两部分 MySQL 问题：访问特定的 MySQL 行和列性能

Question

I have a table with about 150 websites listed in it with the columns "site_name", "visible_name" (basically a formatted name), and "description." For a given page on my site, I want to pull site_name and visible_name for every site in the table, and I want to pull all three columns for the selected site, which comes from the $_GET array (a URL parameter).

Right now I'm using 2 queries to do this, one that says "Get site_name and visible_name for all sites" and another that says "Get all 3 fields for one specific site." I'm guess a better way to do it is:

SELECT * FROM site_list;

thus reducing to 1 query, and then doing the rest post-query, which brings up 2 questions:

1. The "description" field for each site is about 200-300 characters. Is it bad from a performance standpoint to pull this for all 150 sites if I'm only using it for 1 site?
2. How do I reference the specific row from the MySQL result set for the site specificed in the URL? For example, if the URL is "mysite.com/results?site_name=foo" how do I do the post-query equivalent of SELECT * FROM site_list where site_name=foo; ?

I don't know how to get the data for "site_name=foo" without looping through the entire result array and checking to see if site_name matches the URL parameter. Isn't there a more efficient way to do it?

Thanks,

Chris

PS: I noticed a similar question on stackoverflow and read through all the answers but it didn't help in my situation, which is why I'm posting this.

Thanks,

Chris

Answer 1

Generally, it's good practice to have an 'id' field in the table as an auto_increment value. Then, you would:

SELECT id,url,display_name FROM table;

and you'd have the 'id' value there to later:

SELECT * FROM table WHERE id=123;

That's probably your most efficient method if you had WAAAY more entries in the table.

However, with only 150 rows in the table, you're probably just fine doing

SELECT * FROM table;

and only accessing that last field for a matching row based on your criteria.

Answer 2

If you only need the description for the site named foo you could just query the database with SELECT * FROM site_list WHERE site_name = 'foo' LIMIT 1

Otherwise you would have to loop though the result array and do a string comparison on site_name to find the correct description.

Answer 3

I believe what you do now, keeping sperated queries for getting a list of sites with just titles and one detailed view with description for a single given site, is good. You don't pull any unneeded data and both queries being very simple are fast.

It is possible to combine both your queries into one, using left join, something maybe like:

SELECT s1.site_name, s1.visible_name, s2.description
FROM site_list s1
LEFT JOIN
  ( SELECT site_name, description
  FROM site_list
  WHERE site_name = 'this site should go with description' ) s2
ON s2.site_name = s1.site_name

resulting in all sites without matching name having NULL as description, you could even sort it using

ORDER BY description DESC, site_name

to get the site with description as first fetched row, thus eliminating need to iterate through results to find it, but mysql would have to do a lot more work to give you this result, negating any possible gain you could hope for. So basically stick to what you have now, its good.