好吧,我在我正在关注的教程中有这行代码.但是,它没有为我提供有关递归的明确解释.我是cakephp的新手,并搜索了这个"递归".我希望有人可以给我一个外行人对这段代码的解释:
$this->Author->recursive = 1;
谢谢
我在CodeIgniter 2.1中收到以下错误消息:
A PHP Error was encountered
Severity: Notice
Message: Array to string conversion
Filename: database/DB_active_rec.php
Line Number: 1407
我正在尝试更新数据库中的字段.我有这个代码,每个教程对于批量上传都是一样的.无论如何它仍然有效但事情是它显示这样的错误.
这就是我在我的模型中所拥有的:
function update2($data){
$this->db->update_batch('users',$data, "id");
}
这就是我在我的控制器中所拥有的:
public function updateValues(){
$this->load->model('get_db');
$newRow = array(
array(
'id' => '3',
'firstname' => 'Rapphie'
),
array(
'id' => '2',
'firstname' => 'Charmie'
)
);
$this->get_db->update2($newRow);
echo "it has been updated";
}
我目前在phpmyadmin中收到此错误:
我尝试谷歌搜索,并给我尝试的每个解决方案.不幸的是,没有什么对我有用.
顺便说一下,我正在使用wamp服务器.
下面是来自C:\ wamp\apps\phpmyadmin3.2.0.1的config.inc.php的配置文件.
<?php
/* Servers configuration */
$i = 0;
/* Server: localhost [1] */
$i++;
$cfg['Servers'][$i]['verbose'] = 'localhost';
$cfg['Servers'][$i]['host'] = 'localhost';
$cfg['Servers'][$i]['port'] = '';
$cfg['Servers'][$i]['socket'] = '/tmp/mysql.sock';
$cfg['Servers'][$i]['connect_type'] = 'tcp';
$cfg['Servers'][$i]['extension'] = 'mysqli';
$cfg['Servers'][$i]['auth_type'] = 'config';
$cfg['Servers'][$i]['user'] = 'root';
$cfg['Servers'][$i]['password'] = '';
$cfg['Servers'][$i]['AllowNoPassword'] = true;
/* End of servers configuration */
$cfg['DefaultLang'] = 'en-utf-8';
$cfg['ServerDefault'] = 1;
$cfg['UploadDir'] = '';
$cfg['SaveDir'] = '';
?>
我每次尝试发送电子邮件时都会收到大量错误:
hello:
The following SMTP error was encountered:
Failed to send AUTH LOGIN command. Error:
from:
The following SMTP error was encountered:
to:
The following SMTP error was encountered:
data:
The following SMTP error was encountered:
The following SMTP error was encountered:
Unable to send email using PHP SMTP. Your server might not be configured to send mail using this method.
User-Agent: CodeIgniter
Date: Sun, 1 Jul 2012 20:47:47 +0000
From: "Rapphie"
Return-Path:
To: csorila17@gmail.com
Subject: =?iso-8859-1?Q?Email_Test?=
Reply-To: "alinorapjoseph@gmail.com"
X-Sender: …我正在使用codeigniter 2.1.2,现在就是我的情况.我在C:\ wamp\www\cr8v\application\models中有一个名为math.php的模型.我试图从我的控制器从C:\ wamp\www\cr8v\application\controllers\site.php加载它,但我得到此错误:
<?php
class Math_model extends CI_Model(){
public function add(){
return 1+1;
}
}
?>
这是math.php的内容
<?php
class Site extends CI_Controller{
public function index(){
$this->addstuff();
}
public function addstuff(){
$this->load->model("math_model");
echo $this->math->add();
}
}
?>
这是我的site.php的内容
( ! ) Parse error: syntax error, unexpected '(', expecting '{' in C:\wamp\www\cr8v\application\models\math_model.php on line 2
Call Stack
# Time Memory Function Location
1 0.0036 148768 {main}( ) ..\index.php:0
2 0.0062 185072 require_once( 'C:\wamp\www\cr8v\system\core\CodeIgniter.php' ) ..\index.php:202
3 0.0561 784312 call_user_func_array ( ) …我正在研究蛋糕烘焙.我按照从互联网上下载的书中的说明进入这个控制台的东西.但是,由于我在我的控制台中遇到了这个错误,因此无法再推进一步:
PHP Warning: mysql_connect(): MySQL server has gone away in C:\wamp\www\cake\cake\libs\model\datasources\dbo\dbo_mysql.php on line 543
Warning: mysql_connect(): MySQL server has gone away in C:\wamp\www\cake\cake\libs\model\datasources\dbo\dbo_mysql.php on line 543
PHP Warning: mysql_connect(): Error while reading greeting packet. PID=2328 in C:\wamp\www\cake\cake\libs\model\datasources\dbo\dbo_mysql.php on line 543
我的GUI工作正常,因为我正在我的localhost中正确加载Cake PHP的主页.它甚至说我连接到数据库.它与工作路径和应用程序路径有关吗？这个问题让我抓狂.感谢先进的回复:)
我得到这个错误:
Fatal error: Class 'Xml' not found in C:\xampp\htdocs\wifidrivescanportal\app\View\Helper\XmlHelper.php on line 88
我所做的是通过以下代码行在XML帮助器中导入库XML类:
XmlHelper.php
App::import('Core', array('Xml', 'Set'));
但我仍然得到同样的错误.我将xml库存储在此路径中:
C:\xampp\htdocs\wifidrivescanportal\app\Lib\xml.php
顺便说一下,在我的xml.php中,我检查过它是否存在这个类:
class XmlNode extends Object {
// ....
}
好吧,为了简化故事,我需要在cakephp中使用xml的"序列化".Dunno如果我在第一时间做这件事.感谢先进的帮助.
**编辑**
我也尝试在我的控制器中导入XML:
App::import('Xml');
但仍处于相同的情况
我正在尝试创建一个存储过程,它将使用用户在调用SP时传递的参数来搜索我的表.我想使用'LIKE'查询.到目前为止,这就是我得到的.
DROP PROCEDURE IF EXISTS `SEARCH_STUDENT`$$
CREATE
PROCEDURE `test`.`SEARCH_STUDENT`(_textInput VARCHAR(10))
BEGIN
SELECT * FROM records WHERE `lastname` LIKE '%'+_textInput+'%';
END$$
但我得到这个错误:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '@Search varchar(10);
SET @Search=_textInput;
SELECT * FROM records WHERE `last' at line 10
我从谷歌获得的是带有固定字符串的样本.我的取决于用户将在文本框中输入的内容.非常感谢您提供的任何帮助.
我正在创建一个cakephp 2.x应用程序.在开发过程中途,我突然发现自己有一个"找不到表"的错误.
Missing Database Table
Error: Table blocked for model Parental was not found in datasource default.
Notice: If you want to customize this error message, create project\View\Errors\missing_table
我检查了我的数据库,相关的表实际上就在那里.
接下来,我尝试调试模型.该模型可以访问该表.关于cakephp无法找到数据表的错误仍然存在.
这是我在调试模型时得到的结果:
\project\Controller\ParentalsController.php (line 5)
object(Parental) {
useTable => 'blocked'
useDbConfig => 'default'
id => null
data => array()
schemaName => null
table => 'blocked'
primaryKey => 'id'
validate => array()
validationErrors => array()
validationDomain => null
name => 'Parental'
alias => 'Parental'
tableToModel => array(
'blocked' => 'Parental'
)
cacheQueries …在cakephp中有一个规则,在创建表时,它必须是模型的复数形式.问题是,由于一些复杂的情况,我不能在这里重命名一个单数形式的表.这就是cakephp给我一个错误的原因.我试过这个:
class Color_Schema extends AppModel{
var $name = 'ColorSchema';
var $useTable = 'color_schema';
}
但它不起作用.
它仍然给我这个错误......
Error: Table color_schemas for model ColorSchema was not found in datasource default.
好吧,在我进入编码的另一个阶段之前,为了确保我不会在错误的树上吠叫,我可以问是否可以在没有任何按钮点击动作的情况下调用php中的javascript函数来触发它.我也不考虑onload()或onready(),因为即时通讯将在php中的foreach循环中调用javascript函数..这个概念有点像这样:
<script type="text/javascript">
function callMe(){
alert('im called!');
}
</script>
<body>
<?php
.....
foreach($somevar as $var){
// assuming it will loop 5 times
callMe(); // well this is the part where im going to call the javascript function
}
?>
</body>
先感谢您.
******编辑部*****
这是我打算使用的实际代码:
function AddCoordinate( lat, long )
{
var path = LinePath.getPath();
path.push( new google.maps.LatLng( lat, long ) );
}
<?php
foreach($arrayOfPlotPoints as $key => $value){
$longitude = round($value['longitude'],5);
$latitude = round($value['latitude'],5);
$snrLevel = …cakephp ×4
mysql ×4
php ×4
codeigniter ×3
model ×2
sql ×2
connection ×1
database ×1
email ×1
function ×1
javascript ×1
search ×1
xml ×1