更改投票值后,表单将更改POST,然后刷新页面.这在加载时在页面顶部调用:
if (isset($_POST['q'.$question_id]))
{
$user->updateQuestionVotes($question_id, $_POST['q'.$question_id]);
}
为什么每次我第一次刷新后都会更新？我需要以某种方式取消它吗？
dispatch_address_postcode
不是强制性的,即使它是空白的,它仍然会运行:
if (!is_null($_POST['personal_info_first_name']) &&
!is_null($_POST['personal_info_surname']) &&
!is_null($_POST['personal_info_email']) &&
!is_null($_POST['personal_info_telephone']) &&
!is_null($_POST['dispatch_address_country']) &&
!is_null($_POST['dispatch_address_first_name']) &&
!is_null($_POST['dispatch_address_surname']) &&
!is_null($_POST['dispatch_address_address']) &&
!is_null($_POST['dispatch_address_town']) &&
!is_null($_POST['dispatch_address_postcode']) &&
!is_null($_POST['dispatch_address_county']) &&
( ($_POST['payment_method'] == "Pay by credit card.") ||
(
($_POST['payment_method'] == "Pay by new credit card.") &&
!is_null($_POST['card_number']) &&
!is_null($_POST['expiration_date']) &&
!is_null($_POST['security_code'])
)
)
)
是什么赋予了？
试着在这里按照wiki上的示例:扩展模型Auth用户类,并且当它进行验证时我遇到致命错误.
ErrorException [ Fatal Error ]: Class 'Validate' not found
有任何想法吗？这是它失败的代码片段:
class Model_User extends Model_Auth_User
{
public function validate_create(& $array)
{
// Initialise the validation library and setup some rules
$array = Validate::factory($array)
->rules('password', $this->_rules['password'])
->rules('username', $this->_rules['username'])
->rules('email', $this->_rules['email'])
->rules('password_confirm', $this->_rules['password_confirm'])
->filter('username', 'trim')
无法想象我的生活.尝试从客户端证券表中返回列名,然后将结果作为数组返回.谁能指出我离开的地方？
mysql_select_db("HandlerProject", $con); //Selects database
$selectcols = "SELECT * FROM ".$clientname."securitiestable"; //selects all columns from clients security table
$tempcols = mysql_query($selectcols) or die(mysql_error());
$returnedcols = $mysql_fetch_array($tempcols);
$tempsymbol = mysql_query("SHOW COLUMNS FROM".$clientname."securitiestable");
$symbol = $mysql_fetch_array($tempsymbol);
不知道这里出了什么问题.继续投掷......
致命错误:在非对象上调用成员函数prepare()
......每次到达$select = $dbcon->prepare('SELECT * FROM tester1');零件.有人可以说明我做错了什么吗？
function selectall() //returns array $client[][]. first brace indicates the row. second indicates the field
{
global $dbcon;
$select = $dbcon->prepare('SELECT * FROM tester1');
if ($select->execute(array()))
{
$query = $select->fetchall();
$i = 0;
foreach ($query as $row)
{
$client[$i][0] = $row['id'];
$client[$i][1] = $row['name'];
$client[$i][2] = $row['age'];
$i++;
}
}
return $client;
}
$client = selectall();
echo $client[0][0];
第一次玩jquery,我正在努力让一个简单的AJAX设置工作,这样我就能更好地理解它是如何工作的.不幸的是,我不知道很多.这是带脚本的HTML:
<html>
<head>
<title>AJAX attempt with jQuery</title>
<script type="text/javascript" src="jquery.js"></script>
<script type="text/javascript">
function ajax(str){
$("document").ready(function(){
$.post("ajaxjquerytest.php",str,function(){
$("p").html(response);
});
});
</script>
</head>
<body>
<input type="text" onchange="ajax(this.value)"></input>
<p>Age?</p>
</body>
</html>
这是它正在谈论的PHP:
<?php
$age = $_POST['age'];
if ($age < 30)
{
echo "Young";
}
else if ($age > 30)
{
echo "Old";
}
else
{
echo "you're 30";
}
?>
试图做一个简单的切换菜单,我似乎无法使用这个jQuery隐藏/显示子菜单:
$(".topic news").mouseup(function(){
$(".feed groups").hide("fast", function(){
$(".feed messages").hide("fast");
$("ul.feed news").toggle("fast");
});
});
这是相应的HTML:
<div class="topic news">
<span>News Feed</span>
</div>
<ul class="feed news">
<li>News item #1</li>
<li>News item #1</li>
<li>News item #1</li>
<li>News item #1</li>
<li>News item #1</li>
</ul>
有任何想法吗？
认为这非常简单,但我花了半个小时试图弄明白无济于事.
$unique_id = uniqid(microtime(),1);
if (is_null($_COOKIE['client_id']))
{
setcookie("client_id", $unique_id);
}
但是当我通过$ _COOKIE ['client_id']检查cookie时,我得到一个空值.有什么想法吗？
php ×7
jquery ×2
mysql ×2
post ×2
ajax ×1
arrays ×1
cookies ×1
css ×1
html ×1
isnull ×1
javascript ×1
kohana-3 ×1
kohana-auth ×1
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validation ×1