Have been at this for a while so I think it's time for some much needed help. I have the following if statement in an action listener('but' - being the Search button that gets clicked for the following to happen).I have an arrayList(used to save data from database) in my class, calling a specific record works perfectly. I am trying to create a search textField(named 'tf' below). I have been able to search only by typing an exact match to what ever is being searched, but I want to be able to use partial matching on any String in tf. Everything that's commented out is different stuff that I have been testing, therefore can be ignored.
The statement:
if( srch.getText().equals("You have chosen to search by movie: " )
... this is used to make sure that the person is searching only when the panel includes this title, as there is a genre search as well on the same panel. Only the title changes.
if(source.equals(but)){
//String b = a.get(0).get(1);
String result = tf.getText();
Pattern pat = Pattern.compile("[a-z]+");
Matcher m = pat.matcher(result);
//boolean bool = m.matches();
int i = 0;
//al.contains(result) && pat.matcher(a.get(i).get()).matches())
// && z.contains(result)
if( srch.getText().equals("You have chosen to search by movie: " )){
for (ArrayList<String> al : a){
if (pat.m(a.get(i).get(0)).matches()){
System.out.println(a.get(i).get(0)); //movie name
System.out.println(a.get(i).get(1)); //movie desc
}
i++;
}
//ta.setText(b);
}
else{
ta.setText("Please try searching by movie");
}
}
In summary, everything works fine. I just need a regex code to find partial matches as well and a way to add that into my loops. I've added as little code as possible as not to waste anyone's time, so please let me know if any other is needed.Many thanks in advance. Eventually any full or partial matches will display the movie name and description.
I think there is no need of regex.
String.contains(String chars) returns true if the source string have any occurrence of the argument string.
For example:
String str = "ABCDEFG";
if(str.contains("CD")){
System.out.println("Present");
}
Related
I would like to compare and match exactly one word (characters and length) between two strings.
This is what I have:
String wordCompare = "eagle:1,3:7;6\nBasils,45673:ewwsk\nlola:flower:1:2:b";
String lolo = scanner.nextLine();
if ( motCompare.toLowerCase().indexOf(lolo.toLowerCase()) != -1 ) {
System.out.println("Bingo !!!");
} else {
System.out.println("not found !!!");
}
If I type eagle:1,3:7;6 it should display Bingo !!!
If I type eagle:1,3 it still displays Bingo !!! which is wrong, it should display Not found.
If I type eagle:1,3:7;6 Basils,45673:ewwsk or eagle:1,3:7;6\nBasils,45673:ewwsk it should also display Not Found. Length of the typed word should be acknowledged between \n.
If I type Basils,45673:ewwsk, it displays bingo !!!
It looks like what you're wanting is an exact match, with the words being split by the newline character. With that assumption in mind, I would recommend splitting the string out into an array and then loading that into a HashSet like so:
boolean search(String wordDictionary, String search){
String[] options = wordDictionary.split("\n");
HashSet<String> searchSet = new HashSet<String>(Arrays.asList(options));
return searchSet.contains(search);
}
If the search function returns true, it has found whatever word you're searching for, if not, it hasn't.
Installing it in your code will look something like this:
String wordCompare = "eagle:1,3:7;6\nBasils,45673:ewwsk\nlola:flower:1:2:b";
String lolo = scanner.nextLine();
if(search(wordCompare, lolo))
System.out.println("Bingo!!!");
else
System.out.println("Not found.");
(For the record, you'd probably be better off with more clear variable names)
As #Grey has already mentioned within his answer, since you have a newline tag (\n) between your phrases you can Split the String using the String.split() method into a String Array and then compare the elements of that Array for equality with what the User supplies.
The code below is just another example of how this can be done. It also allows for the option to Ignore Letter case:
boolean ignoreCase = false;
String userString = "Basils,45673:ewwsk";
String isInString = "'" + userString + "' Was Not Found !!!";
String wordCompare = "eagle:1,3:7;6\nBasils,45673:ewwsk\nlola:flower:1:2:b";
String[] tmp = wordCompare.split("\n");
for (int i = 0; i < tmp.length; i++) {
// Ternary used for whether or not to ignore letter case.
if (!ignoreCase ? tmp[i].trim().equals(userString) :
tmp[i].trim().equalsIgnoreCase(userString)) {
isInString = "Bingo !!!";
break;
}
}
System.out.println(isInString);
Thank you,
The thing is I am not allowed to use regular expression nor tables.
so basing on your suggestions I made this code :
motCompare.toLowerCase().indexOf(lolo.toLowerCase(), ' ' ) != -1 ||
motCompare.toLowerCase().lastIndexOf(lolo.toLowerCase(),' ' ) != -1)
as a condition for a do while loop.
Could you please confirm if it is correct ?
Thank you.
I'm sure this is fairly simple, however I've tried googling the question but can't find an answer that fits my problem.
I'm playing around with string manipulation and one of the things I'm trying to do is get the first letter of each word. (And then place them all into a string)
I'm having trouble with registering each 'space' so that my If statement will be triggered. Here's what I have so far.
while (scanText.hasNext()) {
boolean isSpace = false;
if (scanText.hasNext(" ")) {isSpace = true;}
String s = scanText.next();
if (isSpace) {firstLetters += s + " ";}
}
Also, if there is a much better way to do this then please let me know
You can also split the original text by spaces, and collect the words.
String input = " Hello world aaa ";
String[] split = input.trim().split("\\s+"); // all types of whitespace; " +" to pick spaces only
// operate on "split" array containing words now: [Hello, world, aaa]
However using regexps here might be overkill.
Assuming that scanText is a Scanner object, you could use something like stated on the documentation:
Scanner s = new Scanner(input).useDelimiter("\\s+"); //regex for spaces
https://docs.oracle.com/javase/7/docs/api/java/util/Scanner.html
I have this code which searches a string array and returns the result if the input string matches the 1st characters of a string:
for (int i = 0; i < countryCode.length; i++) {
if (textlength <= countryCode[i].length()) {
if (etsearch
.getText()
.toString()
.equalsIgnoreCase(
(String) countryCode[i].subSequence(0,
textlength))) {
text_sort.add(countryCode[i]);
image_sort.add(flag[i]);
condition_sort.add(condition[i]);
}
}
}
But i want to get those string also where the input string matches not only in the first characters but also any where in the string? How to do this?
You have three way to search if an string contain substring or not:
String string = "Test, I am Adam";
// Anywhere in string
b = string.indexOf("I am") > 0; // true if contains
// Anywhere in string
b = string.matches("(?i).*i am.*"); // true if contains but ignore case
// Anywhere in string
b = string.contains("AA") ; // true if contains but ignore case
I have not enough 'reputation points' to reply in the comments, but there is an error in the accepted answer. indexOf() returns -1 when it cannot find the substring, so it should be:
b = string.indexOf("I am") >= 0;
Check out the contains(CharSequence) method
Try this-
etsearch.getText().toString().contains((String) countryCode[i]);
You could use contains. As in:
myString.contains("xxx");
See also: http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/String.html, specifically the contains area.
Use the following:
public boolean contains (CharSequence cs)
Since: API Level 1
Determines if this String contains the sequence of characters in the CharSequence passed.
Parameters
cs the character sequence to search for.
Returns
true if the sequence of characters are contained in this string, otherwise false
Actually, the default Arrayadapter class in android only seems to search from the beginning of whole words but by using the custom Arrayadapter class, you can search for parts of the arbitrary string. I have created the whole custom class and released the library. It is very easy to use. You can check the implementation and usage from here or by clicking on the link provided below.
https://github.com/mohitjha727/Advanced-Search-Filter
You can refer to this simple example-
We have names " Mohit " and "Rohan" and if We put " M" only then Mohit shows up in search result, but when we put "oh" then both Mohit and Rohan show up as they have the common letter 'oh'.
Thank You.
So I want to search through a string to see if it contains the substring that I'm looking for. This is the algorithm I wrote up:
//Declares the String to be searched
String temp = "Hello World?";
//A String array is created to store the individual
//substrings in temp
String[] array = temp.split(" ");
//Iterates through String array and determines if the
//substring is present
for(String a : array)
{
if(a.equalsIgnoreCase("hello"))
{
System.out.println("Found");
break;
}
System.out.println("Not Found");
}
This algorithm works for "hello" but I don't know how to get it to work for "world" since it has a question mark attached to it.
Thanks for any help!
Take a look:
http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/String.html#contains(java.lang.CharSequence)
String.contains();
To get a containsIgnoreCase(), you'll have to make your searchword and your String toLowerCase().
Take a look at this answer:
How to check if a String contains another String in a case insensitive manner in Java?
return s1.toLowerCase().contains(s2.toLowerCase());
This will also be true for:
war of the worlds, because it will find world. If you don't want this behavior, youll have to change your method like #Bart Kiers said.
Split on the following instead:
"[\\s?.!,]"
which matches any space char, question mark, dot, exclamation or a comma (add more chars if you like).
Or do a temp = temp.toLowerCase() and then temp.contains("world").
You dont have to do this, it's already implemented:
IndexOf and others
You may want to use :
String string = "Hello World?";
boolean b = string.indexOf("Hello") > 0; // true
To ignore case, regular expressions must be used .
b = string.matches("(?i).*Hello.*");
One more variation to ignore case would be :
// To ignore case
b=string.toLowerCase().indexOf("Hello".toLowerCase()) > 0 // true
Hey everyone,
I'm having a minor difficulty setting up a regular expression that evaluates a sentence entered by a user in a textbox to keyword(s). Essentially, the keywords have to be entered consecutive from one to the other and can have any number of characters or spaces before, between, and after (ie. if the keywords are "crow" and "feet", crow must be somewhere in the sentence before feet. So with that in mind, this statement should be valid "blah blah sccui crow dsj feet "). The characters and to some extent, the spaces (i would like the keywords to have at least one space buffer in the beginning and end) are completely optional, the main concern is whether the keywords were entered in their proper order.
So far, I was able to have my regular expression work in a sentence but failed to work if the answer itself was entered only.
I have the regular expression used in the function below:
// Comparing an answer with the right solution
protected boolean checkAnswer(String a, String s) {
boolean result = false;
//Used to determine if the solution is more than one word
String temp[] = s.split(" ");
//If only one word or letter
if(temp.length == 1)
{
if (s.length() == 1) {
// check multiple choice questions
if (a.equalsIgnoreCase(s)) result = true;
else result = false;
}
else {
// check short answer questions
if ((a.toLowerCase()).matches(".*?\\s*?" + s.toLowerCase() + "\\s*?.*?")) result = true;
else result = false;
}
}
else
{
int count = temp.length;
//Regular expression used to
String regex=".*?\\s*?";
for(int i = 0; i<count;i++)
regex+=temp[i].toLowerCase()+"\\s*?.*?";
//regex+=".*?";
System.out.println(regex);
if ((a.toLowerCase()).matches(regex)) result = true;
else result = false;
}
return result;
Any help would greatly be appreciated.
Thanks.
I would go about this in a different way. Instead of trying to use one regular expression, why not use something similar to:
String answer = ... // get the user's answer
if( answer.indexOf("crow") < answer.indexOf("feet") ) {
// "correct" answer
}
You'll still need to tokenize the words in the correct answer, then check in a loop to see if the index of each word is less than the index of the following word.
I don't think you need to split the result on " ".
If I understand correctly, you should be able to do something like
String regex="^.*crow.*\\s+.*feet.*"
The problem with the above is that it will match "feetcrow feetcrow".
Maybe something like
String regex="^.*\\s+crow.*\\s+feet\\s+.*"
That will enforce that the word is there as opposed to just in a random block of characters.
Depending on the complexity Bill's answer might be the fastest solution. If you'd prefer a regular expression, I wouldn't look for any spaces, but word boundaries instead. That way you won't have to handle commas, dots, etc. as well:
String regex = "\\bcrow(?:\\b.*\\b)?feet\\b"
This should match "crow bla feet" as well as "crowfeet" and "crow, feet".
Having to match multiple words in a specific order you could just join them together using '(?:\b.*\b)?' without requiring any additional sorting or checking.
Following Bill answer, I'd try this:
String input = // get user input
String[] tokens = input.split(" ");
String key1 = "crow";
String key2 = "feet";
String[] tokens = input.split(" ");
List<String> list = Arrays.asList(tokens);
return list.indexOf(key1) < list.indexOf(key2)