小编cra*_*xxx的帖子

对于某些人来说,它可能看起来很简单,但我根本无法得到它.
我一遍又一遍地从同一个表中有多个MS-SQL SELECT查询:

SELECT count(Page) as tAEC
FROM someTable
WHERE Page LIKE '%AEC%'

下一个

SELECT count(Page) as tTOL
FROM someTable
WHERE Page LIKE '%TOL%'

等等...

什么是更有效的方式来编写此查询.我用Google搜索了一堆类似的问题,但我无法让它们中的任何一个起作用.所以任何帮助都非常感谢.

我正在尝试使用Excel将一些旧的SQL Server数据传输到SQL Server.Import/Export Data应用程序似乎自动将大多数数据列设置为NVARCHAR(255).我遇到的问题是,我的一个列应该是一个DATE类型,但其中的所有数据都是这样的18.08.2000 14:48:15.

所以,当我尝试使用此查询时:

SELECT CONVERT(Date, DATE_TIME,  113)
FROM someTable

我收到此错误:

消息9807,级别16,状态0,行1
输入字符串不遵循样式113,更改输入字符串或使用不同的样式.

None of the [styles] from CAST and CONVERT (Transact-SQL) are working in my case.

Any advise or help is greatly appreciated.

SOLVED:

UPDATE myTable
SET columnName = CONVERT(NVARCHAR(255),CONVERT(SMALLDATETIME, columnName,105))
ALTER TABLE myTable
ALTER COLUMN columnName SMALLDATETIME

我试图弄清楚如何输出我的PHP回声我的json文件.

<?php
$sql_query = "SELECT * FROM some_data";
$result = mysql_query($sql_query);
$total_rows = mysql_num_rows($result);
$dataList_desc = '';
$dataList_view = '';
    while($row = mysql_fetch_array($result)) {
            $video_desc = $row["video_desc"];
            $video_views = $row["video_views"];
            $dataList_desc .= '' . $video_desc . '';
            $dataList_view .= '' . $video_views . '';
            echo "[[\" \", \"" . $dataList_desc . "\",],[\" \", \"" . $dataList_view . "\",]";

    }
  mysql_close($con);
?>

我的输出是:

[[" ", "Title 1",],[" ", "1000",][[" ", "Title 1Title 2",],[" ", "1000906",][[" ", "Title 1Title 2Title 3",],[" ", …

我有这个MS-SQL查询在数据库中有数千行记录:

SELECT DISTINCT TOP 7 DATENAME(MM, mydatetime) + ' ' + CAST(DAY(mydatetime) AS VARCHAR(2)) as thedate
    , MONTH(mydatetime)
    , DAY(mydatetime)
    , COUNT(Page) as totalcount
    , count(DISTINCT Page) as visitors
  FROM someTable
  WHERE Page LIKE '%AEC%'
  GROUP BY DATENAME(MM, mydatetime) + ' ' + CAST(DAY(mydatetime) AS VARCHAR(2))
    , MONTH(mydatetime)
    , DAY(mydatetime)
  ORDER BY MONTH(mydatetime) DESC
    , DAY(mydatetime) DESC

它将输出:

thedate     | totalcount | visitors
-----------------------------------
October 17  |     4      |    1
October 15  |     1      |    1
October 12  |     1 …

我有这个HTML选择代码:

?<select name="loc_code">
  <option value="">Select</option>
  <option value="loc1">Loc1 status ON</option>
  <option value="loc2">Loc2 status ON</option>
  <option value="loc3">Loc3 status OFF</option>
  <option value="loc4">Loc4 status ON</option>
</select>????????????????????????????????????????????????

所以,我有一个问题是根据文本名称而不是值来禁用选项下拉列表.我只需要禁用包含状态OFF的选项.

我试图从StackOverflow上的其他类似问题修改这个jQuery代码,但显然这是错误的:

$("select[name="loc_code"] option:selected").text(function() {
  var $thisOption = $(this);
  var valueToCompare = "OFF";

  if($thisOption.val() == valueToCompare) {
    $thisOption.attr("disabled", "disabled");
  }
});?

任何帮助是极大的赞赏.