我在php codeigniter中开发了一个项目.该项目已接近完成,但现在我的经理要我完全删除或隐藏URL中的控制器名称.我当前的网址如下所示:
我的经理希望它看起来像这样:
请注意,我的项目中有10个以上的控制器和许多功能.我想要一个适用于所有人的方法.请帮忙.
该函数FIND_IN_SET()在本地运行,但在服务器上不起作用。当我company_by_category_model通过调用controller(controller name)加载时 ,它将引发错误
FUNCTION firstdial.FIND_IN_SET does not exist
SELECT * FROM (`company_information`, `user_information`)
WHERE `FIND_IN_SET` ('16',company_category)
AND company_information.allowstatus=1
AND company_information.delstatus=0
AND company_information.user_id=user_information.user_id
ORDER BY `company_id` desc
Filename: C:\wamp\www\firstdial\system\database\DB_driver.php
Line Number: 331
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function company_by_category_model($category_id)
{
$this->db->select('*');
$this->db->from('company_information,user_information');
$search="FIND_IN_SET ('$category_id',company_category)
AND company_information.allowstatus=1
AND company_information.delstatus=0
AND company_information.user_id=user_information.user_id";
$this->db->where($search);
$this->db->order_by('company_id','desc');
$query=$this->db->get();
return $query->result();
}
Run Code Online (Sandbox Code Playgroud) 在我下面的图片中,您可以看到我希望能够将删除文件链接的位置更改为内容div区域的外侧.
我尝试过自定义模板,没有运气.
问题如何在内容div区域外部删除文件按钮?而不是图像下方.
我只希望能够在不同的区域中删除图像链接
<div id="content">
<div id="my-dropzone" class="dropzone">
<span class="">Content Div Area</span>
<div class="dz-message">
<h3>Drop files here</h3> or <strong>click</strong> to upload
</div>
</div>
<div class="dropzone_preview" id="dz-preview">
<div class="dropzone_details">
<img data-dz-thumbnail />
</div>
<a id="dz-remove" href="javascript:undefined;" data-dz-remove>Remove <-- here</a>
</div>
</div>
<script src="<?php echo base_url(); ?>vendor/jquery/jquery.min.js"></script>
<script src="<?php echo base_url(); ?>vendor/dropzone/dropzone.min.js"></script>
<script>
Dropzone.autoDiscover = false;
var myDropzone = new Dropzone("#my-dropzone", {
url: "<?php echo site_url("images/upload") ?>",
acceptedFiles: "image/*",
addRemoveLinks: true,
thumbnailWidth: null,
thumbnailHeight: null,
addRemoveLinks: true,
previewTemplate: document.getElementById('dz-preview').innerHTML,
removedfile: function(file) …Run Code Online (Sandbox Code Playgroud) 我很难在我的查询中添加括号.我的查询包括OR,AND和LIKE.这是我的MySQL查询:
SELECT * FROM (`vendor`) JOIN `vendor_detail` ON `vendor`.`id_vendor`
= `vendor_detail`.`id_vendor` JOIN `klasifikasi` ON `vendor_detail`.`id_klasifikasi` = `klasifikasi`.`id_klasifikasi`
WHERE `vendor_detail`.`id_klasifikasi` = 'K0002' AND
`vendor`.`st_aktif` = 1 AND (`vendor`.`id_vendor` LIKE '%eka%' OR
`vendor`.`nama_vendor` LIKE '%eka%' OR `vendor`.`alamat` LIKE '%eka%'
OR `vendor`.`telp` LIKE '%eka%' OR `vendor`.`mobile` LIKE '%eka%' OR
`vendor`.`fax` LIKE '%eka%' OR `vendor`.`cp` LIKE '%eka%' OR
`vendor`.`telp_cp` LIKE '%eka%' OR `vendor`.`catatan` LIKE '%eka%' OR
`vendor`.`email` LIKE '%eka%')
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如您所见,我vendor.st_aktif = 1 AND在查询的最后一行之后和之后添加括号.
如何实现上述查询到Code Igniter Active …
错误:从'来源' http://示例访问' http://www.example.com//assets/global/plugins/font-awesome/fonts/fontawesome-webfont.woff2?v=4.4.0 '中的字体.com '已被CORS策略阻止:请求的资源上没有"Access-Control-Allow-Origin"标头.因此,不允许来源" http://example.com "访问.
解:
<?php
header('Access-Control-Allow-Origin: *');
class Home extends CI_Controller {
public function index()
{
$this->load->view('master');
}
}
?>
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我尝试了这个解决方案,但它无法工作,请你帮帮我如何解决?以及如何从URL中删除index.php?
我正在使用chriskacerguis/codeigniter-restserver来创建我的其他api服务器,我正在尝试启用api密钥,但问题是即使我按照说明操作,我也无法使用任何密钥进行身份验证,即使所有请求都通过,即使他们在请求中没有api密钥头.
这是我目前的配置
rest.php
$config['rest_keys_table'] = 'keys';
$config['rest_enable_keys'] = TRUE;
$config['rest_key_column'] = 'api_key';
$config['rest_limits_method'] = 'ROUTED_URL';
$config['rest_key_name'] = 'X-API-KEY';
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我的表是使用以下sql查询创建的.
CREATE TABLE `keys` (
`id` int(11) NOT NULL,
`user_id` int(11) NOT NULL,
`api_key` varchar(255) NOT NULL,
`level` int(2) NOT NULL,
`ignore_limits` tinyint(1) NOT NULL DEFAULT '0',
`is_private_key` tinyint(1) NOT NULL DEFAULT '0',
`ip_addresses` text,
`date_created` int(11) NOT NULL,
`api_key_activated` enum('yes','no') NOT NULL DEFAULT 'no'
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO `keys` (`id`, `user_id`, `api_key`, `level`, `ignore_limits`,
`is_private_key`, `ip_addresses`, `date_created`, …Run Code Online (Sandbox Code Playgroud) 我无法使用ajax通过表单将数据插入数据库中,没有firebug错误有人可以帮助我
视图:
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Blog</title>
<link rel="stylesheet" href="<?php echo base_url("assets/css/bootstrap.min.css"); ?>">
</head>
<body>
<h3 style="text-align: center;">CODEIGNITER AJAX</h3>
<div class="row">
<div class="alert alert-success" id="message" style="display: none;">
</div>
<div class="col-md-4"></div>
<div class="col-md-4">
<?php echo form_open('blog_c',array('id'=>'myForm'));?>
<div class="form-group">
<label>EMAIL:</label>
<input type="text" name="email" id="email" class="form-control" placeholder="EMAIL">
</div>
<input type="submit" value="ENVOYER" id="btn">
<?php echo form_close()?>
</div>
</div>
<script type="text/javascript" src="<?php echo base_url()?>assets/js/jquery-3.1.1.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$('#btn').click(function(){
var email=$('#email').val();
$.ajax({ //request ajax
url:"<?php echo site_url('blog_c/registre')?>",
type:POST,
data:{email:email},
dataType:json, …Run Code Online (Sandbox Code Playgroud)我正在渲染下面的Property Detail页面(可以在这里访问)

我有一个图像滑块和一个360图像查看器.目前,用户手动上传两种类型的图像,即来自一个界面的普通图像和来自其他界面的360图像.我检查属性是否有360个图像并使用全景查看器显示它们.
我使用以下Controller上传360图像,类似于上传普通图像.
public function upload_360_images()
{
if($this->session->userdata['id'] && $this->session->userdata['type']=='user')
{
if($_FILES)
{
if(isset($_FILES['files'])){
$data['errors']= array();
$extensions = array("jpeg","jpg","png");
foreach($_FILES['files']['tmp_name'] as $key => $tmp_name ){
$file_name = $key.$_FILES['files']['name'][$key];
$file_size =$_FILES['files']['size'][$key];
$file_tmp =$_FILES['files']['tmp_name'][$key];
$file_type=$_FILES['files']['type'][$key];
/*$file_ext=explode('.',$_FILES['image']['name'][$key]) ;
$file_ext=end($file_ext);*/
$i=1;
if($file_size > 7097152){
$data['errors'][$i]='File '.$i.' size must be less than 7 MB';
$i++;
}
$desired_dir="uploads";
if(empty($data['errors'])==true){
if(is_dir($desired_dir)==false){
mkdir("$desired_dir", 0700); // Create directory if it does not exist
}
if(is_dir("$desired_dir/".$file_name)==false){
move_uploaded_file($file_tmp,"uploads/".$file_name);
$this->post_model->add360Image('property_360_images',$file_name,$this->uri->segment(3));
}else{ //rename the file if another one exist
$new_dir="uploads/".$file_name.time(); …Run Code Online (Sandbox Code Playgroud) php codeigniter image 360-panorama-viewer 360-virtual-reality
以下是初始化文件输入的设置.我已成功完成上传文件,但我想编辑,而不能在krajee bootstrap输入的容器中预览图像.如果有人知道请提供删除和预览的完整代码.
$("#userfiles").fileinput({
'dropZoneEnabled': true,
'maxFileCount': totalcount,
'showUpload': false,
'browseLabel': "Click Here or Drag & Drop Images Here",
'browseIcon': "<i class=\"glyphicon glyphicon-picture\"></i> ",
'validateInitialCount': true,
'allowedFileExtensions': ["jpg", "png", "gif", "jpeg"],
'showCaption': true,
'showPreview': true,
'showRemove': true
});
//This is the ajax for get images from database
$.ajax({
type: "POST",
url: site_url+'posting/getpicdata',
data: {pid: url},
dataType: "json",
success: function(response)
{
//console.log(response);
//var result = JSON.parse(response);
$.each(response, function(k, v) {
//display the key and value pair
//console.log(v.url);
image_html = v.imgname;
appendHTML +='<div …Run Code Online (Sandbox Code Playgroud) 我正在为使用codeigniter框架开发的硬件商店的销售点工作.
我想提出一份详细的账户说明,如下图所示
我有两个表是销售和付款,我想从这两个表中获取数据,以便我可以生成特定客户的帐户报表.
这将有所帮助,因为客户能够以现金或信用为基础查看他/她购买的所有物品,并且还显示他们按日期分组的付款额.
还可以计算到期金额.
我可以使用以下代码单独列出(销售和付款),从而计算到期金额.
销售
<?php
$salestotal = 0;
if (is_array($sales) || is_object($sales))
{
foreach ($sales as $key=>$sale): {?>
<tr>
<td>
<?php echo $sale->date; ?>
</td>
<td>
<?php echo $sale->id; ?>
</td>
<td>
<?php echo $sale->grand_total; ?>
</td>
<td>
<?php echo $sale->paid; ?>
</td>
<td></td>
</tr>
<?php
if(isset($sale->grand_total))
$salestotal += $sale->grand_total;
} endforeach ; }?>Run Code Online (Sandbox Code Playgroud)
支付
<?php
$paymentstotal = 0;
if (is_array($payments) || is_object($payments))
{
foreach ($payments as $key=>$payment): {?>
<tr>
<td>
<?php echo $payment->date; ?>
</td>
<td …Run Code Online (Sandbox Code Playgroud)codeigniter ×10
php ×8
jquery ×3
mysql ×2
ajax ×1
api-key ×1
dropzone.js ×1
image ×1
javascript ×1
postman ×1
rest ×1
url ×1