这个让我疯了.它是一个简单而愚蠢的东西,我在俯视.我在表单中有一个多选框.我只是想获取所选的值.在我的循环中,如果我使用警报,那么我没有问题.一旦尝试连接值我得到错误'SelBranch [...] .selected'是null或不是对象
<form name="InventoryList" method="post" action="InventoryList.asp">
<select name="SelBranch" class="bnotes" size="5" multiple="multiple">
<option value="All">All</option>
<option value="001 Renton">001 Renton</option>
<option value="002 Spokane">002 Spokane</option>
<option value="003 Missoula">003 Missoula</option>
<option value="004 Chehalis">004 Chehalis</option>
<option value="005 Portland">005 Portland</option>
<option value="006 Anchorage">006 Anchorage</option>
<option value="018 PDC">018 PDC</option>
</select>
<input type="button" name="ViewReport" value="View" class="bnotes" onclick="GetInventory();">
</form>
<script language="JavaScript">
function GetInventory()
{
var InvForm = document.forms.InventoryList;
var SelBranchVal = "";
var x = 0;
for (x=0;x<=InvForm.SelBranch.length;x++)
{
if (InvForm.SelBranch[x].selected)
{
//alert(InvForm.SelBranch[x].value);
SelBranchVal = SelBranchVal + "," + …Run Code Online (Sandbox Code Playgroud) 我的服务器是Windows 2008服务器.PHP V7.2.7已安装并正在运行.Sql Server 11(64位)已安装并正在运行(有几个asp.net应用程序在运行且已经在使用该数据库)
我从Microsofts网站下载了PHP Sql Server驱动程序,并将.dll文件放在PHP ext目录中.
在我的PHP.ini中,我添加了:extension=php_pdo_sqlsrv_7_nts_x64
在我的.php文件中,我用来测试我的数据库连接:
$SqlServer = "THISSERVER\SQLEXPRESS";
$SqlServerCon = new PDO("sqlsrv:server=$SqlServer;Database=TheDatabase", "DbUName", "DbPassword");
if (!$SqlServerCon) {die('Unable To Connect to Sql Server');}
else
{echo "Connection Successful";}
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我得到:(
PHP Fatal error: Uncaught PDOException: could not find driver in D:\Inetpub\wwwroot\TechStory2\DBtest.php:7第7行是$ SqlServerCon行).
我做错了什么?以及如何建立与Sql Server的连接需要做什么?
我一直在玩Google的DistanceMatrixService.下面的代码可行,但是,如何将另一个参数传递给回调函数或从回调中获取其中一个值?
例如:我有两个div想要在(Results1和Results2)中显示不同的结果,所以我想我需要
将另一个值传递给GoogleMapDistance函数,如GoogleMapDistance(YourLatLong,DestLatLong,TheDiv)
或者
能够抓取ResultStr位于回调文档外部.getElementById ("Results1").innerHTML = ResultStr;
或者
将innerHTM设置为函数document.getElementById("Results1")的返回值.innerHTML = GoogleMapDistance(YourLatLong,DestLatLong);
我被卡住了.我怎么能做到这一点?它现在的样子是我只能运行所有这些代码并且只能写入一个div.
<div id="Results1"></div>
<div id="Results2"></div>
<script src="https://maps.googleapis.com/maps/api/js?v=3.exp&sensor=false"></script>
<script>
function GoogleMapDistance(YourLatLong,DestLatLong)
{
var service = new google.maps.DistanceMatrixService();
service.getDistanceMatrix(
{
origins: [YourLatLong],
destinations: [DestLatLong],
travelMode: google.maps.TravelMode.DRIVING,
unitSystem: google.maps.UnitSystem.IMPERIAL,
avoidHighways: false,
avoidTolls: false
}, callback);
}
function callback(response, status)
{
if (status == google.maps.DistanceMatrixStatus.OK)
{
var origins = response.originAddresses;
var destinations = response.destinationAddresses;
for (var i = 0; i < origins.length; i++)
{
var results = response.rows[i].elements;
for …Run Code Online (Sandbox Code Playgroud) 在尝试阅读有关使用PHP附件发送电子邮件的各种文章后(我使用的是VBScript的ASP),我编写了下面的代码.不幸的是,它根本不起作用.它不仅不会发送带附件的电子邮件,电子邮件似乎根本不发送,即使我的脚本说它确实发送了.我哪里出错了?我没有使用表单上传文件.这是一个静态脚本.
<?php
$EmailTo = "Me@here.com";
$EmailFrom = "You@There.com";
$EmailSubject = "The Email Subject";
$MailBoundary = md5(uniqid(time()));
$Headers = "To: ". $EmailTo . "\r\n";
$Headers .= "From: ". $EmailFrom . "\r\n";
$Headers = "MIME-Version: 1.0\r\n";
$Headers .= "Content-type: multipart/mixed;boundary=\"$MailBoundary \"";
$Headers .= "\r\n\r\n";
$Headers .= "This is a multi-part message in MIME format.";
$Headers .= "\r\n\r\n";
$FileAttachment = "AttachedFile.pdf";
$File = fopen($FileAttachment, "r");
$FileData = fread($File, filesize($FileAttachment));
$FileData = chunk_split(base64_encode($FileData));
$FileName = basename($FileAttachment);
$EmailBody = "--$MailBoundary\r\n";
$EmailBody .= "Content-type: text/html; charset=iso-8859-1\r\n"; …Run Code Online (Sandbox Code Playgroud)