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How can I get this timestamp format with javascript / jquery?

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1

I want to get this format :

2019-03-24 15:05:20

Here is what I have tried:

var today = new Date();
var date = today.getFullYear()+'-'+(today.getMonth()+1)+'-'+today.getDate();
var time = today.getHours() + ":" + today.getMinutes() + ":" + today.getSeconds();
var dateTime = date+' '+time;

I got:

2019-3-24 15:0:20

It's missing leading zeroes.

javascript jquery

share|improve this question

asked Mar 25 at 0:02

JosueJosue

61

add a comment | 

1

I want to get this format :

2019-03-24 15:05:20

Here is what I have tried:

var today = new Date();
var date = today.getFullYear()+'-'+(today.getMonth()+1)+'-'+today.getDate();
var time = today.getHours() + ":" + today.getMinutes() + ":" + today.getSeconds();
var dateTime = date+' '+time;

I got:

2019-3-24 15:0:20

It's missing leading zeroes.

javascript jquery

share|improve this question

asked Mar 25 at 0:02

JosueJosue

61

add a comment | 

I want to get this format :

2019-03-24 15:05:20

Here is what I have tried:

var today = new Date();
var date = today.getFullYear()+'-'+(today.getMonth()+1)+'-'+today.getDate();
var time = today.getHours() + ":" + today.getMinutes() + ":" + today.getSeconds();
var dateTime = date+' '+time;

I got:

2019-3-24 15:0:20

It's missing leading zeroes.

javascript jquery

share|improve this question

asked Mar 25 at 0:02

JosueJosue

61

var today = new Date();
var date = today.getFullYear()+'-'+(today.getMonth()+1)+'-'+today.getDate();
var time = today.getHours() + ":" + today.getMinutes() + ":" + today.getSeconds();
var dateTime = date+' '+time;

share|improve this question

asked Mar 25 at 0:02

JosueJosue

61

share|improve this question

asked Mar 25 at 0:02

JosueJosue

61

asked Mar 25 at 0:02

JosueJosue

61

asked Mar 25 at 0:02

JosueJosue

61

4 Answers
4

active

oldest

votes

2

That's because numbers lower than 10 should be padded with zero but they aren't.

You can use padStart function to achieve the expected result - add a leading 0 character if the length of a number is less than 2. I modified your code a little bit but all you need is to use .toString().padStart(2, '0') on a number.

var today = new Date();

var date = [
 today.getFullYear(),
 today.getMonth() + 1,
 today.getDate(),
].map((value) => value.toString().padStart(2, '0')).join('-');

var time = [
 today.getHours(),
 today.getMinutes(),
 today.getSeconds(),
].map((value) => value.toString().padStart(2, '0')).join(':');;


var dateTime = date + ' ' + time;
console.log(dateTime);

share|improve this answer

edited Mar 25 at 0:14

Ele

26.3k52354

answered Mar 25 at 0:10

Tomasz KajtochTomasz Kajtoch

562513

add a comment | 

0

This is pretty close to ISO 8601, so let's start with that.

const d = new Date();
d
 .toISOString() // Convert date to a string in the format of 2019-03-25T00:07:22.0253Z
 .substr(0, 19) // Strip off the milliseconds and Zulu timezone indication
 .replace('T', ' '); // Replace the T for "time" with a space

This leaves you with a date formatted like 2019-03-25 00:07:22.

share|improve this answer

answered Mar 25 at 0:07

BradBrad

119k29243405

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Why was this downvoted? It's a perfectly reasonable solution to the problem presented, and is well documented. I really don't understand Stack Overflow these days.
  
  – Brad
  Mar 25 at 1:22
  

  
  
  
  

add a comment | 

0

You can use ternary operation for simple use. This should give you leading zero.

 var today = new Date();
 var date = today.getFullYear() + '-' + (today.getMonth() + 1) + '-' + today.getDate();
 var time = today.getHours() + ":" + (today.getMinutes() < 10 ? '0' : '') + today.getMinutes() + ":" + today.getSeconds();
 var dateTime = date + ' ' + time;

 console.log(dateTime, 'result');

share|improve this answer

answered Mar 25 at 0:10

Sundar BanSundar Ban

464414

add a comment | 

0

You can use:

var d = new Date();
d = new Date(d.getTime() - 3000000);
var date_format_str = d.getFullYear().toString()+"-"+((d.getMonth()+1).toString().length==2?(d.getMonth()+1).toString():"0"+(d.getMonth()+1).toString())+"-"+(d.getDate().toString().length==2?d.getDate().toString():"0"+d.getDate().toString())+" "+(d.getHours().toString().length==2?d.getHours().toString():"0"+d.getHours().toString())+":"+((parseInt(d.getMinutes()/5)*5).toString().length==2?(parseInt(d.getMinutes()/5)*5).toString():"0"+(parseInt(d.getMinutes()/5)*5).toString())+":00";
console.log(date_format_str);

share|improve this answer

answered Mar 25 at 0:17

Syrup72Syrup72

2117

add a comment | 

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2

That's because numbers lower than 10 should be padded with zero but they aren't.

You can use padStart function to achieve the expected result - add a leading 0 character if the length of a number is less than 2. I modified your code a little bit but all you need is to use .toString().padStart(2, '0') on a number.

var today = new Date();

var date = [
 today.getFullYear(),
 today.getMonth() + 1,
 today.getDate(),
].map((value) => value.toString().padStart(2, '0')).join('-');

var time = [
 today.getHours(),
 today.getMinutes(),
 today.getSeconds(),
].map((value) => value.toString().padStart(2, '0')).join(':');;


var dateTime = date + ' ' + time;
console.log(dateTime);

share|improve this answer

edited Mar 25 at 0:14

Ele

26.3k52354

answered Mar 25 at 0:10

Tomasz KajtochTomasz Kajtoch

562513

add a comment | 

2

That's because numbers lower than 10 should be padded with zero but they aren't.

You can use padStart function to achieve the expected result - add a leading 0 character if the length of a number is less than 2. I modified your code a little bit but all you need is to use .toString().padStart(2, '0') on a number.

var today = new Date();

var date = [
 today.getFullYear(),
 today.getMonth() + 1,
 today.getDate(),
].map((value) => value.toString().padStart(2, '0')).join('-');

var time = [
 today.getHours(),
 today.getMinutes(),
 today.getSeconds(),
].map((value) => value.toString().padStart(2, '0')).join(':');;


var dateTime = date + ' ' + time;
console.log(dateTime);

share|improve this answer

edited Mar 25 at 0:14

Ele

26.3k52354

answered Mar 25 at 0:10

Tomasz KajtochTomasz Kajtoch

562513

add a comment | 

That's because numbers lower than 10 should be padded with zero but they aren't.

You can use padStart function to achieve the expected result - add a leading 0 character if the length of a number is less than 2. I modified your code a little bit but all you need is to use .toString().padStart(2, '0') on a number.

var today = new Date();

var date = [
 today.getFullYear(),
 today.getMonth() + 1,
 today.getDate(),
].map((value) => value.toString().padStart(2, '0')).join('-');

var time = [
 today.getHours(),
 today.getMinutes(),
 today.getSeconds(),
].map((value) => value.toString().padStart(2, '0')).join(':');;


var dateTime = date + ' ' + time;
console.log(dateTime);

share|improve this answer

edited Mar 25 at 0:14

Ele

26.3k52354

answered Mar 25 at 0:10

Tomasz KajtochTomasz Kajtoch

562513

var today = new Date();

var date = [
 today.getFullYear(),
 today.getMonth() + 1,
 today.getDate(),
].map((value) => value.toString().padStart(2, '0')).join('-');

var time = [
 today.getHours(),
 today.getMinutes(),
 today.getSeconds(),
].map((value) => value.toString().padStart(2, '0')).join(':');;


var dateTime = date + ' ' + time;
console.log(dateTime);

var today = new Date();

var date = [
 today.getFullYear(),
 today.getMonth() + 1,
 today.getDate(),
].map((value) => value.toString().padStart(2, '0')).join('-');

var time = [
 today.getHours(),
 today.getMinutes(),
 today.getSeconds(),
].map((value) => value.toString().padStart(2, '0')).join(':');;


var dateTime = date + ' ' + time;
console.log(dateTime);

var today = new Date();

var date = [
 today.getFullYear(),
 today.getMonth() + 1,
 today.getDate(),
].map((value) => value.toString().padStart(2, '0')).join('-');

var time = [
 today.getHours(),
 today.getMinutes(),
 today.getSeconds(),
].map((value) => value.toString().padStart(2, '0')).join(':');;


var dateTime = date + ' ' + time;
console.log(dateTime);

share|improve this answer

edited Mar 25 at 0:14

Ele

26.3k52354

answered Mar 25 at 0:10

Tomasz KajtochTomasz Kajtoch

562513

edited Mar 25 at 0:14

Ele

26.3k52354

edited Mar 25 at 0:14

Ele

26.3k52354

answered Mar 25 at 0:10

Tomasz KajtochTomasz Kajtoch

562513

answered Mar 25 at 0:10

Tomasz KajtochTomasz Kajtoch

562513

0

This is pretty close to ISO 8601, so let's start with that.

const d = new Date();
d
 .toISOString() // Convert date to a string in the format of 2019-03-25T00:07:22.0253Z
 .substr(0, 19) // Strip off the milliseconds and Zulu timezone indication
 .replace('T', ' '); // Replace the T for "time" with a space

This leaves you with a date formatted like 2019-03-25 00:07:22.

share|improve this answer

answered Mar 25 at 0:07

BradBrad

119k29243405

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Why was this downvoted? It's a perfectly reasonable solution to the problem presented, and is well documented. I really don't understand Stack Overflow these days.
  
  – Brad
  Mar 25 at 1:22
  

  
  
  
  

add a comment | 

0

This is pretty close to ISO 8601, so let's start with that.

const d = new Date();
d
 .toISOString() // Convert date to a string in the format of 2019-03-25T00:07:22.0253Z
 .substr(0, 19) // Strip off the milliseconds and Zulu timezone indication
 .replace('T', ' '); // Replace the T for "time" with a space

This leaves you with a date formatted like 2019-03-25 00:07:22.

share|improve this answer

answered Mar 25 at 0:07

BradBrad

119k29243405

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Why was this downvoted? It's a perfectly reasonable solution to the problem presented, and is well documented. I really don't understand Stack Overflow these days.
  
  – Brad
  Mar 25 at 1:22
  

  
  
  
  

add a comment | 

This is pretty close to ISO 8601, so let's start with that.

const d = new Date();
d
 .toISOString() // Convert date to a string in the format of 2019-03-25T00:07:22.0253Z
 .substr(0, 19) // Strip off the milliseconds and Zulu timezone indication
 .replace('T', ' '); // Replace the T for "time" with a space

This leaves you with a date formatted like 2019-03-25 00:07:22.

share|improve this answer

answered Mar 25 at 0:07

BradBrad

119k29243405

const d = new Date();
d
 .toISOString() // Convert date to a string in the format of 2019-03-25T00:07:22.0253Z
 .substr(0, 19) // Strip off the milliseconds and Zulu timezone indication
 .replace('T', ' '); // Replace the T for "time" with a space

share|improve this answer

answered Mar 25 at 0:07

BradBrad

119k29243405

answered Mar 25 at 0:07

BradBrad

119k29243405

answered Mar 25 at 0:07

BradBrad

119k29243405

0

You can use ternary operation for simple use. This should give you leading zero.

 var today = new Date();
 var date = today.getFullYear() + '-' + (today.getMonth() + 1) + '-' + today.getDate();
 var time = today.getHours() + ":" + (today.getMinutes() < 10 ? '0' : '') + today.getMinutes() + ":" + today.getSeconds();
 var dateTime = date + ' ' + time;

 console.log(dateTime, 'result');

share|improve this answer

answered Mar 25 at 0:10

Sundar BanSundar Ban

464414

add a comment | 

0

You can use ternary operation for simple use. This should give you leading zero.

 var today = new Date();
 var date = today.getFullYear() + '-' + (today.getMonth() + 1) + '-' + today.getDate();
 var time = today.getHours() + ":" + (today.getMinutes() < 10 ? '0' : '') + today.getMinutes() + ":" + today.getSeconds();
 var dateTime = date + ' ' + time;

 console.log(dateTime, 'result');

share|improve this answer

answered Mar 25 at 0:10

Sundar BanSundar Ban

464414

add a comment | 

You can use ternary operation for simple use. This should give you leading zero.

 var today = new Date();
 var date = today.getFullYear() + '-' + (today.getMonth() + 1) + '-' + today.getDate();
 var time = today.getHours() + ":" + (today.getMinutes() < 10 ? '0' : '') + today.getMinutes() + ":" + today.getSeconds();
 var dateTime = date + ' ' + time;

 console.log(dateTime, 'result');

share|improve this answer

answered Mar 25 at 0:10

Sundar BanSundar Ban

464414

 var today = new Date();
 var date = today.getFullYear() + '-' + (today.getMonth() + 1) + '-' + today.getDate();
 var time = today.getHours() + ":" + (today.getMinutes() < 10 ? '0' : '') + today.getMinutes() + ":" + today.getSeconds();
 var dateTime = date + ' ' + time;

 console.log(dateTime, 'result');

share|improve this answer

answered Mar 25 at 0:10

Sundar BanSundar Ban

464414

answered Mar 25 at 0:10

Sundar BanSundar Ban

464414

answered Mar 25 at 0:10

Sundar BanSundar Ban

464414

0

You can use:

var d = new Date();
d = new Date(d.getTime() - 3000000);
var date_format_str = d.getFullYear().toString()+"-"+((d.getMonth()+1).toString().length==2?(d.getMonth()+1).toString():"0"+(d.getMonth()+1).toString())+"-"+(d.getDate().toString().length==2?d.getDate().toString():"0"+d.getDate().toString())+" "+(d.getHours().toString().length==2?d.getHours().toString():"0"+d.getHours().toString())+":"+((parseInt(d.getMinutes()/5)*5).toString().length==2?(parseInt(d.getMinutes()/5)*5).toString():"0"+(parseInt(d.getMinutes()/5)*5).toString())+":00";
console.log(date_format_str);

share|improve this answer

answered Mar 25 at 0:17

Syrup72Syrup72

2117

add a comment | 

0

You can use:

var d = new Date();
d = new Date(d.getTime() - 3000000);
var date_format_str = d.getFullYear().toString()+"-"+((d.getMonth()+1).toString().length==2?(d.getMonth()+1).toString():"0"+(d.getMonth()+1).toString())+"-"+(d.getDate().toString().length==2?d.getDate().toString():"0"+d.getDate().toString())+" "+(d.getHours().toString().length==2?d.getHours().toString():"0"+d.getHours().toString())+":"+((parseInt(d.getMinutes()/5)*5).toString().length==2?(parseInt(d.getMinutes()/5)*5).toString():"0"+(parseInt(d.getMinutes()/5)*5).toString())+":00";
console.log(date_format_str);

share|improve this answer

answered Mar 25 at 0:17

Syrup72Syrup72

2117

add a comment | 

You can use:

var d = new Date();
d = new Date(d.getTime() - 3000000);
var date_format_str = d.getFullYear().toString()+"-"+((d.getMonth()+1).toString().length==2?(d.getMonth()+1).toString():"0"+(d.getMonth()+1).toString())+"-"+(d.getDate().toString().length==2?d.getDate().toString():"0"+d.getDate().toString())+" "+(d.getHours().toString().length==2?d.getHours().toString():"0"+d.getHours().toString())+":"+((parseInt(d.getMinutes()/5)*5).toString().length==2?(parseInt(d.getMinutes()/5)*5).toString():"0"+(parseInt(d.getMinutes()/5)*5).toString())+":00";
console.log(date_format_str);

share|improve this answer

answered Mar 25 at 0:17

Syrup72Syrup72

2117

var d = new Date();
d = new Date(d.getTime() - 3000000);
var date_format_str = d.getFullYear().toString()+"-"+((d.getMonth()+1).toString().length==2?(d.getMonth()+1).toString():"0"+(d.getMonth()+1).toString())+"-"+(d.getDate().toString().length==2?d.getDate().toString():"0"+d.getDate().toString())+" "+(d.getHours().toString().length==2?d.getHours().toString():"0"+d.getHours().toString())+":"+((parseInt(d.getMinutes()/5)*5).toString().length==2?(parseInt(d.getMinutes()/5)*5).toString():"0"+(parseInt(d.getMinutes()/5)*5).toString())+":00";
console.log(date_format_str);

share|improve this answer

answered Mar 25 at 0:17

Syrup72Syrup72

2117

answered Mar 25 at 0:17

Syrup72Syrup72

2117

answered Mar 25 at 0:17

Syrup72Syrup72

2117