小编Ric*_*ich的帖子

当我运行这个时,我收到一个错误'未声明的变量:temp'

<?php 

$maketemp = "CREATE TEMPORARY TABLE temp(`itineraryId` int NOT NULL, `live` varchar(1), `shipCode` varchar(10), `description` text, `duration` varchar(10), PRIMARY KEY(itineraryId))"; 

mysql_query( $maketemp, $connection ) or die ( "Sql error : " . mysql_error ( ) );

$inserttemp = "SELECT live, id AS itineraryId, ship AS shipCode, description AS description, duration AS length FROM cruises WHERE live ='Y' INTO temp";

mysql_query( $inserttemp, $connection ) or die ( "Sql error : " . mysql_error ( ) );

$select = "SELECT intineraryId, …

好吧现在我的滑动打开ul显示所有列表元素,现在我想要点击的标题栏添加一个选定状态,然后在关闭时删除该状态...

UL上面的div有一个.regionHeader类

这是标记的一个例子

<div class="regionHeader">title of the region</div>
<ul class="region"><li>the region i'm hiding/showing</li></ul>

这是javascript

var stockists = {

start: function() {
        $('.region').hide();

            $('.regionTitle').each(function(){

                $(this).click(function(e){
                    e.preventDefault();

                    $(this).parent().next('.region').slideToggle(300);
                });                     
            });
        }            
};

$(stockists.start);

我一直在尝试addClass,但它似乎只是添加类而不是删除它？

我正在运行这个查询,它正在填充我的临时表,然后我从中导出.priceSingle是通过简单地加倍票价来计算的.但是,对于某些行,该类别无论如何都是单一票价,所以我不想加倍.该查询实际上嵌入在PHP中的字符串中,但SQL看起来像:

INSERT INTO temp
    SELECT cruises.code AS sailingId, live, 'USD', 'optioncode', 'rateOptionName', '2',
           category AS cabinCategory, fare*2 AS priceSingle, fare AS priceDouble,
           supplements_usa.adult AS priceAdditional
      FROM fares_usa, supplements_usa, cruises
     WHERE fares_usa.cruise_id = supplements_usa.cruise_id
       AND cruises.id = fares_usa.cruise_id AND cruises.live = 'Y'

fare*2当category等于5或10 时,我怎么能避免做？

我有一个简单的插入记录,用于将项添加到记录集.价格的数据字段设置为十进制,当价格字段发送时,例如9.50,它在数据库中显示为9.00.为什么？

这是php insert语句中的代码

if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form1")) {
$insertSQL = sprintf("INSERT INTO buying (`user`, items, price) VALUES (%s, %s, %s)",
                   GetSQLValueString($_POST['user'], "text"),
                   GetSQLValueString($_POST['items'], "int"),
                   GetSQLValueString($_POST['price'], "int"));