KeepclockSLA clocks for Jira Software and business projects: working hours, pauses, breach emails, JQL search and real compliance reports.
Full name: “Keepclock: SLA Timers for Jira Projects”
From $7 a month for up to 10 users. Billed by Atlassian, with a 30-day free trial.
Put SLA timers on any Jira project, not just service desks. Goals by priority or any JQL, working calendars with lunch breaks and holidays, pauses, at-risk and breach emails, JQL-searchable fields and a fast report. Runs on Atlassian.
What you get
SLAs on any Jira project
Start, pause and stop clocks on status, assignee, resolution or any field. Goals by priority or JQL, counted only in working hours with lunch breaks, time zones and holidays.
Search SLAs with JQL
Every SLA is a real Jira field: find overdue, breached or due-soon work with one query, sort boards and lists by it, and put it in filters, subscriptions and exports.
Reports that load in seconds
Compliance, percentiles, weekly trend and breakdowns by priority, assignee and project, plus live overdue and at-risk lists. Precomputed as work changes, so size doesn't slow it down.
Questions about Keepclock
What does Keepclock do?
SLA clocks for Jira Software and business projects: working hours, pauses, breach emails, JQL search and real compliance reports. Put SLA timers on any Jira project, not just service desks. Goals by priority or any JQL, working calendars with lunch breaks and holidays, pauses, at-risk and breach emails, JQL-searchable fields and a fast report. Runs on Atlassian.
What does Keepclock work with?
Keepclock works in Jira Cloud.
How much does Keepclock cost?
$7 a month for up to 10 users, or $70 a year. From 11 users: $1.85 per user a month, with lower rates for larger sites.
Is there a free trial?
Billed by Atlassian, with a 30-day free trial.
How do I get Keepclock?
Email hello@greatwork.company and we'll help you get Keepclock set up.
Who makes Keepclock?
Great Work LLC, an independent software company. Email hello@greatwork.company with any question about Keepclock and a person will answer.
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