我有一个HTML Ajax网站,它只包含一个HTML文件.如何创建iPhone应用程序呢?HTML文件应该本地存储在iPhone上,以便脱机运行.
我正在尝试编写一段代码来解析IE和Firefox中的javascript.
以下适用于IE,并且功能无需在firefox中抱怨
function XmlDom(sXml){
var oXml;
if (window.ActiveXObject) {
// ie
oXml = new ActiveXObject("Microsoft.XMLDOM");
oXml.resolveExternals = false;
oXml.async = false;
oXml.loadXML(sXml);
}
else if (window.DOMParser){
var parser = new DOMParser();
oXml = parser.parseFromString(sXml, "text/xml");
}
return oXml
}
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以下适用于IE,但在Firefox下会出现错误(因为childNodes不存在)
var oXml = XmlDom(sourceXML);
var listHtml = "";
if (oXml.firstChild != null) {
var childNodes = null;
try {
childNodes = oXml.lastChild.lastChild.firstChild.childNodes;
}
if (childNodes != null && childNodes.length > 0) {
for (var i = 0; i < childNodes.length; …Run Code Online (Sandbox Code Playgroud) 这是我用memcache"存储"mysql查询的代码
$mem = new Memcache;
$mem->connect('127.0.0.1', 11211);
$query = "SELECT * FROM tbl limit 0,20 ";
$key = md5($query);
$get_data = $memcache->get($key);
if($get_data) {
echo 'Data Pulled From Cache';
} else {
$res = mysql_fetch_array(mysql_query($query));
$memcache->set($key, $res, TRUE, 3600);
}
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问题是memcache只存储查询返回的第一行.如何在memcache中的一个键中保存所有20行?
我需要检查浏览器JavaScript是否已关闭,然后显示错误div而不是正文,我该怎么做?
我正在尝试使用codeigniters电子邮件库发送电子邮件,这是我的代码,
$this->load->library('email');
$this->email->initialize(array('mailtype' => 'html'));
$this->email->from('emailaddresshidden'); // TODO - Store this in Config file??
$this->email->bcc('emailaddresshidden');
$this->email->subject('Competition Entry'); // TODO - What does this want to be??
$this->email->message($this->load->view('emails/competition_entry', $this->data, TRUE));
$this->email->send();
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但是我收到了这个错误,
遇到PHP错误
严重性:警告
消息:mail()期望参数1是字符串,给定数组
文件名:libraries/Email.php
行号:1519
我不知道为什么?
我是zend的新手,我想问一下,如果用户闲置10分钟,我怎样才能使用户的会话命名空间的特定密钥失效.我在zend会话中定义了一个命名空间
$session = new Zend_Session_Namespace('loginNamespace');
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现在,当用户登录时,我在会话命名空间中设置了键loggedIn = 1.现在,如果用户保持空闲但只有该密钥,我希望不会使整个会话到期.我怎样才能做到这一点?
请附上我的代码:
<?PHP
session_start();
if (isset($_POST["usernameform"])){
$username = $_POST['usernameform'];
$password1 = $_POST['passwordform'];
$user_name = "XXXX";
$password = "XXXXX";
$database = "XXX";
$server = "XXXX";
$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);
$SQL = "SELECT * FROM login WHERE Username = '$username' AND Password = '$password1' "; //grab all the records from table
$result = mysql_query($SQL)
or die("Error:" . mysql_error());
if (mysql_num_rows($result) > 0){ //if username and password match return number of rows is always 1
$_SESSION['login'] = $username; //by …Run Code Online (Sandbox Code Playgroud) <?php
session_start();
$user = $_SESSION['login'];
$mysql_connect = mysql_connect("localhost", "root", "");
$mysql_select_db = mysql_select_db("site");
$sql = "SELECT * FROM `msg_inbox` WHERE `to` = '$user'";
$result = mysql_query($sql) or die(mysql_error());
while ($row = mysql_fetch_array($result)) {
$mysql_connect = mysql_connect("localhost", "root", "");
$mysql_select_db = mysql_select_db("site");
$query = ("UPDATE msg_inbox SET unread = 0 WHERE id= ".$row['id']);
$result = mysql_query($query);
}
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我收到错误:警告:mysql_fetch_array():提供的参数不是第8行的有效MySQL结果资源你能帮助我吗?
我试图得到php和ajax帖子的悬念.我通过文本框将简单的用户输入发布到php页面.php页面接受用户输入,计算单词并将其作为响应发送回去.但它无法接收数据,回来说.为什么我该如何解决?
这是我的jQuery ajax邮政编码
var dataString = name;
$.ajax({
type: "POST",
url: "test_get.php",
data: dataString,
success: function(response) {
alert(response);
}
});
return false;
});
});
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这是我的PHP代码
// if data are received via POST, with index of 'dataString'
if (isset($_POST['dataString'])) {
$str = $_POST['dataString']; // get data
echo "The string: '<i>".$str."</i>' contains ". strlen($str). ' characters and '. str_word_count($str, 0). ' words.';
}
else echo 'There is no data!';
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