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我有一个带有名称的表users和一个名为source. 该列包含 3 种不同类型的值，它们是字符串（facebook、google、yahoo）。现在我需要计算该列中有多少 Facebook 字符串、google 字符串和 yahoo 字符串。

使用php查询有什么简单的方法吗？

我如何才能显示数字！！！！！！！！！！

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4 回答 4

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您可以使用此解决方案。这会将计数放入单独的列中，因此您可以轻松地引用 php 中不同字符串的计数：

SELECT
    COUNT(CASE WHEN source = 'facebook' THEN 1 END) AS facebook_count,
    COUNT(CASE WHEN source = 'google' THEN 1 END) AS google_count,
    COUNT(CASE WHEN source = 'yahoo' THEN 1 END) AS yahoo_count
FROM
    users

编辑：根据您对此答案的评论...关于如何使用 php 检索这些值，如下所示：

$sql = "
    SELECT
        COUNT(CASE WHEN source = 'facebook' THEN 1 END) AS facebook_count,
        COUNT(CASE WHEN source = 'google' THEN 1 END) AS google_count,
        COUNT(CASE WHEN source = 'yahoo' THEN 1 END) AS yahoo_count
    FROM
        users";

$social = mysql_query($sql);

while($row = mysql_fetch_assoc($social))
{
    echo "FB Count: {$row['facebook_count']}" . '<br />';
    echo "Google Count: {$row['google_count']}" . '<br />';
    echo "Yahoo Count: {$row['yahoo_count']}";
}
于 2012-07-09T08:36:19.317 回答
0

试试这个：

 Select source, count(1) from users where source like 'facebook'
于 2012-07-09T08:11:40.660 回答
0 
select source, count(source) from users group by source
于 2012-07-09T08:11:54.457 回答
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如果您的source列不是可选字段，那么您可以像这样简单地使用，

select source as source_name, count(*) as counts from users group by source

否则你可以使用like，

select source as source_name, count(source) as counts from users group by source
于 2012-07-09T08:26:54.693 回答