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  • HTML/CSS Design for Report

    - by Kevin Brown
    I'm looking for some resources that demonstrate good graphic design for generated (PHP/HTML/CSS) reports. The website I'm designing is essentially a long test. Everything is finished except the report generation, and this part needs to look good! I'd appreciate any advice/resources you can point me to! I know this isn't directly programming related, but my purposes do encompass coding and output.

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  • how to place the value for text area

    - by udaya
    Hi while I am performing the edit function .. i have a textarea where i type some text and submit <tr align="left"> <td class="table_label">Reason </td> <td><textarea cols="17" class="text_box_login_14_width_150" size="5" name="txtReason" id="txtReason"></textarea></td> </tr> when i submit i must receive the value in the edit form <td> <textarea id="txtReason" name="txtReason" value="<?= $row['dReason']?>"></textarea> </td> but i dont receive the value for text area how to receive the value

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  • Empty database output in CI

    - by den-javamaniac
    Hi. I'm building a simple app and trying to test DB result output. But unfortunately all I'm getting is an array of size 0. Here's the controller code excerpt: $data['query'] = $this->db->query('SELECT role_id, role_privilege FROM role'); $this->load->view('welcome_message', $data); And a view code excerpt: <?php echo count($query->result_array())."<br/>"; foreach ($query->result() as $row){ echo $row->role_id . '<br/>'; echo $row->role_privilege . '<br/>'; } echo 'Total result '.$query->num_rows(); ?> And what I get is next: 0 Total result Running query from a command line gives a 2 rowed output. Can someone point out what i'm missing?

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  • populate href of link for next and previous

    - by sea_1987
    Hi There, I am struggling alot with some PHP I am needing to implement a next and previous link, basically I have a search function on my site, that returns multiple results and click on a result navigates to that results page, I want to then be able to click next on that page, and be taken to the next result in the sequence that was returned by the users original search? Is this possible, and how? I have no clue.

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  • Is this a valid url parameter in jquery.ajax()?

    - by udaya
    Is this a valid url parameter in jquery.ajax(), <script type="text/javascript"> $(document).ready(function() { getRecordspage(); }); function getRecordspage() { $.ajax({ type: "POST", url: "http://localhost/codeigniter_cup_myth/index.php/adminController/mainAccount", data: "", contentType: "application/json; charset=utf-8", global:false, async: false, dataType: "json", success: function(jsonObj) { alert(jsonobj); } }); } </script> The url doesn't seem to go to my controller function...

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  • I’m screwing up this LEFT JOIN

    - by Jason Shultz
    I'm doing a left join on several tables. What I want to happen is that it list all the businesses. Then, it looks through the photos, videos, specials and categories. IF there are photos, then the tables shows yes, if there are videos, it shows yes in the table. It does all that without any problems. Except for one thing. For every photo, it shows the business that many times. For example, if there are 5 photos in the DB for a business, it shows the business five times. Obviously, this is not what I want to happen. Can you help? function frontPageList() { $this->db->select('b.id, b.busname, b.busowner, b.webaddress, p.thumb, v.title, c.catname'); $this->db->from ('business AS b'); $this->db->where('b.featured', '1'); $this->db->join('photos AS p', 'p.busid = b.id', 'left'); $this->db->join('video AS v', 'v.busid = b.id', 'left'); $this->db->join('specials AS s', 's.busid = b.id', 'left'); $this->db->join('category As c', 'b.category = c.id', 'left'); return $this->db->get();

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  • many-to-many relationship in CI (not using ORM)

    - by Ross
    I'm implementing a categories system in my CI app and trying to work out the best way of working with many to many relationships. I'm not using an ORM at this stage, but could use say Doctrine if necessary. Each entry may have multiple categories. I have three tables (simplified) Entries: entryID, entryName Categories: categoryID, categoryname Entry_Category: entryID, categoryID my CI code returns a record set like this: entryID, entryName, categoryID, categoryName but, as expected with Many-to-Many relationships, each "entry" is repeated for each "category". What would the best way to "group" the categories so that when I output the results, I am left with something like: Entry Name Appears in Category: Foo, Bar rather than: Entry Name Appears in Category: Foo Entry Name Appears in Category: Bar I believe the option is to track if the post ID matches a previous entry, and if so, store the respective category, and output it as one, rather than several, but am unsure of how to do this in CI. thanks for any pointers (I appreciate this is may be a vague/complex question without a better knowledge of the system).

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  • LIKE and % Wildcard in Doctrine's findBy*()

    - by 01010011
    Hi, How do I write the following MySQL query using Doctrine's findBy*() method?: SELECT column_name1, column_name2 FROM table_name WHERE column_name3 LIKE '%search_key%'; For Example, to fetch multiple rows from a column named "ColumnName" (below) using Doctrine: $users = Doctrine::getTable('User')->findByColumnName('active'); echo $users[0]->username; echo $users[1]->username; I tried: $search_key = 'some value'; $users = Doctrine::getTable('User')->findByColumnName('%$search_key%'); echo $users[0]->username; echo $users[1]->username; and I got no errors, but nothing displayed. Any assistance will be really appreciated. Thanks in advance.

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  • Pagination with multiple searcing criteria in codeigniter

    - by ASD
    By default the pagination links will be 1 and 2 If i use the search criteria, (id name), all the 2 are optional. $config['base_url'] = base_url()."index.php/searchStudent/".$id."/'".$name."'/"; $config['total_rows'] = count($search1); $config['per_page'] = '1';//display perpage $config['num_links'] = 1; $config['full_tag_open'] = ''; $config['full_tag_close'] = ''; $this-pagination-initialize($config); $data['patient_result'] = $this-Student_Model-searchStudent($id,$name,$config['per_page'],$this-uri-segment(3)); But its not working Whats wrong here?

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  • Codeigniter php activerecord orm limit and offset

    - by user2167174
    I am a bit stuck with this problem I have in phpactiverecord. What I am trying to do is a pagination so I need to limit and offset the query results. I am accessing all of the user posts like so: $user-post; How can I query this to limit and offset the results? Thanx in advance. Code: public function office() { if (!$this->session->userdata('username')) { redirect(base_url()); } $data = array(); $data['posts'] = []; $user = User::find('first', array('id' => $this->session->userdata('id'))); if ($user != null) { if ($user->post != null) { foreach ($user->post as $post) { $posts = array($post->name, $post->description, $post->date,'<a href="'.base_url().'Posts/edit/'.$post->id.'">Edit</a> <br /><a href="'.base_url().'Posts/delete/'.$post->id.'">Delete</a>'); array_push($data['posts'], $posts); } $this->table->set_heading('Name', 'Description', 'Date', '<a href="'.base_url().'Posts/create">+Add</a>'); $tmpl = array('table_open' => '<table class="table table-stripped table-bordered user-posts">'); $this->table->set_template($tmpl); $data['table'] = $this->table->generate($data['posts']); } $this->load->view('template/header.php'); $this->load->view('Users/office.php', $data); $this->load->view('template/footer.php'); } else { redirect(base_url()); } }

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  • php codeigniter MySQL search query

    - by kalafun
    I want to create a search query on MySQL database that will consist of 5 different strings typed in from user. I want to query 5 different table columns with these strings. When I for example have input fields like: first name, last name, address, post number, city. How should I query the database that I dont always get all the rows. My query is something like this: SELECT user_id, username from users where a like %?% AND b like %?% AND c like %?% AND d like %?% AND e like %?%; When I exchange the AND for OR I always get all the results which makes sense, and when I use AND I get only the exact matches... Is there any function or statement that would help me with this?

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  • Which Python Framework and CMS coming from PHP - Codeigniter+ExpresionEngine?

    - by Joshua Fricke
    We are currently developing most of our applications in PHP using CodeIgniter (CI) and ExpressionEngine (EE) and are looking to try our hands at Python. So we are looking for a Framework and ideally a CMS that work well together like the CI+EE combo does. Have done a bit of research, it looks like these are some good suggestions (though we are not limiting to these): Frameworks - http://wiki.python.org/moin/WebFrameworks Django Web2py CMS - http://wiki.python.org/moin/ContentManagementSystems Below picked because they are developed with a Framework (my only frame of reference using CI+EE) Merengue Mezzanine Django CMS Input would be great in helping us decide.

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  • Burning Linux ISO to DVD and making it bootable.

    - by toc777
    Hi everyone, I just downloaded the Fedora 14 Live-Desktop ISO and used CDBurnerXP to burn the image to a DVD. For some reason the first time I burned the image nothing showed up on the DVD when I accessed it even though CDBurnerXP said it had successfully burned to the disk. I did it again and the ISO shows up on the disk (I don't think this is right, should it be the files inside the image that show up on disk or the image file??). The problem now is my dell PC can't find the ISO when I try to boot from it. I get an error saying it can't boot from the CD. I have verified the ISO image as directed from the Fedora website. My question is how do I make a bootable CD from a Fedora Live-Desktop ISO? How can I verify that the ISO was written to the CD correctly and has anyone had any issues booting from a CD using a Dell desktop (I'm not at home at the moment so I can't check what model it is but its old enough, I've had it for about 5 years). EDIT: All that needed to be done was to burn the image to CD as an image and not a data file. The first three times failed, I'm not sure if this was because of faulty DVD's or if the write speed was too high (16x). I put in a new DVD and changed the write speed to 8x, the image was then properly burned to the disk without any errors. Thanks.

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  • How to convince boss to start using Codeigniter or YII at work?

    - by mahen23
    Hello, i work for a web development company and during the one year i have spent here, there were no improvements in the technologies we used to built our websites. I introduced jquery to them (buying the Novice to Ninja by Sitepoint) and now, i want to get rid of all these crappy PHP from scratch and use a PHP framework instead. So what reasoning i can use to convince my boss to switch, and how to convice the other developers too?

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  • How to downgrade from PHP 5.4 to 5.3

    - by Burning the Codeigniter
    Recently I upgraded PHP to 5.4, but it seems to be failing on me because APC isn't working with it, however this gave me no choice but to downgrade to 5.3, how can I do this? I executed apt-get remove php5 and apt-get purge php5 then I removed the 5.4 repo from apt-get then executed apt-get install php5, it says it was installing 5.3, and I restarted php-fastcgi and php5-fpm but when I run php -v it says 5.4. How can I downgrade from 5.4 to 5.3? My server is running Ubuntu 11.04.

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  • System that splits passwords across two servers

    - by Burning the Codeigniter
    I stumbled upon this news article on BBC, RSA splits passwords in two to foil hackers' attacks tl;dr - a (randomized) password is split in half and is stored across two separate servers, to foil hackers that gained access to either server upon a security breach. Now the main question is, how would this kind of system would be made... codespeaking, for PHP which I commonly develop on my web applications, the database password is normally stored in a configuration file, i.e. config.php with the username and password, in that case it is understandable that the passwords can be stolen if the security was compromised. However when splitting and sending the other half to the other server, how would this go on when making a communication to the other server (keeping in mind with PHP) since the other server password would be stored in a configuration file, wouldn't it? In terms of security is to keep the other server password away from the main one, just exactly how would the main server communicate, without exposing any other password, apart from the first server. This certainly makes me think...

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  • Cant burn the iso file on disc and usb will not startup

    - by Jason
    I am having very big problems trying to get this going for my old laptop. I tried burning the iso image with 5 different iso burning programs and none of the disks worked none started up. Then I tried to do the USB way used the program that puts it on the usb for you it starts up on my laptop fine but will not start up on my compaq presario 2178cl. If any1 can help me with this problem I would be much appreciative ty for your time.

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  • PHP: Ideal Folder Structure of MVC framework

    - by Sarfraz
    Hello guys, I would like to ask what is the ideal folder structure for a MVC framework that should be able to support multiple installations. For example, I install xyz framework and i run two or more sites based on this single installation of xyz framework rather than installing the framework for each site. This is probably done in Codeigniter too but i don't know much about CodeIgniter, so i need your suggestions. I know some of you might even have better idea than what is done by CodeIgniter, so please share.

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  • Why is the value not passed to my controller page in codeigniter?

    - by udaya
    I am selecting state from country and city from state. This is my country select box: <td width=""><select name="country" onChange="getState(this.value)" class="text_box_width_190"> <option value="0">Select Country</option> <? foreach($country as $row) { ?> <option value="<?=$row['dCountry_id']?>"><?=$row['dCountryName']?></option> <? } ?> </select></td> This is my state select box: <select name="state" id="state" class="text_box_width_190" > <option value="0">Select State</option> </select> This is my city select box: <td width=""><div id="citydiv"><select name="city" class="text_box_width_190"> <option>Select City</option> </select></div></td> This is my script: <script type ="text/javascript"> function getXMLHTTP() { //fuction to return the xml http object var xmlhttp=false; try{ xmlhttp=new XMLHttpRequest(); } catch(e) { try{ xmlhttp= new ActiveXObject("Microsoft.XMLHTTP"); } catch(e){ try{ xmlhttp = new ActiveXObject("Msxml2.XMLHTTP"); } catch(e1){ xmlhttp=false; } } } return xmlhttp; } function getState(countryId) { var strURL="http://localhost/ssit/system/application/views/findState.php?country="+countryId; var req = getXMLHTTP(); if (req) { req.onreadystatechange = function() { if (req.readyState == 4) { // only if "OK" if (req.status == 200) { document.getElementById('statediv').innerHTML=req.responseText; } else { alert("There was a problem while using XMLHTTP:\n" + req.statusText); } } } req.open("GET", strURL, true); req.send(null); } } function getCity(countryId,stateId) { var strURL="http://localhost/ssit/system/application/views/findCity.php?country="+countryId+"&state="+stateId; var req = getXMLHTTP(); if (req) { req.onreadystatechange = function() { if (req.readyState == 4) { // only if "OK" if (req.status == 200) { document.getElementById('citydiv').innerHTML=req.responseText; } else { alert("There was a problem while using XMLHTTP:\n" + req.statusText); } } } req.open("GET", strURL, true); req.send(null); } } </script> This is my find state page: <? $country=intval($_GET['country']); $link = mysql_connect('localhost', 'root', ''); //changet the configuration in required if (!$link) { die('Could not connect: ' . mysql_error()); } mysql_select_db('ssit'); $query="Select dStateName,dState_id FROM tbl_state Where dCountry_id='1'"; $result=mysql_query($query); ?> <select name="state" onchange="getCity(<?=$country?>,this.value)"> <option value="0">Select State</option> <? while($row=mysql_fetch_array($result)) { ?> <option value=<?=$row['dState_id']?>><?=$row['dStateName']?></option> <? } ?> </select> This is my find city page: <? $countryId=intval($_GET['country']); $stateId=intval($_GET['state']); $link = mysql_connect('localhost', 'root', ''); //changet the configuration in required if (!$link) { die('Could not connect: ' . mysql_error()); } mysql_select_db('ssit'); $query="Select dCityName,dCity_id FROM tbl_city Where dState_id='30'"; $result=mysql_query($query); ?> <select name="city"> <option>Select City</option> <? while($row=mysql_fetch_array($result)) { ?> <option value><?=$row['dCityName']?></option> <? } ?> </select> When I post a country, I can receive it but I can't receive my state and city. How to receive them?

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  • why the value is not passed to my contrller page in codeigniter?

    - by udaya
    Hi I am selecting state from country and city from state This is my select country Select box <td width=""><select name="country" onChange="getState(this.value)" class="text_box_width_190"> <option value="0">Select Country</option> <? foreach($country as $row) { ?> <option value="<?=$row['dCountry_id']?>"><?=$row['dCountryName']?></option> <? } ?> </select></td> This is my select state select box <select name="state" id="state" class="text_box_width_190" > <option value="0">Select State</option> </select> This is my select city selectbox <td width=""><div id="citydiv"><select name="city" class="text_box_width_190"> <option>Select City</option> </select></div></td> this is my script <script type ="text/javascript"> function getXMLHTTP() { //fuction to return the xml http object var xmlhttp=false; try{ xmlhttp=new XMLHttpRequest(); } catch(e) { try{ xmlhttp= new ActiveXObject("Microsoft.XMLHTTP"); } catch(e){ try{ xmlhttp = new ActiveXObject("Msxml2.XMLHTTP"); } catch(e1){ xmlhttp=false; } } } return xmlhttp; } function getState(countryId) { var strURL="http://localhost/ssit/system/application/views/findState.php?country="+countryId; var req = getXMLHTTP(); if (req) { req.onreadystatechange = function() { if (req.readyState == 4) { // only if "OK" if (req.status == 200) { document.getElementById('statediv').innerHTML=req.responseText; } else { alert("There was a problem while using XMLHTTP:\n" + req.statusText); } } } req.open("GET", strURL, true); req.send(null); } } function getCity(countryId,stateId) { var strURL="http://localhost/ssit/system/application/views/findCity.php?country="+countryId+"&state="+stateId; var req = getXMLHTTP(); if (req) { req.onreadystatechange = function() { if (req.readyState == 4) { // only if "OK" if (req.status == 200) { document.getElementById('citydiv').innerHTML=req.responseText; } else { alert("There was a problem while using XMLHTTP:\n" + req.statusText); } } } req.open("GET", strURL, true); req.send(null); } } </script> This is my findstate page <? $country=intval($_GET['country']); $link = mysql_connect('localhost', 'root', ''); //changet the configuration in required if (!$link) { die('Could not connect: ' . mysql_error()); } mysql_select_db('ssit'); $query="Select dStateName,dState_id FROM tbl_state Where dCountry_id='1'"; $result=mysql_query($query); ?> <select name="state" onchange="getCity(<?=$country?>,this.value)"> <option value="0">Select State</option> <? while($row=mysql_fetch_array($result)) { ?> <option value=<?=$row['dState_id']?>><?=$row['dStateName']?></option> <? } ?> </select> This is my find city page <? $countryId=intval($_GET['country']); $stateId=intval($_GET['state']); $link = mysql_connect('localhost', 'root', ''); //changet the configuration in required if (!$link) { die('Could not connect: ' . mysql_error()); } mysql_select_db('ssit'); $query="Select dCityName,dCity_id FROM tbl_city Where dState_id='30'"; $result=mysql_query($query); ?> <select na me="city" Select City when i post country i can receive it but i cant receive my state and city How to receive it

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  • Ideal Folder Structure of MVC framework

    - by Sarfraz
    Hello guys, I would like to ask what is the ideal folder structure for a MVC framework that should be able to support multiple installations. For example, I install xyz framework and i run two or more sites based on this single installation of xyz framework rather than installing the framework for each site. This is probably done in Codeigniter too but i don't know much about CodeIgniter, so i need your suggestions. I know some of you might even have better idea than what is done by CodeIgniter, so please share.

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  • CentOS ISO DVD self-disc-check failing

    - by Jakobud
    I downloaded a CentOS 5.4 DVD ISO from one of the official mirrors. The MD5 sum is correct for the iso file that was downloaded. I burned the ISO onto a DVD+R using ImgBurn. I got no errors during the burning process and it says it finished burning successfully. When booting a server using this DVD to install CentOS on it, the first thing I did was run the self-disc-check (where it checks the files to make sure its genuine, etc). It failed. Didn't really say why. Am I doing something wrong here? Or is there someplace I can see why its failing? If the MD5 is correct and I don't get any burning errors, how can it be failing its own self-check mechanism?

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  • connect() failed (111: Connection refused) while connecting to upstream

    - by Burning the Codeigniter
    I'm experiencing 502 gateway errors when accessing a PHP file in a directory (http://domain.com/dev/index.php), the logs simply says this: 2011/09/30 23:47:54 [error] 31160#0: *35 connect() failed (111: Connection refused) while connecting to upstream, client: xx.xx.xx.xx, server: domain.com, request: "GET /dev/ HTTP/1.1", upstream: "fastcgi://127.0.0.1:9000", host: "domain.com" I've never experienced this before, how do I do a solution for this type of 502 gateway error? This is the nginx.conf: user www-data; worker_processes 4; pid /var/run/nginx.pid; events { worker_connections 768; # multi_accept on; } http { ## # Basic Settings ## sendfile on; tcp_nopush on; tcp_nodelay on; keepalive_timeout 65; types_hash_max_size 2048; # server_tokens off; # server_names_hash_bucket_size 64; # server_name_in_redirect off; include /etc/nginx/mime.types; default_type application/octet-stream; ## # Logging Settings ## access_log /var/log/nginx/access.log; error_log /var/log/nginx/error.log; ## # Gzip Settings ## gzip on; gzip_disable "msie6"; # gzip_vary on; # gzip_proxied any; # gzip_comp_level 6; # gzip_buffers 16 8k; # gzip_http_version 1.1; # gzip_types text/plain text/css application/json application/x-javascript text/xml application/xml application/xml+rss text/javascript; ## # Virtual Host Configs ## include /etc/nginx/conf.d/*.conf; include /etc/nginx/sites-enabled/*; } #mail { # # See sample authentication script at: # # http://wiki.nginx.org/ImapAuthenticateWithApachePhpScript # # # auth_http localhost/auth.php; # # pop3_capabilities "TOP" "USER"; # # imap_capabilities "IMAP4rev1" "UIDPLUS"; # # server { # listen localhost:110; # protocol pop3; # proxy on; # } # # server { # listen localhost:143; # protocol imap; # proxy on; # } #}

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  • How to install nginx and install the configuration files too

    - by Burning the Codeigniter
    I've just completely uninstalled nginx 1.0.6 from my server (Ubuntu 11.04) using apt-get remove nginx rm -rf /etc/nginx/ rm -rf /usr/sbin/nginx rm /usr/share/man/man1/nginx.1.gz apt-get remove nginx* Now I want to install it again, however when starting nginx, I get errors such as: Restarting nginx: nginx: [emerg] open() "/etc/nginx/nginx.conf" failed (2: No such file or directory) Then I placed my own conf file, then I get a new error: Restarting nginx: nginx: [emerg] open() "/etc/nginx/mime.types" failed (2: No such file or directory) in /etc/nginx/nginx.conf:12 Now it seems that apt-get install nginx doesn't install it completely, I cleared the apt-get cache, doesn't seem to help. How can I get a full installation of nginx using apt-get?

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  • Redirecting wildcard emails to one email with postfix

    - by Burning the Codeigniter
    I'm creating a bounce email system where emails can reply to messages on my site. However when the emails are sent to the user containing the previous message, the Reply-To field contains an email something like this [email protected] (which contains the ID at the end). If the user replies, the reply message will be sent back to [email protected] which of course, doesn't have its own mailbox, except the [email protected]. How would I redirect all incoming messages coming from a specific wildcard notification-message-*@mysite.com to [email protected]? I did some research, but no solid part worked, including the luser_relay = [email protected] and putting notification-message-* in the postfix aliases table, the notification@ has a Maildir, so the emails would go into it. I am using Ubuntu 11.04.

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