Question

我有数据库＆＃34; db2＆＃34;与表＆＃34; menjava＆＃34;

在表格中，menjava有＆＃34; id＆＃34;，＆＃34; author＆＃34;和＆＃34; date_submitted＆＃34;字段

id - auto_increment
author - int（11）
date_submitted - datetime

我想基于名为＆＃39; date_submitted＆＃39;的DATETIME字段计算今天日期的所有行和昨天日期的所有行（因此将有两个具有条件的代码）。它包含每个记录创建的日期和时间。

在文件result.php中，显示了此计数，但它不起作用。在同一个文件（result.php）中我有一些其他代码来显示来自不同数据库的数据，所以我认为povezava.php工作正常。

我的代码：

 <?
    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());
}

    ?>

我的文件povezava.php看起来像这样：

<?
$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;
}
    ?>

我得到的错误：

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

修正：

<?
    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"];
}

    ?>

谢谢！

Answer 1

尝试：

$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.';

Answer 2

请尝试以下SQL命令：

$sqlToday = "Select COUNT(*) FROM menjava WHERE DATE(date_submitted) = CURRENT_DATE()";

$sqlYesterday = "Select COUNT(*) FROM menjava WHERE DATE(dc_created) = CURDATE() - INTERVAL 1 DAY";

我想根据DATETIME - mysql + php计算今天/昨天日期的所有行

2 个答案: