I want to test in Selenium (Java, Testng)
If my actual value is equal to one of the two values because my value can be Value1 or Value2 and both values will be correct.
If I want to assert only one equality I use this construction:
String expectedPopUpV1 = "Value1";
String expectedPopUpV2 = "Value2";
String actualPopUp = driver.findElement(By.cssSelector(Value)).getText();
Assert.assertEquals(actualPopUp,expectedPopUp);
but what should I do if I want to make something like
Assert.assertEquals(actualPopUp,expectedPopUp1||expectedPopUp1);
You can use assertTrue(boolean condition, String message) for this and give the details in the message
boolean isEqual = actualPopUp.equals(expectedPopUpV1) || actualPopUp.equals(expectedPopUpV2);
Assert.assertTrue(isEqual, "The text " + actualPopUp + " is not " + expectedPopUpV1 + " or " + expectedPopUpV2);
below option should also work using ternary operator:
Assert.assertEquals(expectedPopUpV1.equalsIgnoreCase(actualPopUp )?expectedPopUpV1:expectedPopUpV2 , actualPopUp );
There isn't any such method available. The two closest options I can think of:
if(!actualPopUp.equals(expectedPopUp1) || !actualPopUp.equals(expectedPopUp2) {
Assert.fail("insert failure messsage here");
}
or
if(actualPopUp.equals(expectedPopUp1) || actualPopUp.equals(expectedPopUp2) {
Assert.assertTrue("insert message");
}
The other option would be to really extend the Assert capability by constructing your own with a Builder pattern to be in the tune of:
AssertBuilder.with(actualPopUp).equals(expectedPopUp1).or(expectedPopUp2);
But that may be too complicated for a simple use case.
Can someone help me here? I dont understand where's the problem...
I need check if a String have more than 1 char like 'a', if so i need replace all 'a' for a empty space, but i still want only one 'a'.
String text = "aaaasomethingsomethingaaaa";
for (char c: text.toCharArray()) {
if (c == 'a') {
count_A++;//8
if (count_A > 1) {//yes
//app crash at this point
do {
text.replace("a", "");
} while (count_A != 1);
}
}
}
the application stops working when it enters the while loop. Any suggestion? Thank you very much!
If you want to replace every a in the string except for the last one then you may try the following regex option:
String text = "aaaasomethingsomethingaaaa";
text = text.replaceAll("a(?=.*a)", " ");
somethingsomething a
Demo
Edit:
If you really want to remove every a except for the last one, then use this:
String text = "aaaasomethingsomethingaaaa";
text = text.replaceAll("a(?=.*a)", "");
You can also do it like
String str = new String ("asomethingsomethingaaaa");
int firstIndex = str.indexOf("a");
firstIndex++;
String firstPart = str.substring(0, firstIndex);
String secondPart = str.substring(firstIndex);
System.out.println(firstPart + secondPart.replace("a", ""));
Maybe I'm wrong here but I have a feeling your talking about runs of any single character within a string. If this is the case then you can just use a little method like this:
public String removeCharacterRuns(String inputString) {
return inputString.replaceAll("([a-zA-Z])\\1{2,}", "$1");
}
To use this method:
String text = "aaaasomethingsomethingaaaa";
System.out.println(removeCharacterRuns(text));
The console output is:
asomethingsomethinga
Or perhaps even:
String text = "FFFFFFFourrrrrrrrrrrty TTTTTwwwwwwooo --> is the answer to: "
+ "The Meeeeeaniiiing of liiiiife, The UUUniveeeerse and "
+ "Evvvvverything.";
System.out.println(removeCharacterRuns(text));
The console output is........
Fourty Two --> is the answer to: The Meaning of life, The Universe and Everything.
The Regular Expression used within the provided removeCharacterRuns() method was actually borrowed from the answers provided within this SO Post.
Regular Expression Explanation:
Hi that println giving me error but it working fine if I just print them separately. Any Idea why this happening?
class StringTesting
{
public static void main(String me[])
{
String s1="Varun";
String s2="varun";
String s3="Varun";
String s4=new String("Varun");
String s5=new String("Varun");
System.out.println(" "+s1==s3+" "+s1==s2+" ");//here its giving me error
}
}
And also thanks in Advance :)
In terms of operators, + has the precedence on ==.
It means that this code :
" "+s1==s3+" "+s1==s2+" "
is translated by the compiler by something like that :
" Varun"=="Varun Varun"=="Varun "
the first part " Varun"=="Varun Varun" produces a boolean
and applying to it the "Varun " String with the == comparator, it results to :
booleanExpression == "Varun "
Comparing boolean with a String is not valid.
Anyway, even if it was, you don't want to as you want to output the boolean result of the == comparison between the objects.
So just put between parentheses the == comparisons to indicate to the compiler that operations in have to be evaluated together :
" "+(s1==s3)+" "+(s1==s2)+" "
It will produce a valid concatenation :
String + boolean + String + boolean + String
According to the Java documentation on String, the proper way to compare two strings is by using the .equals method.
System.out.println("s1 == s2 ?: " + s1.equals(s2))
By using the == operator between two Objects you are comparing memory locations, not the actual primitive value(s) contained in the object.
You can't compare booleans and strings using == and this is what is indeed happening there because of the order of operators being applied.
You can fix that using simple brackets:
System.out.println(" " + (s1 == s3) + " " + (s1 == s2) + " ");
Of course, if you do not combine results of two different operators, that will work just fine:
System.out.println(s1 == s3);
So, it was enough to just read the compilation error.
How would I check to ensure that a string does not contain any permutations of the space character in Java?
i.e., I want to check if my string equals "", " ", " ", etc.
I do not want to just check if my string contains a space, because if I had a string "My code works", it should not be parsed. I want to check only for strings that contain exclusively spaces, and no letters or other characters whatsoever.
Is this possible to do in a single if-statement?
Use String#matches() with the pattern \s*. This pattern will match empty string or any amount of continuous whitespace.
String input = " ";
if (input.matches("\\s*")) {
System.out.println("match");
}
If possible, you can also use StringUtils.isBlank(aString) provided by apache commons.
Checks if a CharSequence is whitespace, empty ("") or null.
StringUtils.isBlank(null) = true
StringUtils.isBlank("") = true
StringUtils.isBlank(" ") = true
StringUtils.isBlank(" ") = true
StringUtils.isBlank("bob") = false
StringUtils.isBlank(" bob ") = false
one more way
if(input.trim().length() == 0) {
...
}
This worked best for me:
for (i = 0; i < string.length(); i++) {
string = string.get(i).replaceAll(" ", "");
}
if (string.equals("")) {
//do something
}
I am trying to concatenate strings in Java. Why isn't this working?
public class StackOverflowTest {
public static void main(String args[]) {
int theNumber = 42;
System.out.println("Your number is " . theNumber . "!");
}
}
You can concatenate Strings using the + operator:
System.out.println("Your number is " + theNumber + "!");
theNumber is implicitly converted to the String "42".
The concatenation operator in java is +, not .
Read this (including all subsections) before you start. Try to stop thinking the php way ;)
To broaden your view on using strings in Java - the + operator for strings is actually transformed (by the compiler) into something similar to:
new StringBuilder().append("firstString").append("secondString").toString()
There are two basic answers to this question:
[simple] Use the + operator (string concatenation). "your number is" + theNumber + "!" (as noted elsewhere)
[less simple]: Use StringBuilder (or StringBuffer).
StringBuilder value;
value.append("your number is");
value.append(theNumber);
value.append("!");
value.toString();
I recommend against stacking operations like this:
new StringBuilder().append("I").append("like to write").append("confusing code");
Edit: starting in java 5 the string concatenation operator is translated into StringBuilder calls by the compiler. Because of this, both methods above are equal.
Note: Spaceisavaluablecommodity,asthissentancedemonstrates.
Caveat: Example 1 below generates multiple StringBuilder instances and is less efficient than example 2 below
Example 1
String Blam = one + two;
Blam += three + four;
Blam += five + six;
Example 2
String Blam = one + two + three + four + five + six;
Out of the box you have 3 ways to inject the value of a variable into a String as you try to achieve:
1. The simplest way
You can simply use the operator + between a String and any object or primitive type, it will automatically concatenate the String and
In case of an object, the value of String.valueOf(obj) corresponding to the String "null" if obj is null otherwise the value of obj.toString().
In case of a primitive type, the equivalent of String.valueOf(<primitive-type>).
Example with a non null object:
Integer theNumber = 42;
System.out.println("Your number is " + theNumber + "!");
Output:
Your number is 42!
Example with a null object:
Integer theNumber = null;
System.out.println("Your number is " + theNumber + "!");
Output:
Your number is null!
Example with a primitive type:
int theNumber = 42;
System.out.println("Your number is " + theNumber + "!");
Output:
Your number is 42!
2. The explicit way and potentially the most efficient one
You can use StringBuilder (or StringBuffer the thread-safe outdated counterpart) to build your String using the append methods.
Example:
int theNumber = 42;
StringBuilder buffer = new StringBuilder()
.append("Your number is ").append(theNumber).append('!');
System.out.println(buffer.toString()); // or simply System.out.println(buffer)
Output:
Your number is 42!
Behind the scene, this is actually how recent java compilers convert all the String concatenations done with the operator +, the only difference with the previous way is that you have the full control.
Indeed, the compilers will use the default constructor so the default capacity (16) as they have no idea what would be the final length of the String to build, which means that if the final length is greater than 16, the capacity will be necessarily extended which has price in term of performances.
So if you know in advance that the size of your final String will be greater than 16, it will be much more efficient to use this approach to provide a better initial capacity. For instance, in our example we create a String whose length is greater than 16, so for better performances it should be rewritten as next:
Example optimized :
int theNumber = 42;
StringBuilder buffer = new StringBuilder(18)
.append("Your number is ").append(theNumber).append('!');
System.out.println(buffer)
Output:
Your number is 42!
3. The most readable way
You can use the methods String.format(locale, format, args) or String.format(format, args) that both rely on a Formatter to build your String. This allows you to specify the format of your final String by using place holders that will be replaced by the value of the arguments.
Example:
int theNumber = 42;
System.out.println(String.format("Your number is %d!", theNumber));
// Or if we need to print only we can use printf
System.out.printf("Your number is still %d with printf!%n", theNumber);
Output:
Your number is 42!
Your number is still 42 with printf!
The most interesting aspect with this approach is the fact that we have a clear idea of what will be the final String because it is much more easy to read so it is much more easy to maintain.
The java 8 way:
StringJoiner sj1 = new StringJoiner(", ");
String joined = sj1.add("one").add("two").toString();
// one, two
System.out.println(joined);
StringJoiner sj2 = new StringJoiner(", ","{", "}");
String joined2 = sj2.add("Jake").add("John").add("Carl").toString();
// {Jake, John, Carl}
System.out.println(joined2);
You must be a PHP programmer.
Use a + sign.
System.out.println("Your number is " + theNumber + "!");
"+" instead of "."
Use + for string concatenation.
"Your number is " + theNumber + "!"
This should work
public class StackOverflowTest
{
public static void main(String args[])
{
int theNumber = 42;
System.out.println("Your number is " + theNumber + "!");
}
}
For exact concatenation operation of two string please use:
file_names = file_names.concat(file_names1);
In your case use + instead of .
For better performance use str1.concat(str2) where str1 and str2 are string variables.
String.join( delimiter , stringA , stringB , … )
As of Java 8 and later, we can use String.join.
Caveat: You must pass all String or CharSequence objects. So your int variable 42 does not work directly. One alternative is using an object rather than primitive, and then calling toString.
Integer theNumber = 42;
String output =
String // `String` class in Java 8 and later gained the new `join` method.
.join( // Static method on the `String` class.
"" , // Delimiter.
"Your number is " , theNumber.toString() , "!" ) ; // A series of `String` or `CharSequence` objects that you want to join.
) // Returns a `String` object of all the objects joined together separated by the delimiter.
;
Dump to console.
System.out.println( output ) ;
See this code run live at IdeOne.com.
In java concatenate symbol is "+".
If you are trying to concatenate two or three strings while using jdbc then use this:
String u = t1.getString();
String v = t2.getString();
String w = t3.getString();
String X = u + "" + v + "" + w;
st.setString(1, X);
Here "" is used for space only.
In Java, the concatenation symbol is "+", not ".".
"+" not "."
But be careful with String concatenation. Here's a link introducing some thoughts from IBM DeveloperWorks.
You can concatenate Strings using the + operator:
String a="hello ";
String b="world.";
System.out.println(a+b);
Output:
hello world.
That's it
So from the able answer's you might have got the answer for why your snippet is not working. Now I'll add my suggestions on how to do it effectively. This article is a good place where the author speaks about different way to concatenate the string and also given the time comparison results between various results.
Different ways by which Strings could be concatenated in Java
By using + operator (20 + "")
By using concat method in String class
Using StringBuffer
By using StringBuilder
Method 1:
This is a non-recommended way of doing. Why? When you use it with integers and characters you should be explicitly very conscious of transforming the integer to toString() before appending the string or else it would treat the characters to ASCI int's and would perform addition on the top.
String temp = "" + 200 + 'B';
//This is translated internally into,
new StringBuilder().append( "" ).append( 200 ).append('B').toString();
Method 2:
This is the inner concat method's implementation
public String concat(String str) {
int olen = str.length();
if (olen == 0) {
return this;
}
if (coder() == str.coder()) {
byte[] val = this.value;
byte[] oval = str.value;
int len = val.length + oval.length;
byte[] buf = Arrays.copyOf(val, len);
System.arraycopy(oval, 0, buf, val.length, oval.length);
return new String(buf, coder);
}
int len = length();
byte[] buf = StringUTF16.newBytesFor(len + olen);
getBytes(buf, 0, UTF16);
str.getBytes(buf, len, UTF16);
return new String(buf, UTF16);
}
This creates a new buffer each time and copies the old content to the newly allocated buffer. So, this is would be too slow when you do it on more Strings.
Method 3:
This is thread safe and comparatively fast compared to (1) and (2). This uses StringBuilder internally and when it allocates new memory for the buffer (say it's current size is 10) it would increment it's 2*size + 2 (which is 22). So when the array becomes bigger and bigger this would really perform better as it need not allocate buffer size each and every time for every append call.
private int newCapacity(int minCapacity) {
// overflow-conscious code
int oldCapacity = value.length >> coder;
int newCapacity = (oldCapacity << 1) + 2;
if (newCapacity - minCapacity < 0) {
newCapacity = minCapacity;
}
int SAFE_BOUND = MAX_ARRAY_SIZE >> coder;
return (newCapacity <= 0 || SAFE_BOUND - newCapacity < 0)
? hugeCapacity(minCapacity)
: newCapacity;
}
private int hugeCapacity(int minCapacity) {
int SAFE_BOUND = MAX_ARRAY_SIZE >> coder;
int UNSAFE_BOUND = Integer.MAX_VALUE >> coder;
if (UNSAFE_BOUND - minCapacity < 0) { // overflow
throw new OutOfMemoryError();
}
return (minCapacity > SAFE_BOUND)
? minCapacity : SAFE_BOUND;
}
Method 4
StringBuilder would be the fastest one for String concatenation since it's not thread safe. Unless you are very sure that your class which uses this is single ton I would highly recommend not to use this one.
In short, use StringBuffer until you are not sure that your code could be used by multiple threads. If you are damn sure, that your class is singleton then go ahead with StringBuilder for concatenation.
First method: You could use "+" sign for concatenating strings, but this always happens in print.
Another way: The String class includes a method for concatenating two strings: string1.concat(string2);
import com.google.common.base.Joiner;
String delimiter = "";
Joiner.on(delimiter).join(Lists.newArrayList("Your number is ", 47, "!"));
This may be overkill to answer the op's question, but it is good to know about for more complex join operations. This stackoverflow question ranks highly in general google searches in this area, so good to know.
you can use stringbuffer, stringbuilder, and as everyone before me mentioned, "+". I'm not sure how fast "+" is (I think it is the fastest for shorter strings), but for longer I think builder and buffer are about equal (builder is slightly faster because it's not synchronized).
here is an example to read and concatenate 2 string without using 3rd variable:
public class Demo {
public static void main(String args[]) throws Exception {
InputStreamReader r=new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(r);
System.out.println("enter your first string");
String str1 = br.readLine();
System.out.println("enter your second string");
String str2 = br.readLine();
System.out.println("concatenated string is:" + str1 + str2);
}
}
There are multiple ways to do so, but Oracle and IBM say that using +, is a bad practice, because essentially every time you concatenate String, you end up creating additional objects in memory. It will utilize extra space in JVM, and your program may be out of space, or slow down.
Using StringBuilder or StringBuffer is best way to go with it. Please look at Nicolas Fillato's comment above for example related to StringBuffer.
String first = "I eat"; String second = "all the rats.";
System.out.println(first+second);
Using "+" symbol u can concatenate strings.
String a="I";
String b="Love.";
String c="Java.";
System.out.println(a+b+c);