bound exception for this. I don't know how to solve this. I'm getting exception on "String temp =" line. I have a picture link of the console output to see better picture.
public static void bitStringToText (String binText){
String bs = binText;
int from =0;
int to = 1;
while(bs != null){
String temp = bs.substring(from, to);
if (main.newMapDecoding.containsKey(temp)){
main.decodedTxt += main.newMapDecoding.get(temp);
from =to;
to = from +1;
} else {
to = to + 1;
}
}
}
Image of console exception is here
First of all there is no need to check if bs is null because no part of your code changes the value of bs. Your current code will cross the possible index of your binText at some point. It's better to loop just binText and check if you find something within it. After all you have to traverse the complete string anyways. Change your code as follows
public static void bitStringToText (String binText){
//no need to do this if you are not modifying the contents of binText
//String bs = binText;
int from =0;
int to = 1;
int size = binText.length();
String temp = "";
while(to <= size ){
temp = binText.substring(from, to);
if (main.newMapDecoding.containsKey(temp)){
main.decodedTxt += main.newMapDecoding.get(temp);
from =to;
to = from +1;
} else {
to = to + 1;
}
}
}
Hope it helps.
First, give it a try to practice debugging. It is an easy case. Either use run in debug mode (place break point on String temp = bs.substring(from, to); line) or print values of from and to before the same line. It will help to understand what is going on.
Solution:
If bs is not null you will always have StringIndexOutOfBoundsException. Because you are not checking if to is pointing to not existed index of bs String. Easiest example of the first one will be empty String: bs == "".
One of the solution could be to replace condition in while to while (to <= bs.length()).
So recently I got invited to this google foo.bar challenge and I believe the code runs the way it should be. To be precise what I need to find is the number of occurrences of "abc" in a String. When I verify my code with them, I pass 3/10 test cases. I'm starting to feel bad because I don't know what I am doing wrong. I have written the code which I will share with you guys. Also the string needs to be less than 200 characters. When I run this from their website, I pass 3 tests and fail 7. Basically 7 things need to be right.
The actual question:
Write a function called answer(s) that, given a non-empty string less
than 200 characters in length describing the sequence of M&Ms. returns the maximum number of equal parts that can be cut from the cake without leaving any leftovers.
Example : Input : (string) s = "abccbaabccba"
output : (int) 2
Input: (string) s = "abcabcabcabc"
output : (int) 4
public static int answer(String s) {
int counter = 0;
int index;
String findWord ="ABC";
if(s!=null && s.length()<200){
s = s.toUpperCase();
while (s.contains(findWord))
{
index = s.indexOf(findWord);
s = s.substring(index + findWord.length(), s.length());
counter++;
}
}
return counter;
}
I see a couple of things in your code snippet:
1.
if(s.length()<200){
Why are you checking for the length to be lesser than 200? Is that a requirement? If not, you can skip checking the length.
2.
String findWord ="abc";
...
s.contains(findWord)
Can the test program be checking for upper case alphabets? Example: "ABC"? If so, you might need to consider changing your logic for the s.contains() line.
Update:
You should also consider putting a null check for the input string. This will ensure that the test cases will not fail for null inputs.
The logic of your code is well but on the other hand i found that you didn't check for if input string is empty or null.
I belief that google foo.bar wants to see the logic and the way of coding in a proper manner.
so don't be feel bad
I would go for a simpler approach
int beforeLen = s.length ();
String after = s.replace (findWord, "");
int afterLen = after.length ();
return (beforeLen - afterLen) / findWord.length ();
String pattern = "abc";
String line="<input text here>";
int i=0;
Pattern TokenPattern=Pattern.compile(pattern);
if(line!=null){
Matcher m=TokenPattern.matcher(line);
while(m.find()){
i++;
}}
System.out.println("No of occurences : "+ " "+i);
put declaration of index out before while block, isn't never good re-declare the same variable n time.
int index;
while (s.contains(findWord))
{
index = s.indexOf(findWord);
....
}
I hope this help
Update:
try to compact your code
public static int answer(String s) {
int counter = 0;
int index;
String findWord = "ABC";
if (s != null && s.length() < 200) {
s = s.toUpperCase();
while ((index = s.indexOf(findWord)) > -1) {
s = s.substring(index + findWord.length(), s.length());
counter++;
}
}
return counter;
}
Update:
The logic seems good to me, I'm still try to improve the performance, if you can try this
while ((index = s.indexOf(findWord, index)) > -1) {
//s = s.substring(index + findWord.length(), s.length());
index+=findWord.length();
counter++;
}
I would like to ask you for your help, regarding this code. I am trying to do a kind of encoding of particular words, such as "Microsoft" etc. (random ones, just to learn the technique). I've suceeded in doing everything, but to make this kinf of searching for words case insesitive. Here is the code:
public class BannedWords {
public static String returnStars(int length){
String stars = "";
String addStar = "*";
for (int i = 1; i<=length; i++){
stars += addStar;
}
return stars;
}
public static void main(String[] args) {
String textString = "Microsoft announced its next generation Java compiler today."
+ " It uses advanced parser and special optimizer for the Microsoft JVM.";
StringBuilder text = new StringBuilder(textString);
String bannedWords = "Java, JVM, Microsoft";
String [] bWordsArr = bannedWords.split("[, ]+");
for(int i = 0; i<bWordsArr.length; i++){
int index = textString.indexOf(bWordsArr[i]);
while(index != -1){
text = text.replace(index, index+bWordsArr[i].length(), returnStars(bWordsArr[i].length()));
index = textString.indexOf(bWordsArr[i], index +1);
}
}
System.out.println(text.toString());
}
}
I need to search for "Java", "JVM" and "Microsoft" words regardless of their case, even if we try "MiCrosoFt" it should work, but after a few hours thinking and trying to do it with using toUpperCase(), toLowerCase(), I couldn't find out how to do that. Do you have any ideas ?
Thank you beforehand ! :)
When using indexOf(), toLowerCase() will convert checked text to lowercase. Then, you must put your search terms in lowercase.
String text = "Java is a good programming language.";
int index = text.toLowerCase().indexOf("java");
You can also use toUpperCase(), simply put your search terms in uppercase.
Actually you should be using equalsIgnoreCase. This is the correct way of comparing strings irrespective of their case. And also you don't have to modify the original string to upper or lower case to perform check. I hope it helps :)
yourString.equalsIgnoreCase(anotherString)
This may be a simple question, but I have been Googling for over an hour and haven't found an answer yet.
I'm trying to simply use the String.split() method with a small Android application to split an input string. The input string will be something along the lines of: "Launch ip:192.168.1.101;port:5900". I'm doing this in two iterations to ensure that all of the required parameters are there. I'm first trying to do a split on spaces and semicolons to get the individual tokens sorted out. Next, I'm trying to split on colons in order to strip off the identification tags of each piece of information.
So, for example, I would expect the first round of split to give me the following data from the above example string:
(1) Launch
(2) ip:192.168.1.101
(3) port:5900
Then the second round would give me the following:
(1) 192.168.1.101
(2) 5900
However, the following code that I wrote doesn't give me what's expected:
private String[] splitString(String inputString)
{
String[] parsedString;
String[] orderedString = new String[SOSLauncherConstants.SOCKET_INPUT_STRING_PARSE_VALUE];
parsedString = inputString.trim().split("; ");
Log.i("info", "The parsed data is as follows for the initially parsed string of size " + parsedString.length + ": ");
for (int i = 0; i < parsedString.length; ++i)
{
Log.i("info", parsedString[i]);
}
for (int i = 0; i < parsedString.length; ++i )
{
if (parsedString[i].toLowerCase().contains(SOSLauncherConstants.PARSED_LAUNCH_COMMAND_VALUE))
{
orderedString[SOSLauncherConstants.PARSED_COMMAND_WORD] = parsedString[i];
}
if (parsedString[i].toLowerCase().contains("ip"))
{
orderedString[SOSLauncherConstants.PARSED_IP_VALUE] = parsedString[i].split(":")[1];
}
else if (parsedString[i].toLowerCase().contains("port"))
{
orderedString[SOSLauncherConstants.PARSED_PORT_VALUE] = parsedString[i].split(":")[1];
}
else if (parsedString[i].toLowerCase().contains("username"))
{
orderedString[SOSLauncherConstants.PARSED_USERNAME_VALUE] = parsedString[i].split(":")[1];
}
else if (parsedString[i].toLowerCase().contains("password"))
{
orderedString[SOSLauncherConstants.PARSED_PASSWORD_VALUE] = parsedString[i].split(":")[1];
}
else if (parsedString[i].toLowerCase().contains("color"))
{
orderedString[SOSLauncherConstants.PARSED_COLOR_VALUE] = parsedString[i].split(":")[1];
}
}
Log.i("info", "The parsed data is as follows for the second parsed string of size " + orderedString.length + ": ");
for (int i = 0; i < orderedString.length; ++i)
{
Log.i("info", orderedString[i]);
}
return orderedString;
}
For a result, I'm getting the following:
The parsed data is as follows for the parsed string of size 1:
launch ip:192.168.1.106;port:5900
The parsed data is as follows for the second parsed string of size 6:
launch ip:192.168.1.106;port:5900
192.168.1.106;port
And then, of course, it crashes because the for loop runs into a null string.
Side Note:
The following snippet is from the constants class that defines all of the string indexes --
public static final int SOCKET_INPUT_STRING_PARSE_VALUE = 6;
public static final int PARSED_COMMAND_WORD = 0;
public static final String PARSED_LAUNCH_COMMAND_VALUE = "launch";
public static final int PARSED_IP_VALUE = 1;
public static final int PARSED_PORT_VALUE = 2;
public static final int PARSED_USERNAME_VALUE = 3;
public static final int PARSED_PASSWORD_VALUE = 4;
public static final int PARSED_COLOR_VALUE = 5;
I looked into needing a possible escape (by inserting a \\ before the semicolon) on the semicolon delimiter, and even tried using it, but that didn't work. The odd part is that neither the space nor the semicolon function as a delimiter, yet the colon works on the second time around. Does anybody have any ideas what would cause this?
Thanks for your time!
EDIT: I should also add that I'm receiving the string over a WiFi socket connection. I don't think this should make a difference, but I'd like you to have all of the information that you need.
String.split(String) takes a regex. Use "[; ]". eg:
"foo;bar baz".split("[; ]")
will return an array containing "foo", "bar" and "baz".
If you need groups of spaces to work as a single delimiter, you can use something like:
"foo;bar baz".split("(;| +)")
I believe String.split() tries to split on each of the characters you specify together (or on a regex), not each character individually. That is, split(";.") would not split "a;b.c" at all, but would split "a;.b".
You may have better luck with Guava's Splitter, which is meant to be slightly less unpredictable than java.lang.String.split.
I would write something like
Iterable<String> splits = Splitter.on(CharMatcher.anyOf("; ")).split(string);
but Splitter also provides fluent-style customization like "trim results" or "skip over empty strings."
Is there a reason why you are using String.split(), but not using Regular Expressions? This is a perfect candidate for regex'es, esp if the string format is consistent.
I'm not sure if your format is fixed, and if it is, then the following regex should break it down for you (am sure that someone can come up with an even more elegant regex). If you have several command strings that follow, then you can use a more flexible regex and loop over all the groups:
Pattern p = Pattern.compile("([\w]*)[ ;](([\w]*):([^ ;]*))*");
Matcher m = p.match( <input string>);
if( m.find() )
command = m.group(1);
do{
id = m.group(3);
value = m.group(4);
} while( m.find() );
A great place to test out regex'es online is http://www.regexplanet.com/simple/index.html. It allows you to play with the regex without having to compile and launch you app every time if you just want to get the regex correct.