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1,860 PointsQuiz keeps telling me I'm wrong about logging numbers, but the preview works fine
With this question: Create a for loop that logs the numbers 4 to 156 to the console. To log a value to the console use the console.log( ) method.
I used this script:
``javascript var html = '';
for (var i = 4; i <= 156; i += 1) { html += i + ' '; } console.log(html); `` In my log it shows the numbers from 4 to 156, but I keep getting this message:
Bummer! You need to log out EVERY number from 4 to 156 to the console. Your loop calls the console.log() method 1 times
Anybody can tell me where I'm going wrong?
var html = '';
for (var i = 4; i <= 156; i += 1) {
html += i + ' ';
}
console.log(html);
2 Answers
Michal Janek
Front End Web Development Techdegree Graduate 30,654 PointsThey need you to log out the numbers during the loop as a part of it. What you are doing is concatenating a string and storing it in a variable and after the loop is done then you console.log it out just once. While visually it may seem similar to what you are expected to do, functionally it is very different. So think about it again where is the difference or check the solution bellow.
This is what they are after:
for (var i = 4; i <= 156; i +=1 ) {
console.log(i);
}
or
for (var i = 4; i < 157; i++ ) {
console.log(i);
}
frits vd Velden
1,860 PointsSorry, I wish I could, but I haven't got the code anymore. Didn't save it.
frits vd Velden
1,860 Pointsfrits vd Velden
1,860 PointsHi Michal, I tried something very similar (without adding the variable) but it didn't work, that's why I thought; 'hey, this might work' so I ended up with the code above. Thanks for showing the correct solution!
Michal Janek
Front End Web Development Techdegree Graduate 30,654 PointsMichal Janek
Front End Web Development Techdegree Graduate 30,654 PointsWould you show us then what was your previous try? Someone (or me) may tell you why that did not work out either.