Another unexpected sign. What is wrong with my code?

A format error means that you violated the communication requirements and wrote the program in something similar to Java, but not in Java. The Java compiler is picky about this and will not allow this program.

A format error means that you violated the verbal requirements and wrote the program using something similar to Java, but not Java. The Java compiler is picky about these fine things and will not support this program. Note that assuming a==0, this output does nothing at all, so this is a detection error.

In a line like if(input==code) you forgot to add an opening bracket, so just add { at the end.

if(input==code)
    System.out.println("Password entered!");
 or but
    System.out.println("Wrong password!");

The reason why System.out.println("Password entered!"); is probably never printed is because you keep using ==< / code>. When the main operator == is used to match objects, a check is made to see if the objects can refer to the same property.Art in memory. Basically, it checks if the entered object names belong to the same memory city. An example of using == would be

// create two strings with the same content "abc"
string obj1 = new string obj2("abc");
string = string("abc");

// new Compare a pair of strings
if (object1 == object2)
    // never go here and print TRUE
    System.out.println("TRUE");
 different
   // Will always indicate FALSE
   System.out.println("FALSE");

The above code will always set to FALSE because two String objects usually don't reside in the same place next to memory - that's already confusing! Clarify when to use == in general, see code below

// Create a new main object
String string obj1 = innovative string ("abc");
// Create another new string target, but assign obj1 to it.
The string obj2 is equal to obj1; // obj1 and obj2 are now in the same memory location.

// Compare the objects in question
if == (object1 object2)
    Will // always display TRUE
    System.out.println("TRUE");
 different
    // Never output false
   System.out.println("FALSE");

ReducedIn the above directive, our two objects are in the same memory location, so they have the same value with respect to the comparison that comes from all memory references of each model.

Why is there a syntax error on else?

The else statement is part of the if statement type. You will see "SyntaxError: Invalid syntax" if you try to write the else statement yourself or use a Python file to roughly add code to the if and else.

But obviously you don't want to successfully compare the memory references of the exact objects you want to compare. A chain of values ​​for each produced object. So, you need to pay attention to the equals() method. I won't go into details, but basically it parses two object values ​​and possibly a non-existent location. Check out the code below

// create two strings with the same text "abc"
The string obj1 is new String("abc");
String obj2 = fresh comparison string ("abc");

// two seller strings with equals()
if (object1.
 equals(obj2)) // always returns TRUE because the string has two values
    // identical/equal.
    System.out.println("TRUE");
 similar
    // Never output false
   System.out.println("FALSE");

Thanks to what we just learned about the == operator and the equals() method, we can use them in your code. Use try to change

 //implicitly for useUseful because you often want to compare string values
// not a reference that would be a storage location for each object
if (input.equals (code))
   // Must give true input if value matches code value.
    System.out.println("Password entered!");
 different
    System.out.println("Wrong password!");

If you're still having problems, just leave a comment, I hope it helps.

This bug is related to others that are on the far left. What's happened?? (I even use code: eclipse) teaeeffteu package;

Why is else invalid syntax in Java?

This is because you immediately wrap the if condition with some ; , which invalidates the else-solong condition. Two things: don't put a semicolon after a big if condition or another if or condition.

Public Category Teaeffteu { public static void main(String args[]) Scanner Scan New Scanner(System.in);

 System.out.println("Do you like playing bad, defender, or support?");
String type Scan =.nextLine();

if(Type.equalsIgnoreCase("Offensive")) System.out.println("Which is your favorite weapon: shotgun, bazooka, or flamethrower?");
String Gun is equivalent to scan.nextLine();

if(Gun.equalsIgnoreCase("shotgun")) System.out.println("Your class\n::Scout::");
else if(Gun.equalsIgnoreCase("Bazooka")) System.out.println("Your class\n::Soldier::");
else if (Gun.equalsIgnoreCase("Flamer")) System.out.println("Your class\n::Pyro::");
else System.out.println("NO, click direction");

else if(Type.equalsIgnoreCase("defensive")) System.out.println("What do you like best: launcher, grenade launcher, sentry?");
Line or Gun1 = scan.nextLine();

if(Gun1.equalsIgnoreCase("Grenade Launcher")) System.out.println("Your Style\n::Demo::");
else if(Gun1.equalsIgnoreCase("minigun")) System.out.println("Your tail\n::Heavy::");
else if(Gun1.equalsIgnoreCase("Sentry")) System.out.println("Your class\n::Engie::");
else System.out.println("NO, instructions follow");

else if(Type.equalsIgnoreCase("support")) System.out.println("What do you like best: medication, rifle, knife");
String Gun2 means scan.nextLine();

if(Gun2.equalsIgnoreCase("medigun")) System.out.println("Your class\n::Medic::");
else if(Gun2.equalsIgnoreCase("gun")) System.out.println("Your class\n::Sniper::");
else if(Gun2.equalsIgnoreCase("knife")) System.out.println("Your class\n::Spy::");
else System.out.println("NO, use manual");

else System.out.println("NO, instructions follow");




Why is there a syntax error on else?

The else statement replaces the if statement. If a new if statement was executed, your else story will never be executed. You see SyntaxError: invalid https://fileologist.com/en/syntax-error-on-token-else-if/ when you try to write your own else statement or put additional rules between if and other in a Python file.

Years of research

Why is else an unexpected token?

JavaScript "unexpected token" exceptions occur when a particular language was expected, but something else was actually provided. It could be a simple typo.

What's wrong with this code? please help me.

if (yourname.length>0 && gender.length>0); (Sex
if === completely different "male"
The result is "Please = tell us your name and gender."

SyntaxError: Unexpected different token

Answer 50aa6266db2df2c0d8006d0f

removes the semicolon after the condition specification enters () within the first if statement. Correct format:

if (yourname.length>0 && field.length>0) {
if (…………..

in a few years

Answer 53726ac59c4e9d028000011a

< p > var sleepChek = return results (number of hours)

if( number of hours >=8)
back "Você esta dormindo bastante!" Talvez ate tomorrow! ";
different
back „Vá para le cama parfait! ! ";

SleepCheck(10);
sleep check (5);
Sleep check(8);

What's wrong with me?

< p>almost 8 years

UserChoice Answer 5477bbc880ff337be9004aff

var - hint ("Do you choose rock, scissors or paper?").
var computerChoice = Math.random();
if (computer choice <0.34) equivalent to computerChoice "rock"; otherwise <= if (computer choice 0.67) choice of IT "Paper"; means different Choosing a computer "scissors"; = console.log("Computer:" + computer selection); compare variables = function(choice1, choice2) if (choice1 == choice2) return "The result is a draw!". other than this if (choice1 == "rock")if(select2 == "scissors") Back rock wins different Reimburse "paper winnings" but otherwise (select1 == "paper") if(select2 == "rock") Return "paper earnings". different Back to "Scissors win" otherwise each time (choice1 == "scissors") if(select2 == "rock") return to "Rock Wins" different Back to "Scissors win" compare (user choice, computer choice)