从SQL表中选择行的百分比? [英] Select percentage of rows from SQL table?

问题描述

我有一个带有PHP脚本的网站，该脚本内部有一个SQL查询，返回由JavaScript文件访问的数据.该数据是一个庞大的航班数据列表，我需要能够在指定的任何给定日期中随机选择(比如说)全部航班的40％.为了论证，让我们这样说:

I've got a site with a PHP script, this script has an SQL query inside returning data that is accessed by a JavaScript file. The data is a huge list of flight data, and I need to be able to select (let's say) a random 40% of the total flights for any given day specified. For arguments sake lets put it like this:

$query = "SELECT * FROM `Flight_Data` WHERE DepDateTimeUTC LIKE '%1/1/14%' ";

我知道要获得随机的行数，您只需使用ORDER BY RAND() LIMIT 40'，理想情况下我想说LIMIT 40%，但这行不通.

I understand that to get a random number of rows you simply use ORDER BY RAND() LIMIT 40' and ideally I want to say LIMIT 40% but that doesn't work.

$query = "SELECT * FROM `Flight_Data` WHERE DepDateTimeUTC LIKE '%1/1/14%' ";
$row = mysqli_fetch_row($result);
$total = $row[0];
$percent = $total * 0.40;
$query = "SELECT * FROM `Flight_Data` WHERE DepDateTimeUTC LIKE '%1/1/14%' LIMIT . $percent ";

推荐答案

您可以COUNT所有记录，然后像这样计算所需的%:

You can COUNT all records and then calculate the % you need like this:

$query = "SELECT COUNT(*) FROM `Flight_Data` WHERE DepDateTimeUTC LIKE '%1/1/14%' ";
$result = mysqli_query($connection,$query);
$row = mysqli_fetch_row($result));

$total = $row[0];
$percent = intval($total * 0.40);

$query = "SELECT * FROM `Flight_Data` WHERE DepDateTimeUTC LIKE '%1/1/14%' LIMIT ". $percent;
//execute your query....

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