我有以下代码,我得到例外:
已经有一个
DataReader与此相关的开放Connection,必须先关闭.
我在这个项目中使用Visual Studio 2010/.Net 4.0和MySQL.基本上我在尝试使用数据读取器执行其他任务时运行另一个SQL语句.我正在排队cmdInserttblProductFrance.ExecuteNonQuery();
SQL = "Select * from tblProduct";
//Create Connection/Command/MySQLDataReader
MySqlConnection myConnection = new MySqlConnection(cf.GetConnectionString());
myConnection.Open();
MySqlCommand myCommand = new MySqlCommand(SQL, myConnection);
MySqlDataReader myReader = myCommand.ExecuteReader();
myCommand.Dispose();
if (myReader.HasRows)
{
int i = 0;
// Always call Read before accessing data.
while (myReader.Read())
{
if (myReader["frProductid"].ToString() == "") //there is no productid exist for this item
{
strInsertSQL = "Insert Into tblProduct_temp (Productid) Values('this istest') ";
MySqlCommand cmdInserttblProductFrance = new MySqlCommand(strInsertSQL, …我有一个如下代码,我不知道什么类型的数据变量$ACTIVITYGROUPS[]有什么,我怎么读它？
$ACTIVITYGROUPS[] = saprfc_table_read ($fce, "ACTIVITYGROUPS", $i);
当我这样做时,print_r(saprfc_table_read ($fce, "ACTIVITYGROUPS", $i);我得到了一堆没有任何分隔符的数组,并且不确定如何精确数据.谁可以告诉我它在上面的句子中做了什么？
print_r(saprfc_table_read ($fce, "ACTIVITYGROUPS", $i);结果给出了以下结果:
Array (
[AGR_NAME] => Y:SECURITY_DISPLAY
[FROM_DAT] => 20080813
[TO_DAT] => 99991231
[AGR_TEXT] => Security Display - Users & Roles
[ORG_FLAG] => C
)
Array (
[AGR_NAME] => Y:SECURITY_ADMIN_COMMON
[FROM_DAT] => 20080813
[TO_DAT] => 99991231
[AGR_TEXT] => Security Administrator
[ORG_FLAG] => C
)
Array (
[AGR_NAME] => Y:LOCAL_TRANSPORT
[FROM_DAT] => 20090810
[TO_DAT] => 99991231
[AGR_TEXT] => Transport into target client - …我有下面的代码和java脚本抛出错误"无法获取未定义或空引用的属性'值'".我究竟做错了什么 ？下面是我试图执行以验证一个输入字段的示例代码.
<html lang="en" xml:lang="en" xmlns="http://www.w3.org/1999/xhtml">
<head>
<title></title>
<meta content="text/html;charset=iso-8859-2" http-equiv="content-type" />
<script type="text/javascript">
function validate_frm_new_user_request()
{
alert('test');
valid = true;
if ( document.frm_new_user_request.u_isid.value == '' )
{
alert ( "Please enter your valid ISID Information." );
document.frm_new_user_request.u_isid.focus();
valid = false;
}
return valid;
</script>
</head>
<body
<form method="post" action="" name='frm_new_user_request' id="frm_new_user_request" onsubmit="return validate_frm_new_user_request();">
<center>
<table>
<tbody>
<tr align="left">
<td><Label>ISID<em>*:</Label><input maxlength="15" id="u_userid" name="u_userid" size="20" type="text"/></td>
<td>
<tr>
<td align="center" colspan="4">
<input type="image" src="images/button/btn_create_request.gif" border="0" ALT="Create New Request">
</td>
</tr>
</table> …我有下面的代码，但出现错误：
警告：ldap_bind()：无法绑定到服务器：无法联系第 17 行 C:\xampp\htdocs\ldap.php 中的 LDAP 服务器
<?php
$ldapconfig['host'] = "dsua1.company.com";
$ldapconfig['port'] = 636;
$ldapconfig['basedn'] = "cn=userid,ou=Applications,ou=Company,ou=Services,dc=iM-2,dc=com";
$ldapconfig['binddn'] = "userid";
$ldapconfig['bindpw'] = "password";
$ldapconn=ldap_connect($ldapconfig['host'],$ldapconfig['port']);
ldap_bind($ldapconn, $ldapconfig['binddn'], $ldapconfig['bindpw']);
?>