js时间戳格式化方法

怼烎@ 2023-06-18 12:56 122阅读 0赞

## js时间戳格式化方法 ##

### 方法一 ###

function formatTime(value) { 
    	if(value) { 
    		let date = new Date(value * 1000)	// 时间戳为秒：10位数
    		//let date = new Date(value) // 时间戳为毫秒：13位数
    		let year = date.getFullYear()
    		let month = date.getMonth() + 1 < 10 ? `0${ date.getMonth() + 1}` : date.getMonth() + 1
    		let day = date.getDate() < 10 ? `0${ date.getDate()}` : date.getDate()
    		let hour = date.getHours() < 10 ? `0${ date.getHours()}` : date.getHours()
    		let minute = date.getMinutes() < 10 ? `0${ date.getMinutes()}` : date.getMinutes()
    		let second = date.getSeconds() < 10 ? `0${ date.getSeconds()}` : date.getSeconds()
    		return `${ year}-${ month}-${ day} ${ hour}:${ minute}:${ second}`
    	} else { 
    		return ''
    	}
    }
    
    console.log(formatTime(1575277007))

### 方法二 ###

// 时间戳转xxxx-xx-xx格式的日期
    export const timestampToDate = (timestamp, type = false) => { 
      const fullTime = new Date(timestamp)
      const year = fullTime.getFullYear()
      const month = (fullTime.getMonth() + 1) < 10 ? '0' + (fullTime.getMonth() + 1) : (fullTime.getMonth() + 1)
      const day = (fullTime.getDate()) < 10 ? '0' + (fullTime.getDate()) : (fullTime.getDate())
      const hours = (fullTime.getHours()) < 10 ? '0' + (fullTime.getHours()) : (fullTime.getHours())
      const minutes = (fullTime.getMinutes()) < 10 ? '0' + (fullTime.getMinutes()) : (fullTime.getMinutes())
      const seconds = (fullTime.getSeconds()) < 10 ? '0' + (fullTime.getSeconds()) : (fullTime.getSeconds())
      return type === false ? `${ year}-${ month}-${ day}` : `${ year}-${ month}-${ day} ${ hours}:${ minutes}:${ seconds}`
    }

--------------------

### mysql中 时间戳生成方法 ###

select unix_timestamp(now())

<table> 
 <thead> 
  <tr> 
   <th>unix_timestamp(now())</th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td>1575277371</td> 
  </tr> 
 </tbody> 
</table>

### js 生成时间戳 ###

> Date.parse(new Date().toString())
    < 1575277771000
    # 13位，但是末尾3位位000，可截取前10位
    > Date.parse(new Date())/1000
    < 1575278320
    
    > (new Date()).valueOf()
    < 1575277823412
    
    > new Date().getTime()
    < 1575277839379
    
    > Math.round(new Date().getTime()/1000)
    < 1575277858