每天添加的mysql计数记录
我是mysql新手。我想知道如何输出每天有多少条记录被添加到一个表中 此数据 loggedID datelogged 1 2015-10-03 2 2015-10-03 3 2015-10-05 4 2015-10-05 5 2015-10-06 6 2015-10-06 7 2015-10-06 日志记录 1 2015-10-03 2 2015-10-03 3 2015-10-05 4 2015-10-05 5 2015-10-06 6 2015-10-06 7 2015-10-06 将输出 2015-10-03 2 2015-10-04 0 2015-10-05 2 2015-10-06 3 2015-10-03 2 2015-10-04 0 2015-10-05 2 2015-10-06 3每天添加的mysql计数记录,mysql,Mysql,我是mysql新手。我想知道如何输出每天有多少条记录被添加到一个表中 此数据 loggedID datelogged 1 2015-10-03 2 2015-10-03 3 2015-10-05 4 2015-10-05 5 2015-10-06 6
提前感谢我不知道单用SQL怎么做,所以我用了一点PHP。输出将转到图形。这就是为什么我需要没有记录呼叫的日期显示为0
SELECT datelogged, COUNT(*)
FROM your_table
GROUP BY datelogged
<?php
# Open the database
require 'config.php';
require 'database_connect.php';
function datediff($date1) {
$date2 = "2015-09-14";
$date1 = date_create($date1);
$date2 = date_create($date2);
$diff=date_diff($date1,$date2);
$realdiff = $diff->format("%a");
return $realdiff;
}
$query = "select datelogged, count(*) as daycalls from rti_loggedcalls group by datelogged";
try
{
$sth = $db->query($query);
while ($row = $sth->fetch (PDO::FETCH_ASSOC))
{
$date = ($row["datelogged"]);
$mdate = datediff($date);
$output[$mdate] = ($row["daycalls"]);
}
}
catch (PDOException $e)
{
printf("We had a problem: %s\n", $e->getMessage());
}
for ($i = 0; $i < ($mdate + 1) ; $i++) {
echo $i." ".(0 + $output[$i])."<br>";
}
?>
该去读SQL教程了。这是非常基本的东西。答案很接近,但输出中没有“2015-10-04 0”。