I have database "db2" with table "menjava"
In table menjava have "id", "author" and "date_submitted" field
I want to count all the rows for todays date and all the rows for yesterdays date (so there will be two codes with conditions) based on a DATETIME field called 'date_submitted' that holds the date and time of each record's creation.
In the file result.php, there is this count displayed, but it does not work. In the same file (result.php) I have some other code to display data from different database, so I think that povezava.php is working ok.
My code:
<?
require "povezava.php";
$q=mysql_query(" SELECT COUNT(*) AS total_number FROM menjava
WHERE date_submitted >= DATE_SUB(CURRENT_DATE(), INTERVAL 1 DAY)",$link2);
// now you can
if ( $nt = mysql_fetch_array($q)){
echo $nt["total_number"];
$q=mysql_query($nt) or die(mysql_error());
}
?>
my file povezava.php looks like this:
<?
$servername='localhost';
$dbusername='user';
$dbpassword='pass';
$dbname1='db1';
$dbname2='db2';
$link1 = connecttodb($servername,$dbname1,$dbusername,$dbpassword);
$link2 = connecttodb($servername,$dbname2,$dbusername,$dbpassword);
function connecttodb($servername,$dbname,$dbusername,$dbpassword)
{
$link=mysql_connect ("$servername","$dbusername","$dbpassword",TRUE);
if(!$link){die("Could not connect to MySQL");}
mysql_select_db("$dbname",$link) or die ("could not open db".mysql_error());
return $link;
}
?>
Error that I get:
A PHP Error was encountered
Severity: NoticeMessage: Array to string conversionFilename: templates/master.phpLine Number: 231 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Array' at line 1
Fixed:
<?
require "povezava.php";
$q=mysql_query("SELECT COUNT(*) AS total_number FROM menjava WHERE date_submitted >= DATE_SUB(CURRENT_DATE(), INTERVAL 0 DAY)",$link2);
// working
if ( $nt = mysql_fetch_array($q)){
echo $nt["total_number"];
}
?>
Thank you!
Try :
$q = 'SELECT COUNT(*) FROM menjava
WHERE date_submitted >= DATE_SUB(CURRENT_DATE(), INTERVAL 1 DAY)';
$result = mysql_query($q);
$total_rows = mysql_fetch_row($result);
print $total_rows[0] . ' authors have been submitted today and yesterday.';
Please try the following SQL commands:
$sqlToday = "Select COUNT(*) FROM menjava WHERE DATE(date_submitted) = CURRENT_DATE()";
$sqlYesterday = "Select COUNT(*) FROM menjava WHERE DATE(dc_created) = CURDATE() - INTERVAL 1 DAY";
I use this function it's very simple
function statsofdays($database,$tableName,$coloneDate,$past_days_to_count=0)
{
$datenow = date("Y-m-d H:i:s");
$stringDays = "(SELECT COUNT(Id) FROM `".$tableName."` WHERE DAY(".$coloneDate.") = '".add_date_to_date($datenow,'0 days','d')."') as day1,";
for ($i=2; $i <= $past_days_to_count; $i++)
$stringDays .= "(SELECT COUNT(Id) FROM `".$tableName."` WHERE DAY(".$coloneDate.") = '".add_date_to_date($datenow,'-'.$i.' days','d')."') as day".$i.",";
$row = $database->query("SELECT ".$stringDays." (SELECT COUNT(Id) FROM `".$tableName."`) as total");
$stringReturn[0] = $row['total'];
$stringReturn[1] = $row['day1'];
for ($c=2; $c <= $past_days_to_count; $c++)
$stringReturn[$c] = $row['day'.$c];
return $stringReturn;
}
Params:
$database :
You can modify the function has your database structure, for me i use a class load to use ->query()
$tableName :
The name of the table you want to get data from
$coloneDate
Name of date format fields
$past_days_to_count
Number of days to go back to the past (if == 0 you get a count of total rows and today rows)
Example
$stats_of_year = statsofdays( $database , "table_name" , "creation_date" , 365 );
return
total rows for each day for 365 days like example (the values >= 0)
using
echo $stats_of_year['total'];
echo $stats_of_year['day1'];
...
echo $stats_of_year['day365'];
you need this also:
function add_date_to_date($stringDate, $days, $stringFormat)
{
$date = date_create($stringDate);
date_add($date,date_interval_create_from_date_string($days));
return date_format($date,$stringFormat);
}
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