# From seconds to days - how?

``day = sec / hr = sec / 60 / 60; min = (sec - horas * 3600) / 60; seg = sec % 60;` `

Tell me how to connect the days?

• @extazys, are you a little perverted over UNIX TimeStamp? - Salivan

1 day = 24 hours
1 hour = 60 minutes, i.e. 1 day = 24 * 60 minutes
1 minute = 60 seconds, i.e. 1 day = 24 * 60 * 60 seconds,

those. in `X` seconds `X / ( 24 * 60 * 60 )` days

usually you need to set the time in a specific format like dd: hh: mm: ss, and not just the number of minutes or seconds

Is logical Then we calculate the remains:

`D = 24 * 60 * 60`
`H = 60 * 60`
`M = 60`

`XD = X / D` days
`XH = ( X - XD * D ) / H` hours
`XM = ( X - XD * D - XH * H ) / M` minutes
`XS = ( X - XD * D - XH * H - XM * M )` seconds

Given 100 seconds, find: minutes, hours, days. Decision:

` `1min = 60 sec => 100 sec = 100/60 1 hour = 60 min = 60*60 sec => 100/60/60 1 day = 24 hour = 24*60 min = 24*60*60 sec = 100/24/60/60` `

not?

• not. And who will count the remains? - gecube
• @gecube is required? - lampa
• For sure! Because usually you need to set the time in a format like dd: hh: mm: ss, and not just the number of minutes or seconds ... - gecube
• @gecube and what if I do in this format, but it turns out that it was not necessary? I think it should be unfinished, than to redo. Especially in the TK nothing about the translation in what format does not say. - lampa