I am looping Text Field hundreds times
for (var i = 0; i < 100; i++)
my label Text is look like this labelText: '${i + 1}',
I want to set Condition if i is below 10 then add extra 0 before number
I did try with like this
'${i < 10 ? "0"i + 1 : i + 1}',
getting error Expected to Find '}'
Here is a another solution by using String's padLeft method.
for (var i = 0 ; i < 100 ; i++) {
print(i.toString().padLeft(2, '0'));
}
for (var i = 0; i < 100; i++){
print('${i+1 < 10 ? "0${i + 1}" : i + 1}');
}
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Im trying to figure out to print star pattern using Dart language which implementing logic code. The existing code I use as indent so that the star have some space. Is this the right method to do so?
void main(){
for(int i = 0 ; i< 7; i++){
var stars='';
for(int j = (7-i); j > 1 ;j--) {
stars += ' ';
}
for(int j = 0; j <= i ;j++){
stars += '* ';
}
print(stars);
}
}
Here is my answer write in dartpad, change starWidth to adjust star size.
Idea is get string of the star and it padding per row then printing its.
EDIT: Updated description comment for each functional
void main() {
const starWidth = 7;
// return `*` or `space` if even/odd
starGenerator(i) => i % 2 == 0 ? "*" : " ";
// just return a string full of `space`
printPad(w) => " " * w;
// cause we need `space` between `*`, length is `w * 2 - 1`,
// return a string build of star
printStars(int w) => List.generate(w * 2 - 1, starGenerator).join('');
for (int row = 1; row <= starWidth; row++) {
// cause our width with space is `starWidth * 2`,
// padding left is `padding = (our width with space - star with with space) / 2`,
// but we only need print left side (/2) so the math is simple
// `padding = width - star with without space`
var padding = starWidth - row;
print("$row:" + printPad(padding) + printStars(row));
}
}
void main(){
for(int i = 0 ; i< 7; i++){
var stars='';
for(int j = (7-i); j > 1 ;j--) {
stars += ' ';
}
for(int j = 0; j <= i ;j++){
stars += '* ';
}
print(stars);
}
}
I want to print that in the console:
*
**
***
**
*
so, my code is:
Scanner input = new Scanner(System.in);
int n = input.nextInt();
char c = '*';
if (1 < n && n < 20) {
for (int row = 1; row <= n; row++) {
for (int col = 1; col <= row; col++) {
System.out.print(c);
}
System.out.println();
}
any proposals how to finish?
You want to print "* **". Write this:
System.out.print("* **");
If my answer is not that what you excpected, than give us more information. What is your actual problem. See StackOverflowFAQ
If n is your "*" length,below code:
if (1 < n && n < 20) {
for (int row = 1; row <= n; row++) {
for (int col = 1; col <= row; col++) {
System.out.print(c);
}
System.out.println();
}
for (int row =1; row <= n; row++) {
for (int col = n-row; col >0 ; col--) {
System.out.print(c);
}
System.out.println();
}
}
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I'm looking for a way to solve this crypt arithmetic problem of:
ROBERT + GERALD = DONALD
and potentially others as well, where each letter represents a digit.
How would you go about solving this by hand and how does that relate to solving it programmatically?
Thank you in advance
You can actually work this out as a sum:
robert
+ gerald
------
= donald
and use basic mathematical knowledge.
For example, there's no carry possible in the right column (6) and we have T + D = D. That means T must be zero.
Similarly, for column 5, there's no carry from column 6 and R + L = L means R is zero as well, and no carry to column 4.
Same with column 4, E + A = A so E is zero.
So we now have:
0ob000
+ g00ald
------
= donald
From there, we can infer from columns 3 and 1 that b==n and g==d and they (along with o/a/l/d) can be any value since every digit is being added to zero so there is no chance of carry anywhere. So let's just make them all one:
011000
+ 100111
------
= 111111
In fact, you could make them all zero and end up with 000000 + 000000 = 000000.
But that's hardly programming related, so let's make it so:
#include <stdio.h>
int main (void) {
int robert, gerald, donald;
for (int r = 0; r < 10; r++) {
for (int o = 0; o < 10; o++) {
for (int b = 0; b < 10; b++) {
for (int e = 0; e < 10; e++) {
for (int t = 0; t < 10; t++) {
for (int g = 0; g < 10; g++) {
for (int a = 0; a < 10; a++) {
for (int l = 0; l < 10; l++) {
for (int d = 0; d < 10; d++) {
for (int n = 0; n < 10; n++) {
robert = r * 100000 + o * 10000 + b * 1000 + e * 100 + r * 10 + t;
gerald = g * 100000 + e * 10000 + r * 1000 + a * 100 + l * 10 + d;
donald = d * 100000 + o * 10000 + n * 1000 + a * 100 + l * 10 + d;
if (robert + gerald == donald) {
printf (" %06d\n", robert);
printf ("+ %06d\n", gerald);
printf (" ------\n");
printf ("= %06d\n", donald);
printf ("........\n");
}
}
}
}
}
}
}
}
}
}
}
return 0;
}
That will give you a whole host of solutions.
And, before you complain that you cannot have repeated digits, there is no solution if that's the case, since mathematically both T and R must be zero, as shown in the original reasoning above. And you can prove this empirically with:
#include <stdio.h>
int main (void) {
int robert, gerald, donald;
for (int r = 0; r < 10; r++) {
for (int o = 0; o < 10; o++) {
if (o==r) continue;
for (int b = 0; b < 10; b++) {
if ((b==r) || (b==o)) continue;
for (int e = 0; e < 10; e++) {
if ((e==r) || (e==o) || (e==b)) continue;
for (int t = 0; t < 10; t++) {
if ((t==r) || (t==o) || (t==b) || (t==e)) continue;
for (int g = 0; g < 10; g++) {
if ((g==r) || (g==o) || (g==b) || (g==e) || (g==t)) continue;
for (int a = 0; a < 10; a++) {
if ((a==r) || (a==o) || (a==b) || (a==e) || (a==t) || (a==g)) continue;
for (int l = 0; l < 10; l++) {
if ((l==r) || (l==o) || (l==b) || (l==e) || (l==t) || (l==g) || (l==a)) continue;
for (int d = 0; d < 10; d++) {
if ((d==r) || (d==o) || (d==b) || (d==e) || (d==t) || (d==g) || (d==a) || (d==l)) continue;
for (int n = 0; n < 10; n++) {
if ((n==r) || (n==o) || (n==b) || (n==e) || (n==t) || (n==g) || (n==a) || (n==l) || (n==d)) continue;
robert = r * 100000 + o * 10000 + b * 1000 + e * 100 + r * 10 + t;
gerald = g * 100000 + e * 10000 + r * 1000 + a * 100 + l * 10 + d;
donald = d * 100000 + o * 10000 + n * 1000 + a * 100 + l * 10 + d;
if (robert + gerald == donald) {
printf (" %06d\n", robert);
printf ("+ %06d\n", gerald);
printf (" ------\n");
printf ("= %06d\n", donald);
printf ("........\n");
}
}
}
}
}
}
}
}
}
}
}
return 0;
}
which outputs no solutions.
Now DONALD + GERALD = ROBERT, that's a different matter but you can solve that simply by modifying the code above slightly, making the if statement into:
if (donald + gerald == robert) {
printf (" %06d\n", donald);
printf ("+ %06d\n", gerald);
printf (" ------\n");
printf ("= %06d\n", robert);
printf ("........\n");
}
and you get the single solution:
526485
+ 197485
------
= 723970
Using Highstock, I've a serie timestamp / value
with different rangeselectors (hour, day, week, month,...) or zoomX
I want to display the average value for the displayed time period.
Now, I can compute the average of the overall series data:
for (i = 0; i < chart.series[0].yData.length; i++) {
total += chart.series[0].yData[i];
}
seriesAvg = (total / chart.series[0].yData.length).toFixed(4); // fix decimal to 4 places
$('#report1').html('<b>Average:</b>: '+ seriesAvg);
How to compute the average only on the displayed datapoints ? And to refresh automatically after zoom ?
Thank you
Use series.processedYData or series.points.
I implemented the same but I notice that processedYdata is loading also one data more than the one showed at the screen.
for that reason i added one additional command to remove the first processedYData:
processedYData.splice(0, 1);
below is the final result of the function
function showStat(objHighStockchart) {
for (j = 0; j < (json_data.length); j++) {
var seriesAvg = 0,
processedYData = objHighStockchart.series[j].processedYData;
processedYData.splice(0, 1);
var seriesMin = Math.min.apply(null, processedYData);
var seriesMax = Math.max.apply(null, processedYData);
var i = 0
var total = 0;
console.log(processedYData);
for (i = 1; i < processedYData.length; i++) {
total += processedYData[i];
}
seriesAvg = (total / processedYData.length).toFixed(2); // fix decimal to 4 places
$('#container_stat' + j).html(
'<br>Statistics for ' + objHighStockchart.series[j].name + '<br>' +
'Total: ' + total + ' logs<br>' +
'Min: ' + seriesMin + ' logs<br>' +
'Avg: ' + seriesAvg + ' logs<br>' +
'Max: ' + seriesMax + ' logs<br>'
+ '---' + processedYData
);
}
};
On coffeescript loop 'for'
eg.
if 1 < x, code like below:
console.debug i for i in [1..0]
Generated code is:
var i;
for (i = 1; i >= 0; i--) {
console.debug(i);
}
if 1 > x,code like below:
console.debug i for i in [1..2]
Generated code is:
var i;
for (i = 1; i <= 2; i++) {
console.debug(i);
}
If i want write that javascript.How to ?
for(var i=1;i<=0;i++){
console.debug(i);
}
Because i don't know the condition is greater than left side or less than left side.
But i just want it i++
What's wrong with me?
EDIT BELOW:
For coffeescript's feature,I add condition before the loop or add condition on for loop.
eg:
if x - y >=1
console.debug i for i in [1..x-y]
or
console.debug i for i in [1..x-y] and x-y >=1
That's my way.Some one have good advice?
It looks like you want to do this:
console.debug i for i in [1..x-y] by 1
Which gets compiled to:
var i, _i, _ref;
for (i = _i = 1, _ref = x - y; _i <= _ref; i = _i += 1) {
console.debug(i);
}
for(var i=1;i<=0;i++){
console.debug(i);
}
is equivalent to
var i = 1;
while(true) {
console.debug(i);
i++;
}
which in coffeescript is written as
i = 1
while true
console.debug(i);
i++;