Eclipse Else Syntax Error
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Eclipse Syntax Error Parameterized Types
Join them; it only takes a minute: Sign up Syntax error on token “else”, delete this [closed] up vote 0 down vote favorite I keep on getting this error and I tried mixing it around. But then when
Eclipse Syntax Error On Token(s) Misplaced Construct(s)
I choose the option, it does the option, but then said "you did not enter 1, 2 or 3". This is the the full code. How to fix it? The error is at } else { System.out.println("You did not enter 1, 2 or 3"); } else { System.out.println("The Pin You Entered Was Wrong"); } java if-statement syntax share|improve this question edited Nov 17 '14 at 2:41 Andrew T. 3,62051636 asked Aug 12 '13 at 10:45 user2674614 1113 closed eclipse syntax error on tokens delete these tokens as off-topic by devnull, Tala, Toto, Beryllium, fedorqui Aug 12 '13 at 14:08 This question appears to be off-topic. The users who voted to close gave this specific reason:"Questions asking for code must demonstrate a minimal understanding of the problem being solved. Include attempted solutions, why they didn't work, and the expected results. See also: Stack Overflow question checklist" – devnull, Tala, Toto, Beryllium, fedorquiIf this question can be reworded to fit the rules in the help center, please edit the question. 1 Please format your code (to remove all of the blank space) and then post the entire block, not an external link. –chrylis Aug 12 '13 at 10:50 look at my answer, you do not use the option parameter right! –No Idea For Name Aug 12 '13 at 10:53 1 The if-then and if-then-else Statements –Ravi Thapliyal Aug 12 '13 at 10:53 add a comment| 6 Answers 6 active oldest votes up vote 1 down vote accepted The problem is in the code you provided on your paste bin. You use two else statements, so Java complains as it doesn't know which to go to after the initial if statement. You need to enter in another conditional statement using else if, then else. For example: if (option == 1){ Option_1 Optionone = new Option_1(); Optionone.Withdraw(); } etc }else if (nothing entered) { System.out.println("You did not ent
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Eclipse Syntax Error Insert To Complete Classbody
Meta Discuss the workings and policies of this site About Us eclipse javascript syntax error Learn more about Stack Overflow the company Business Learn more about hiring developers or posting ads with eclipse php syntax error us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just http://stackoverflow.com/questions/18185023/syntax-error-on-token-else-delete-this like you, helping each other. Join them; it only takes a minute: Sign up Syntax error on token “else” up vote 0 down vote favorite Please help, I want to make the button visible only if the checkbox is checked. But the token "else" gives me a syntax error. What do you think it could be? http://stackoverflow.com/questions/19546179/syntax-error-on-token-else For a while I thought it could be some bracket but I don't really know. package com.example.holaamigos; import android.app.Activity; import android.content.Intent; import android.os.Bundle; import android.view.Menu; import android.view.View; import android.view.View.OnClickListener; import android.widget.Button; import android.widget.CheckBox; import android.widget.CompoundButton; import android.widget.CompoundButton.OnCheckedChangeListener; import android.widget.EditText; import android.widget.Toast; public class MainActivity extends Activity { public final static String EXTRA_SALUDO = "com.example.holaamigos.SALUDO"; @Override protected void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.activity_main); final EditText txtNombre = (EditText)findViewById(R.id.TxtNombre); final Button btnHola = (Button)findViewById(R.id.BtnHola); final CheckBox checkbox1 =(CheckBox)findViewById(R.id.checkBox1); checkbox1.setOnCheckedChangeListener(new OnCheckedChangeListener(){ @Override public void onCheckedChanged(CompoundButton arg0, boolean checked) { if (checked) Toast.makeText(checkbox1.getContext(), "Activo", Toast.LENGTH_LONG).show(); btnHola.setVisibility(0); btnHola.setOnClickListener(new OnClickListener() { @Override public void onClick(View v) { Intent intent = new Intent(MainActivity.this, ActivitySaludo.class); String saludo = txtNombre.getText().toString(); intent.putExtra(EXTRA_SALUDO, saludo); startActivity(intent); else Toast.makeText(checkbox1.getContext(), "Inactivo", Toast.LENGTH_SHORT).show(); } }); } }); } @Override public boolean onCreateOptionsMenu(Menu menu) { // Inflate the menu; this adds items to the action bar if it is present. getMenuInflater().inflate(R.menu.main, menu); return true; } } java android eclipse share|improve this question edited Oct 23 '13 at 17:19 Zong Zheng Li 4,4
Programming Forum View Course 3683 points Submitted by Judy about 3 years ago if / else errors - learn how to fix these If you have a question that isn't answered here, please use the green "Ask a question" button. Copy/paste all https://www.codecademy.com/en/forum_questions/52373a75548c3515940000dc of your code in with your question, include the exercise number you are working on and the https://coderanch.com/t/551633/java/java/syntax-error-token exact error message you are seeing. Typically the following error messages are caused by faulty if/else syntax "syntax error" "missing operand; found else" "Unexpected token else" "expected an identifier and instead saw "else"" First off, a quick review of what if statements look like. A typical if () statement looks like this. Depending on what you are checking for, the else if and else blocks syntax error may not be needed. Note how the { }, ( ) and ; are used. if (this condition is true) { do something; } else if { do this other thing; } else if { do this awesome thing; } else { do something different; } Do a quick check to see if you haven't misspelled or mis-capitalized if or else. Using any of these is going to earn you an error: IF, If, ELSE, Else All good with your spelling and capitalization? Okay, syntax error on on to the next check. Is there an indicator complaining about your else? In JavaScript semicolons mark the end of statements. When you put a semicolon at the end of a condition, like this: if (choice1 === choice2); you are telling the interpreter to consider that as the end of your if statement. Any code following that will be processed without regard for the outcome of that if condition. If you are getting an error about the else it is because you've told the interpreter that the ; was the end of your if statement so when it finds the else a few lines later it starts complaining. A few examples of where not to put a semicolon: if (age < 18); if ( 9 > 10 ); if ("Yogi Bear".length < 3); if ("Jon".length * 2 / (2+1) === 4 ); if (userAnswer === "yes"); if (feedback > 8); if (income >= 100); I think you can see what I'm getting at. Checked for all those and still haven't found the bug yet? Have a look and make sure you don't have any semicolons on your else or your else if ( ), doing that will also end your if statement sooner than you had planned. These are both going to give you errors: } else if (computerChoice < 2/3); { else; Curious about other situations where semicolons should be used/not used? 34 votes permalink else is not allowed to have a condition @Albion
This Site Careers Other all forums Forum: Beginning Java syntax error on token "else" Abel Gonza Greenhorn Posts: 3 posted 5 years ago I can't seem to figure out why this is happening. Any tips at all will be very appreciated. syntax error on token "else", delete this token. I'll post all the code if requested. public class WhoWins { Pickrps picked = new Pickrps(); String theanswer = picked.getAnswer(); void whowon(){ if ("rock".equals(theanswer)||"Rock".equals(theanswer));{ System.out.println("rock"); } else if ("paper".equals(theanswer)||"Paper".equals(theanswer));{ System.out.println("paper"); } else ("scissors".equals(theanswer)||"Scissors".equals(theanswer));{ System.out.println("scissors"); } } } Rob Spoor Sheriff Posts: 20678 65 I like... posted 5 years ago Welcome to the Ranch! After the word "else" there shouldn't be any guard anymore - just the { or statement. That statement can be another "if", which is what I think you want: else if ("scissors".equals(theanswer)||"Scissors".equals(theanswer));{Also, are you aware of method equalsIgnoreCase? You can use that to shorten your code: else ("scissors".equalsIgnoreCase(theanswer));{ // likewise for the others SCJP 1.4 - SCJP 6 - SCWCD 5 - OCEEJBD 6 - OCEJPAD 6 How To Ask Questions How To Answer Questions Abel Gonza Greenhorn Posts: 3 posted 5 years ago Rob Spoor wrote:Welcome to the Ranch! After the word "else" there shouldn't be any guard anymore - just the { or statement. That statement can be another "if", which is what I think you want: else if ("scissors".equals(theanswer)||"Scissors".equals(theanswer));{Also, are you aware of method equalsIgnoreCase? You can use that to shorten your code: else ("scissors".equalsIgnoreCase(theanswer));{ // likewise for the others What do you mean guard? fred rosenberger lowercase baba Bartender Posts: 12219 36 I like... posted 5 years ago and else statement doesn't have a condition on it. The idea is that on the IF and IF-ELSE lines, you have handled most conditions. the ELSE is for everything that wasn't caught by the previous statements. the gene