Supposed I have this kind of loop in java (not an exact code, neither it's a truly valid code, but just to give you a general perspective of situation):
print "<ul>";
while (res = fetch(database)) {
print ("<li>" + res.col['data'] + "</li>");
}
print "</ul>";
and I have this CSS, which makes the last item on the list has red color.
ul li:last-child { color: red; }
this works fine on most browsers. the problems are:
I need to make this work on IE8 too.
IE8 doesn't support last-child.
I cannot test whether current iteration of "while" is entering last iteration or not. or let's say, there's no way I could check when the loop would end. this is not an ordinary loop, but let's just say like that. so I can't give the last <li> a class, say class="lastchild".
I also tried with javascript and jquery, and both can't select "last-child" either, as they depends on the css for the selection, I think.
how is the best approach in this situation? thanks.
print "<ul>";
String liText=null;
while (res = fetch(database)) {
if (liText!=null) {
print ("<li>" + liText + "</li>");
}
liText=res.col['data'];
}
if (liText!=null) {
print ("<li style='the last element style'>" + liText + "</li>");
}
print "</ul>";
Move the adding of the last line out of the while and apply desired style there
Assuming that the problem is in Java, I am wondering why there is no way to detect last record. Probably you need to reconsider you data structure first, populate the data in a Map or a List and then iterate over.
And, if that is not possible, you can possibly opt for a StringBuilder instnace to build the entire String html tags first by looping over and then replace the last li element with the desired style and after that put it in the print statement for rendering.
The above points are valid even if you are trying to code in javascript, with little bit of change.
Related
I am trying to count an element in an array of objects.
long number = Stream.of(jobTitle).count();
System.out.println("There are " + number + " employees.");
What happens is that it will print out the message as many times as many employees have the same job title. Yet "number" stays always 1.
Any guiding would be much appreciated.
long number = Stream.of(jobTitle).count();
Counts the elements in a stream that contains one element.
It is not surprising that this operation always ends up with the exact same result.
Your code is equivalent to:
List<Whatever> titels = new ArrayList<>();
titels.put(oneEntry);
... print titels.size()
Long story short: that statement is nonsensical. What you probably meant was:
if (arbetstitel.equalsIgnoreCase(jobCount)){
g++;
or something alikw. Of course g is a rather bad name for a counter.
But the real answer here is: step back. Think what the problem is you intend to solve, and what the elements are you need to look at. The code you are showing here is simply not making (much) sense. I can't tell you how to fix it, because, as said: it is not clear what you try to achieve here.
A streamish way of counting:
long usersWithMatchingTitle = Arrays.stream(employees).filter(e -> e.getJobTitle().equalsIgnoreCase(jobTitleFromUser)).count();
Meaning: instead of manually iterating your array, you can turn the whole array into a stream, and then filter/count whatever you want to.
Please note: your code seems to only care about the first 30 elements in that array. If that is really what you want, you will need ...stream(employees).limit(30)...
You need to change the stream of command to define a proper Predicate for filter option.
Stream.of(employees).filter(e -> e.getJobTitle().equals(jobTitle)).count();
im working on a problem where i have to obtain all permutations of an arraylist of numbers. The only restriction is that any number cant start with 0, so if we have [0,1,2] we would obtain
[1,2,0]
[1,0,2]
[2,0,1]
[2,1,0]
i know how to do this with 3 loops but the thing is that i have to repeat this to different sets of numbers with differentes sizes, so i need one method that i can apply to different sets of numbers but i have no clue on how to do this. I imagine i have to used some kind of recursive function but i dont know how to implement it so the numbers cant start with a 0. Any ideas? please dont just post the code i want to understand the problem, thank you in advantage!!
Curious question! Interesting code kata.
I naively think I would have a recursive method that takes:
a list of the items currently chosen by the caller
a set of the items available for the callee
The method would iterate over the set to chose 1 more item and call itself with the list extended by this item, and the set reduced by this item. Upon return, remove from list, add back to set and go on with next item (take a defensive copy of the set of course).
If the current list is empty, the selected first item cannot be 0, as per your rules. If you must collect the permutations somewhere (not just print), a 3rd argument would be required for a collection or an observer.
The recursion obvioulsy stops when the available set is empty, at which point the permutation is sent to the collection or observer.
If items can repeat, you may have benefit from sorting them first in order to skip electing the same item again at a given position.
Beware this quires a recursion depth of N, for N items. But the danger is minimal because even with N=10000, it may not stackoverflow, but the CPU time to complete would be order(N!) (probably end of universe...)
You could solve this recursively as described here: Permutation of an ArrayList of numbers using recursion.
The only thing that is missing there is your restriction with the zeros, which could be solved somehow like this (the loop is taken from the example above):
for (List<Integer> al : myLists) {
// The part you need to add:
if (al.get(0) == 0) {
continue;
}
String appender = "";
for (Integer i : al) {
System.out.print(appender + i);
appender = " ";
}
System.out.println();
}
You basically check the first element of each permutation and skip the ones with a leading zero. The continue jumps to the next iteration of the loop and therefore to the next permutation.
I have a string that needs to be compared to the names that are on the website. So the first thing I do is get the number of rows (because some arrays have more or fewer than 2 people in them) and then put that size into an int. String[] names come from the names that selenium is supposed to find when it goes to the website to execute this statement assertTrue(assertion.getText().contains(names[i-1])); The problem is: if the names do not appear in the order in which they appear in the array it breaks. In other words, if Mick Jagger is in li[1] and Keith Richards is in li[2], everything runs as expected. But if Keith Richards appears in li[1] it breaks. Furthermore, I am supposed to use the assertTrue command to do this. I have tried sorting, pushing whats on the web into a new ArrayList and I keep getting errors. Anyone know a good way to ensure the order isn't important and still use the assertTrue command?
Thanks,
Scott
WebElement assertion = null;
List<WebElement> assignees = driver.findElements(By.xpath(".//*[#id='assignee']/li"));
int count = assignees.size();
String[] names = {"Mick", "Keith"};
for (int i = 1; i < count; i++)
{
assertion = driver.findElement(By.xpath(".//*[#id='assignee']/li["+i+"]"));
assertTrue(assertion.getText().contains(names[i-1]));
If names represents the full string, you can just flip it. Make sure the text in your assertion (probably should be named something like assignee instead of assertion) is contained in your collection:
assertTrue(Arrays.asList(names).contains(assertion.getText());
Let me know if this won't work because a name is actually a subset of the text in assertion and I'll adjust the answer.
If they don't exactly match (which you have indicated they don't), you could use linq in c# to match this. Since you're using java you can use an additional loop. There may be a more efficient way to do this in java that I'm not aware of.
String assigneeText = assertion.getText();
boolean nameFound = false;
for(String name: names)
{
nameFound = assigneeText.contains(name);
if(nameFound)
{
break;
}
}
assertTrue(nameFound, "None of the expected names were found in the following assignee text: " + assigneeText);
I was having an error in which I couldn't cast an incoming variable to int. Changed it to Integer, and voilá: it's now working. I tried making a for loop with that variable (just testing purposes, you know), and it began throwing hundreds of errors, some of them NullPointerException.
Check the code:
<%
Integer number = (Integer)request.getAttribute("num");
System.out.println(number);
for(int i=0;i<number;i++){
System.out.println(i);
}
%>
In the very beginning, I was trying to replicate some <p>'s so I could see this works (I pass the variable from this same view to the controller, and retrieve it back to here). If I comment the for, the correct results appear in the System.out.println.
This doesn't work either if I put a static value (like 5 or so) in the conditional operator of the for loop.
Why can't I use a for in my view.jsp? Is there a reason for this?
The only place where you can get null objects in the code that you post is in the line
Integer number = (Integer)request.getAttribute("num");
If your request doesn't have this attribute set, you should either do nothing, e.g. wrap the rest of your code in an if(number == null) { ... } block or initialize it explicitly, e.g. if(number == null) number = 0;
However, it would be most elegant to not have any scripting blocks in your JSP - rather use taglibs and EL - but that wouldn't answer your exact question, just guide you towards better (maintainable) code.
Problem:
The problem here is that you are using System.out.println().
Explanation:
If you want to print results in the JSP page, you should use :
out.println(i);
Because if you write System.out.println(i); the results will be printed to the console.
Trying to perform a binary search on a sorted array of Book objects.
Its not working well, it returns the correct results for some of the objects, but not all.
I went through the loop on paper and it seems that a number can get missed out due to rounding #.5 upwards.
Any ideas how to make this work?
Book found = null;
/*
* Search at the center of the collection. If the reference is less than that,
* search in the upper half of the collection, else, search in the lower half.
* Loop until found else return null.
*/
int top = numberOfBooks()-1;
int bottom = 0;
int middle;
while (bottom <= top && found == null){
middle = (bottom + top)/2;
if (givenRef.compareTo(bookCollection.get(middle).getReference()) == 0) {
found = bookCollection.get(middle);
} else if (givenRef.compareTo(bookCollection.get(middle).getReference()) < 0){
bottom = middle + 1;
} else if (givenRef.compareTo(bookCollection.get(middle).getReference()) > 0){
top = middle - 1;
}
}
return found;
A couple suggestions for you:
there's no need to keep a Book variable. In your loop, just return the book when it's found, and at the end return null. And you can also remove the boolean check for the variable in the while condition.
the middle variable can be scoped inside the loop, no need to have it live longer.
you're doing bookCollection.get(middle).getReference() three times. Consider creating a variable and then using it.
the middle = (bottom + top)/2 is a classic mistake in binary search implementation algorithms. Even Joshua Bloch, who wrote the Java Collection classes, made that error (see this interesting blog post about it). Instead, use (bottom+top) >>> 1, to avoid integer overflow for very large values (you probably wouldn't encounter this error, but it's for the principle).
As for your actual problem statement, rounding would be downwards (integer division), not upwards. To troubleshoot the problem:
are you sure the numberOfBooks() method corresponds to the length of your collection?
are you sure the compareTo() method works as expected for the types you are using (in your code example we do not know what the getReference() return type is)
are you sure your collection is properly sorted according to getReference()?
and finally, are you sure that using givenRef.compareTo(bookCollection.get(middle).getReference()) < 0 is correct? In standard binary search implementations it would be reversed, e.g. bookCollection.get(middle).getReference().compareTo(givenRef) < 0. This might be what donroby mentions, not sure.
In any case, the way to find the error would be to try out different values and see for which the output is correct and for which it isn't, and thus infer what the problem is. You can also use your debugger to help you step through the algorithm, rather than using pencil and paper if you have to run many tests. Even better, as donroby said, write a unit test.
What about Collections.binarySearch()?
All of JRL's suggestions are right, but the actual fail is that your compares are reversed.
I didn't see this immediately myself, but replicating your code into a function (using strings instead of Books), writing a some simple Junit tests and then running them in the debugger made it really obvious.
Write unit tests!
I found the problem.
It turns out i was binary searching my bookCollection arrayList, and NOT the new sroted array i had created - sortedLib.
Silly mistake at my end, but thanks for the input and suggestions!