当我在phpMyAdmin中运行此查询时,它会加载所有行。但是当我通过PHP SQL运行它时,我只得到第一行。我不确定我做错了什么,我真的压力很大。
下面是我的代码:
function getRows($user)
{
$q = mysql_query("SELECT day(closed) AS day, COUNT( closed ) AS c FROM ost_ticket WHERE year(closed) = '2013' AND monthname(closed) = 'January' AND source = '
我现在有这样的代码:
$rawsql2 = "SELECT
*
FROM
_erc_foffices n
INNER JOIN
_erc_openings o ON n.id = o.branch_id
INNER JOIN
_erc_openings_times t ON o.id = t.opening_id
WHERE
(
n.id = %d
);";
$sql2 = sprintf($rawsql2, mysql
我有一张桌子上有事件和日期,另一张桌子上有每个晚上的客人名单。我可以打印这样的夜晚:
$day = date("l");
$date = getFullDateString($day);
$result = mysql_query("SELECT * FROM nights WHERE day = '$day' AND promoter = 'blah'") or die(mysql_error);
while ($row = mysql_fetch_array($result)) {
echo "<h2
好的,我有一个需要搜索的日期字段,但我需要像在mysql查询中那样按天搜索它
search_conditions << ["DAY(open_date) != ?", event.thursday.day] if options[:thur].blank?
我需要在思考Sphinx的情况下这样做,所以我尝试了这个
attr_accessor :event_day
def event_day
self.start_date.day
end
#thinking sphinx configurations for the e
isset()函数可以用来检查输入类型submit是否被按下,但是有没有办法检查输入类型按钮是否被按下了呢?
在我的代码中,按钮除了在.Onclick()事件上调用一个函数之外什么也不做,该函数随后刷新页面并在PHP...and内创建一个数据库条目。我希望它只在按钮为pressed...and后才创建条目。我不能对其他reasons...following使用submit类型。下面是一些代码:
function send()
{
var events = document.getElementById("event").value;
location.href
我经常看到人们用这样的查询来回答MySQL问题:
SELECT DAY(date), other columns
FROM table
GROUP BY DAY(date);
SELECT somecolumn, COUNT(*)
FROM table
HAVING COUNT(*) > 1;
我总是喜欢给列一个别名,并引用GROUP BY或HAVING子句中的别名。
SELECT DAY(date) AS day, other columns
FROM table
GROUP BY day;
SELECT somecolumn, COUNT(*) AS c
FROM table
H
我有一个网站,在那里我可以显示mysql表中的信息,例如上次更改的日期。在mysql中,字段类型是'TIMESTAMP‘...我有一个这样的代码:
$postdate = date( "j F", strtotime( $row['insert_date'] ) ); // Getting the date from the database
$posthour = date( "H:i", strtotime( $row['insert_date'] ) ); // Getting the time from the d
我有代码:
@Id
@Column(name = "id")
@GeneratedValue
private int id;
@Formula(value = "(SELECT count(history.city_id) FROM history where history.ts > (now() - INTERVAL 30 DAY) and history.city_id = id)")
private int last30daysUpdates;
因此,hiberante将此公式解析为:
...where
history.
我使用sqoop将数据从MySQL导入到hdfs,并将其作为片材文件,在该文件中由Impala使用。将MySQL日期类型转换为Impala时间戳存在问题。
执行compute stats table或select *时的Impala错误消息是:
File 'hdfs://....parquet'
has an incompatible type with the table schema for column 'day'.
Expected type: INT32. Actual type: INT64
将日期列的数据类型更改为BIGINT或STRING将使错
我有一个php表,它在表中显示月份的天数:
//count up the days, untill we've done all of them in the month
while ( $day_num <= $days_in_month )
{
echo "<td> $day_num </td>";
$day_num++;
$day_count++;
我有第二个表,它为每个月的每一天调用一个单独的查询:
<tr>
<td> </td>
当我尝试在已创建的数据库中创建表时,我以A link to the server could not be established in /var/www/Test.php on line 25和mysql_select_db(): Access denied for user 'pass'@'localhost' (using password: NO) in /var/www/Test.php on line 25的身份收到错误。
我的代码:
<?php
$con = mysqli_connect("localhost",&
我的sql查询有问题。
此代码用于发布日期为一天的显示文章。
示例:
BLOG ONE: day of pubblication every 16th of mounth
BLOG TWO: day of pubblication every 2nd of mounth
BLOG THREE: day of pubblication every 23th of mounth
所以今天24号我必须看博客,发布日期是24,25,26,27,28,1,2,3,4,5.................
问题是合并两个查询
( DAY FROM TODAY TO 31th) + (DAY FROM 1s