Database Administrators Stack Exchange is a question and answer site for database professionals who wish to improve their database skills and learn from others in the community. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm using MySql to store a basic table of this format:

    id      int(11)          //Auto-Incrementing ID
    data    varchar(5120)    //Random input data, compressed by a program, not mysql.
    Row size ø  916 B

That's kind of troubling, because right now I have about 5,000 records each month, and I'd like to optimize that better than ending up averaging out at 1mb/1000 records.

I set it up this way originally because I wanted to be able to capture the larger sets, but they're very infrequent, as you can see here by this graph:

enter image description here

Link to graph API

count   n*128 bytes
1       28
1       26
1       24
2       22
8       21
4       20
13      19
12      18
16      17
27      16
43      15
58      14
69      13
114     12
184     11
262     10
399     9
588     8
807     7
1224    6
1245    5
546     4
73      3
9       2
6       1
1       0

My main concern here is how much space I've wasted just to have accommodated the tail end of that chart, and what happens when I end up with even larger outliers? I'll have to bump up my row size again, when roughly 80% of the data fits very nicely into a varchar 1024 block, 1/5th the size of what I'm using now.

So how should I have built this thing in the first place?

share|improve this question
up vote 4 down vote accepted

Since it is built already, let's look at what you have. There is an interesting way to plan columns definitions for data currently present. If the table name is mydata, try running this query:


This will not display any of your data. This will examine the column data itself, calculate statistics based on the first 256 distinct values (by default, you can specify the distinct value count) and recommend the proper column type the table should have.

As long as the table is MyISAM, don't worry too much about the rows size because the default row format is DYNAMIC. If the table is InnoDB (and I sure hope it is not), please make sure the data is not part of the PRIMARY KEY. Your clustered index will grow like pouring undiluted MiracleGro in your backyard.

You may need to break up the data column into chunks of 128 and run either MD5 (32 character output) or SHA1 (40 character output) and concatenate those MD5 or SHA1 outputs and store them. That would save at 75% of storage. Have fun trying to code that breakup. This was just a rambling suggestion off the top of my head.

You may want to consider using Sphinx Indexing as well.

share|improve this answer

For MyISAM, if you never UPDATE or DELETE records, a record with N bytes for the blob will take N+8 bytes in the .MYD file.

The MyISAM PK will be found in the .MYI file; it will be pretty small -- about 11 bytes/row, plus a little overhead, rounded up to 1KB blocks.

If you are using InnoDB, the layout is quite complicated. There are 16KB data blocks which are never quite full, there are blob extents (even for VARCHAR), which kick in after 767 bytes (maybe sooner, depending on version), etc. The extents are allocated in 1MB chunks. Plan on something like 2N bytes per row average.

The InnoDB PK is stored with the data, so it is completely included above. And, being AUTO_INCREMENT, your blocks will tend to be relatively full. Note: blocks are 16KB in size, allowing you at least 20 per block.

Don't play games with your own chunking; it will slow down things, complicate your code, etc. Disk is cheap. (Yeah, it is fun to play such games.)

Looks like half your records won't spill past the 767 cutoff.

An aside: If the blob is compressed data, you should really use BLOB, not VARCHAR. VARCHAR assumes CHARACTER SET attributes, which would really burn you if you switched to utf8. If you must use VAR-xx, use VARBINARY.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.