How much does a database architect make in Orlando?
$143,500a year
Median yearly pay for database architects in the Orlando-Kissimmee-Sanford, FL metro area. That is $69.01 an hour or $11,960 a month.
- Hourly
- $69.01
- Vs. national median
- +3%
- Rank among cities
- 32 of 132
- Jobs in the area
- 270
Pay range
Lowest 10%: $87,030Median: $143,500Highest 10%: $215,200
| Percentile | Yearly | Hourly | |
|---|---|---|---|
| 10th | $87,030 | $41.84 | |
| 25th | $109,800 | $52.80 | |
| Median | $143,500 | $69.01 | |
| 75th | $170,700 | $82.07 | |
| 90th | $215,200 | $103.46 |
Other cities in Florida
Median yearly pay for database architects.
| Palm Bay, FL | $150,300 |
| Miami, FL | $141,200 |
| Jacksonville, FL | $136,600 |
| Tampa, FL | $133,300 |
Related jobs in Orlando
Same field, median yearly pay.
| Computer Network Architect | $156,600 |
| Software Developer | $129,600 |
| Information Security Analyst | $129,400 |
| Actuary | $122,300 |
| Computer and Information Research Scientist | $116,500 |
| Statistician | $116,100 |
| Database Administrator | $115,100 |
| Data Scientist | $106,600 |
| Computer Occupation | $106,300 |
| Software Quality Assurance Analyst and Tester | $104,300 |
Frequently asked questions
How much does a database architect make an hour in Orlando?
The median hourly pay in the Orlando-Kissimmee-Sanford, FL metro area is $69.01 ($143,500 a year).
Do database architects make more in Orlando than the national average?
Yes: $143,500 against $139,500 nationally, 3% more.
How does Orlando compare with the rest of Florida?
The median in Florida is $138,300, so Orlando pays 4% more.
Where does Orlando rank for database architect pay?
32 of 132 metro areas with data.
How much does a database architect make starting out in Orlando?
The lowest-paid 10% make $87,030 or less a year; the top 10% make $215,200 or more.
Source: U.S. Bureau of Labor Statistics, Occupational Employment and Wage Statistics (OEWS): pay in May 2025, published by BLS in 2026. Hourly pay is the yearly figure divided by 2,080 hours.