Having weird problems with Part 1 solution and Part 2 solution conflicting. Any help is greatly appreciated!

Question

I'm having such a weird problem with this one.

To do part 1, I put a break in the last IF statement behind the print getReport function in order to exit the loop. This worked fine when printing any name or the 'sorry' message when the name is wrong, but it interferes with how the code runs when printing names that show up twice, so I removed it.

For part 2, I moved the creation of the 'report' variable outside of the getReport function, so it would exist globally, and just created a line at the end of the while loop to clear out the 'report' variable with each time the loop ran the prompt. This is where it gets weird. It seemed to work just fine when printing the name that shows up twice in the array (Jared), but for some reason now if you type in any other name from the directly that shows up only once in the array, it defaults to the "sorry that is not a student" text which is created with the ELSE statement.

If I remove the ELSE statement at the very end, everything works perfectly, but then it won't use the 'sorry' text, which means part 1 doesn't work anymore. I just can't figure it out. Not sure if it's and problem with the if/else, the loop, how I set up my functions...Any help?

var students = [
  {
    name: 'Patrick',
    track: 'Javascript',
    achievements: 24,
    points: 1587
  },
  {
    name: 'Jon',
    track: 'CSS',
    achievements: 8,
    points: 956
  },
  {
    name: 'Mike',
    track: 'Ruby',
    achievements: 10,
    points: 1254
  },
  {
    name: 'Jared',
    track: 'HTML',
    achievements: 4,
    points: 687
  },
  {
    name: 'Jared',
    track: 'Python',
    achievements: 7,
    points: 894
  },
];


//Search engine

var message = '';
var student;
var search;
var report = '';

//Creates a function that converts array into a string
function getReport(student){
  for ( var i = 0; i < students.length; i+= 1 ){ //Loop through all students and update report
    report += '<h2>Name: ' + student.name + '</h2>';
    report += '<p>Track: ' + student.track + '</p>';
    report += '<p>Achievements: ' + student.achievements + '</p>';
    report += '<p>Points: ' + student.points + '</p>';
    return report; //Add data to report and return report
  } 
};

//Creates a function to print to ID output
function print(message){
  var outputDiv = document.getElementById('output');
  outputDiv.innerHTML = message;
};

//Prompts user for response
while (true) {
  search = prompt('Search for a student. Type "quit" to exit.');
  if (search === null || search.toLowerCase() === 'quit'){
    break;
  } //If 'quit" exit the function
  for ( var i = 0; i < students.length; i += 1 ) { //Loop through all students to check names
    student = students[i];
    if (search.toLowerCase() === student.name.toLowerCase()) { //If name is correct print getReport function
      print(getReport(student));

      // break; ---- If I add this just this break, the program will print either the name or 'sorry' message correctly, but won't be able to print multiple names
    } 
      // else {print('Sorry, that is not a student.');}  ---- If I add just this else statement, the program will work fine if you look for the person listed twice, but it displays the sorry message for any other name regardless of wether they are in the array

  } report = ''; //Refresh the report with each prompt
}

Answer 1 · 2016-01-02T17:41:42Z

January 2, 2016 5:41pm

Okay, so this part gets a bit tricky.

When using your code with both the break and the else code, everything works fine apart from not printing both records when you have the same name more than once. This is because you're printing the first student record you find, and then immediately exiting.

The easiest way to fix this is a bit lazy. You can remove the break and the else, use the global message variable you declared (which you're not using right now), and then add a check at the end to see if message is empty. There is probably a more elegant way to do this, but this would be my first thought.

var message = '';
var student;
var search;
var report = '';

function getReport(student){
  for (var i = 0; i < students.length; i+= 1) {
    report = '<h2>Name: ' + student.name + '</h2>'; // You'll need to generate a new report.
    report += '<p>Track: ' + student.track + '</p>';
    report += '<p>Achievements: ' + student.achievements + '</p>';
    report += '<p>Points: ' + student.points + '</p>';
    return report;
  }; 
};

function print(message) {
  var outputDiv = document.getElementById('output');
  outputDiv.innerHTML = message;
};

while (true) {
  search = prompt('Search for a student. Type "quit" to exit.');

  if (search === null || search.toLowerCase() === 'quit'){
    break;
  };

  for (var i = 0; i < students.length; i += 1) {
    student = students[i];
    if (search.toLowerCase() === student.name.toLowerCase()) {
       message += getReport(student); // If there is a matching student, add it to the message.
    };
  }; 

  // This way you can print multiple students with the same name.
  if (message === '') {
    // If the message is empty, there was no student with a matching name.
    print('Sorry, that is not a student.');
  } else {
    print(message);
  };

  // Reset message instead of report.
  message = ''; 
};

Hopefully that helps give you an idea of what's going on. Try thinking of a way to make it more robust, elegant, and less lazy! Please let me know if you have any questions. Happy coding! :)

Answer 2 · 2016-01-14T21:05:36Z

January 14, 2016 9:05pm

Ooh ok, that makes a little more sense. I still can't say I fully grasp why it will only print the last name on the list though. If you leave out the break, and the loop does in fact evaluate every student, why does it only print it if you searched the last one? In other words, why is it that the match is over-written by the non-match unless you're searching for the last student on the list? What is the significance of it being the last item on the list?

Also, if you search for another name that is in fact on the list (though not the last name), how is it that it's overwritten at all? Maybe I'm confused about how if-else statements work, but if the "if" condition is met (which it should be if you type in any name from the list), why is it even running the "else" statement at all? I wasn't even sure why I'd need the break at all; I just tried it thinking it might work (which it sorta did, though it wouldn't work with printing multiple names), though I'm not sure why the break would even be necessary to exit the if-else statement. Thanks again for all your help. I'm sure eventually this will all sink in, and I'll understand what's going on.

Answer 3 · 2016-01-15T19:32:41Z

January 15, 2016 7:32pm

Okay, let's add some code so you can see what's happening as the program runs. Try using the following code I've modified from your original version. This does not include a break, and does have the else clause.

// For EACH (every) student in the list, do the following...
for (var i=0; i < students.length; i++) {
  student = students[i];
  if (search.toLowerCase() === student.name.toLowerCase()) {
    print(getReport(student));
    console.log("This is iteration number " + i + " and you found student: " + student);
  } else {
    print("Sorry, that is not a student.");
    console.log("This is iteration number " + i + " and you didn't find a student.");
  };
};

Then use the developer tools in your browser to take a look at the console as the program runs. Keep in mind that if the print function is called anytime after you printed getReport(student), then it will be overwritten with the else phrase.

I hope that being able to watch what's happening inside your for block will help make things clearer, but please let me know if you have any other lingering questions. It's great that you want to understand the way these blocks work thoroughly. :)

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Patrick Mead

Patrick Mead

Having weird problems with Part 1 solution and Part 2 solution conflicting. Any help is greatly appreciated!

3 Answers

rydavim

rydavim

Patrick Mead

Patrick Mead

rydavim

rydavim

Patrick Mead

Patrick Mead

rydavim

rydavim