javascript use console.log without moving to new line - javascript

I am new to js and hope this is not too trivial, but I am unable to find any help on the net.
I wish to output to console.log and prevent moving to a new line, so the next time the output will be appended to the same line. ie,
"use strict";
for (let i = 0; i<=9;i++){
console.log(i); // here i would like to freeze the output so the result is 0123456789 on one line, rather than those digits in a column.
}
I have seen fixes involving assigning the outputs to a string and printing in 1 hit, but that seems incredibly crude. Even in Fortran 4 as I recall in the '70s, you could prevent moving to a new line before printing again, so I think I am missing something fundamental. Also I cannot find any general help on formatting numerical output in javascript. Can someone point me in the right direction?
Thanks

Unfortunately, the console.log() method will only write out a string to a single line and doesn't support the appending behavior you are looking for.
As you detailed in your original post, you could accomplish writing the final result out through the use of a variable (i.e. displaying the final concatenated string), but not continually appending to the same line within the console itself as the loop is being iterated over.
Alternative Grouping Option
The concept of grouping entries is supported, which is obviously very different than your original ask, but it may be worth considering as mentioned in the documentation for console.group() and might look something like this:
var rollingConcatenation = '';
console.group("Looping Group Example");
for (let i = 0; i<=9;i++){
rollingConcatenation += i;
console.log(rollingConcatenation);
}
console.groupEnd();
This can give your console the following appearance, which can help with readability (depending on your use cases):
Do It Yourself Implementation
Another option might be to store your current console value within a variable and at clear it and rewrite the updated values out. Depending on your very specific use cases, you could achieve the behavior you are looking for using something like this crude implementation:
// Define a custom console
var customConsole = {
// Store a reference to your backing value
tempValue: '',
// Always write out the most recent value
log: function(msg) {
this.tempValue += msg;
console.clear();
console.log(this.tempValue);
},
// A clear method to clear the backing console
clear: function() {
this.tempValue = '';
console.clear();
}
}
for (var i = 0; i < 10; i++) {
// Use your custom console instead of the normal one
customConsole.log(i);
}

Take a new variable outside the loop and then prepare that string inside the loop and then you can console.log() outside the loop.
var str = '';
for (let i = 0; i <= 9; i++) {
str += i;
}
console.log(str);

Related

How do i fix the wrong integration of an automatic transfer of rows between sheets?

What I'm trying to accomplish is:
have a data input sheet called 'data' (its data is fed by a form)
script moves the information from data to sheet1/sheet2/.../sheetn (according to a string that is to be found in column 3)
script also deletes moved rows
I think the deleteRow command works fine, i suspect the culprit being the detection of the string in the array.
I've already used the search a lot, tried a few codes and I've identified this as the most probable candidate (its by cooper), as it's almost doing what i need it to do.
I tried logging a bit, but unfortunately i dont know too much about coding yet. Currently im learning by trial and error.
If i log for vals[i][2] i only get 1 string, instead of a few from my example input.
When i set only one targetsheet (sh1) and target-term it works. but when i extend it it doesnt work anymore.
{
var ss=SpreadsheetApp.getActive();
var sh0=ss.getSheetByName('Data');
var rg0=sh0.getDataRange();
var sh1=ss.getSheetByName('Applesheet');
var sh2=ss.getSheetByName('Banana');
var sh3=ss.getSheetByName('Cherry');
var vals=rg0.getValues();
Logger.log(vals)
for(var i=vals.length-1;i>0;i--)
{
if(vals[i][2]=='Apple')
Logger.log("PV Abfrage -", vals[i][2])
{
sh1.appendRow(vals[i]);
sh0.deleteRow(i+1);
}
if(vals[i][2]=='Banana') //also tried with else if here
{
sh2.appendRow(vals[i]);
sh0.deleteRow(i+1);
}
if(vals[i][2]=='Cherry')
{
sh3.appendRow(vals[i]);
sh0.deleteRow(i+1);
}
}
}
My code moves rows that dont contain any of the terms.
It's also supposed to only move rows that contain this term, but its doing so super unrealiably.
I think all rows get appended to Applesheet, rows that contain banana are moved to banana but the ones with cherry wont.
Im definitely not experienced enough to judge, but this code seems a bit unreliable, because even my test version with just one if fails to perform the way i want it to.
Issue:
Your first if statement is forced to return true by the Logger.log() you've included between if and {. As soon as you remove it, your code functions exactly as you're expecting.
Example:
If we run the following script:
var check = ['Apple', 'Pear', 'Fruit'];
for (i = 0; i < check.length; i++) {
if (check[i] === 'Apple') {
console.log('Found!');
}
}
We're looping through an array, and logging "Found!" for every time the item in the array is found. This is the same way your script works. It works as expected, "Apple" is only found once in the array, so the log looks like this:
Found!
As soon as we put a log between the if and the {, like so:
var check = ['Apple', 'Pear', 'Fruit'];
for (i = 0; i < check.length; i++) {
if (check[i] === 'Apple')
console.log("Oops!")
{
console.log('Found!');
}
}
We get the following:
Oops!
Found!
Found!
Found!
Summary:
Make sure to only include your conditions for the if statement between your if and {, adding anything else can return false positives like you've experienced today.
Reference:
JavaScript if Statements

object with array type string causing a typeError

The best way to explain this is just to show you.
var condition = 70;
var formnames = new Array("wheelcheckbox1", "wheelcheckbox2","spokecheckbox","spokecheckbox2","tirecheckbox","tirecheckbox2","tirecheckbox3");
formnames.forEach(function(entry) {
console.log(obj.entry);
if(obj.entry == "") {
condition = condition - 10;
}
});
as you can see I used the console log to show how it needs to work
as that works perfect, however, using the array causes an error as
they're strings and not what the obj wants, it wants text without it being a string.
Any ideas?
for..in should not be used to iterate over an array. Consider using forEach instead.

Trying to make sense of "this" in my javascript code (one thing works, the other doesn't)

I've been trying to learn javascript by refactoring some Jquery examples in a book into javascript. In the following code I add a click listener to a tab and make it change to active when the user clicks on the tab.
var tabs = document.querySelectorAll(".tabs a span");
var content = document.querySelectorAll("main .content li");
for (var tabNumber = 0; tabNumber <= 2; tabNumber++) {
tabs[tabNumber].addEventListener("click", function (event) {
for (var i = 0; i < tabs.length; i++) {
tabs[i].classList.remove("active");
}
tabs[tabNumber].classList.add("active");
for (var i = 0; i < content.length; i++) {
content[i].innerHTML = "";
}
event.preventDefault();
});
}
This returns an undefined error when I run it. However, I tried replacing tabs[tabNumber].classList.add("active") with this.classList.add("active") and it worked.
Why doesn't the previous code work? As far as I can see they are referring to the same thing, and tabs[tabNumber] should work since at that point in the code it is tabs[0].
If use this, I think it's better and a more polished solution. If you still want to use tabNumber, it's probably evaluating to 3 in every click callback, because it's the number after the last iteration, and you don't have a tabs[3] position.
So, you just have to make a closure of the tabNumber variable.
I guess other answers told you why tabs[tabNumber] does not work (because it comes from the score of the for loop and so, is always equal to the greater value of tabNumber).
That's why I would recommend using a .forEach loop. Careful though because it doesn't work on arrays of DOM nodes produced by document.querySelectorAll(), but you can use:
// ES6
Array.from(document.querySelectorAll('...'))
// ES5
[].slice.call(document.querySelectorAll('...'))
Anyway, I made a simplified working demo of your code.
Note that I save the currently active tab in a variable, to prevent another for loop. You could also do:
document.querySelector('.active').classList.remove('active')
But I like to reduce the amount of DOM reading.
Good luck for your apprentissage, re-writing some jQuery code into Vanilla JS seems like a good method, and you might acquire a far better comprehension of JavaScript.

Why the eval() is not working with array

What's wrong with this code? Can anyone help out?
var read=new Array("i=10","j=20","k=i*j");
for(var i=0;i<read.length;i++)
{
alert(eval(read[i]));
}
Expecting output:
alert three times with values 10,20,200.
But actual output:
But alert Once with value 10.
When the loop executes the first time, you are setting i = 10, with eval. So the loop breaks out immediately.
So, you might want to change the loop variable to something else, like this
var read = new Array("i=10","j=20","k=i*j");
for(var idx=0; idx < read.length; idx++)
{
console.log(eval(read[idx]));
}
Output
10
20
200
Note: Please make sure that you read this question and the answers to that question, before using eval in your code.
Try this code
var read=new Array("i=10","j=20","k=i*j");
for(var index=0;index<read.length;index++)
{
alert(eval(read[index]));
}
while the loop is executing, at the first execution of i=10 the i variable is set to 10; so loop will be terminated because of the condition i<read.length (here... 10<3) remains false.
see the eval() tutorial.

Javascript: TypeError variable is undefined

I am currently building a small web application with similar functionality across all modules. I want to code small generic functions so that all programmers next to me, call these functions and these functions return necessary but important data for them to implement their functionality. In this example, I am trying to deal with the typical "choose true or false" exercise. So from the template.php they call this function:
function checkAnswers(){
var radiobuttons = document.form1.exer1;
var correctAnswers = answers(); //this is an array of string
var checkedAnswers = checkExerciseRB(radiobuttons, 2, correctAnswers);
for(i=0; i<checkedAnswers.length; i++){
alert(checkedAnswers[i]);
}
}
Function checkExerciseRB is my generic function, it is called from checkAnswers.
function checkExerciseRB(rbuttons, opciones, correct){
var answers = new Array();
var control = 0;
for(i=0; i<rbuttons.length; i++){
var noPick="true";
for(j=0; j<opciones; j++){
if(rbuttons[control+j].checked){
if(rbuttons[control+j].value==correct[i]){
answers[i]= 1;
noPick="false";
break;
}
else{
answers[i]=2;
noPick="false";
break;
}
}
}
if(noPick=="true")
answers[i]=0;
control=control+opciones;
}
return answers;
}
It works great but while looking at my favorite browsers (FireFox, Chrome) error log it says:
TypeError: rbuttons[control + j] is undefined
Any clue on how to deal with this matter?
This probably means that control + j is greater than or equal to the length of the array rbuttons. There's no such array element as rbuttons[control + j].
You should learn how to use the JavaScript debugger in your favorite browsers! Debuggers are great. They let you watch this code run, line by line, as fast or as slow as you want, and watch how the value of control changes as you go.
You’ll watch it, and you’ll think “Oh! That line of code is wrong!”
You're looping through rbuttons.length times, but in each loop you're adding 2 to control. Using control to index your array, you're going to run past the end.
Does the index specified by control + j exist in the array? i.e: If that evaluates to 4, is there at least 5 items in the array?
Also, you should be using var i, var j, etc inside your for loop. Without it your variables are leaking into the scope this code is executed in (most likely the global scope, and that's not good) :)

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