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  • mysql result for pagination

    - by Reteras Remus
    The query is: SELECT * FROM `news` ORDER BY `id` LIMIT ($curr_page * 5), ( ($curr_page * 5) + 5 ) Where $curr_page is a php variable which is getting a value from $_GET['page'] I want to make a pagination (5 news on each page), but I don't know why the mysql is returning me extra values. On the first page the result ok: $curr_page = 0 The query would be: SELECT * FROM `news` ORDER BY `id` LIMIT 0, 5 But on the second page, the result from the query is adding extra news, 10 instead of 5. The query on the second page: SELECT * FROM `news` ORDER BY `id` LIMIT 5, 10 Whats wrong? Why the result has 10 values instead of 5? Thank you!

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  • MySQL pivot tables - covert rows to columns

    - by user2723490
    This is the structure of my table: Then I run a query SELECT `date`,`index_name`,`results` FROM `mst_ind` WHERE `index_name` IN ('MSCI EAFE Mid NR USD', 'Alerian MLP PR USD') AND `time_period`='M1' and get a table like How can I convert "index_name" rows to columns like: date | MSCI EAFE Mid NR USD | Alerian MLP PR USD etc In other words I need each column to represent an index and rows to represent date-result. I understand that MySQL doesn't have pivot table functions. What is the easiest way of doing this? I've tried this code, but it generates an error: SELECT `date`, MAX(IF(index_name = 'Alerian MLP PR USD' AND `time_period`='M1', results, NULL)) AS res1, MAX(IF(index_name = 'MSCI EAFE Mid NR USD' AND `time_period`='M1', results, NULL)) AS res2 FROM `mst_ind` GROUP BY `date I need to make the conversion on the query level - not PHP. Please suggest a nice and elegant solution. Thanks!

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  • Order mysql results without identifier

    - by Alex Crooks
    Usually I would have a table field called ID on auto increment. That way I could order using this field etc. However I have no control over the structure of a table, and wondered how to get the results in reverse order to default. I'm currently using $q = mysql_query("SELECT * FROM ServerChat LIMIT 15"); However like I said there is no field I can order on, so is there a way to tell mysql to reverse the order it gets the results? I.e last row to first row instead of the default.

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  • MySql database design for a quiz

    - by Mark
    I'm making an online quiz with php and mysql and need a bit of help deciding how to design the database for optimal insert of questions/answers and to select questions for the quiz. The table will hold 80 questions each with 4 possible options plus the correct answer. When retrieving the questions and options from the database I will randomly select 25 questions and their options. Is it better to make a single column for all questions, options, and correct answers? For example: ID | Q | OPT1 | OPT2 | OPT3 | OPT4 | ANS Or would it be better to make a column for each individual question, option, and correct answer? For example: Q1 | Q1_OPT1 | Q1_OPT2 | Q1_OPT3 | Q1_OPT5 | Q1_ANS | Q2 | Q2_OPT1 | Q2_OPT2...

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  • Bible reference books (PHP / MySQL / Unix)

    - by Josh K
    I'm looking for some nice heavy books to liter around my desk and make it look like I'm a hard core programmer. On the occasion that I might want to look something up they will also need to be useful dependable books. I'm looking for the equivalent bible in PHP, MySQL, and Unix. Should be laid out with some chapters I can actually read, along with having an in-depth reference to that particular subject. I know that the majority of this can be found on Google, but I would prefer it in book form.

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  • PHP MySQL query string

    - by user1174762
    I am newer to PHP and MySQL and I am having trouble understanding join. I think, for me, the problem lies with actually understanding the logic of the query. What I am trying to do Is select all of the status updates from a table named "post", but only ones from users I am "following", and then display them In order by date. So, I have two databases which are set up like so: posts |post_id|user_id|post_body|date_upload| | 1 | 4 | hey. | 01/2/2012 | follows |relation_id|user_id|followee_id| | 1 | 4 | 2 | Could someone please explain how I should syntactically and Logically set this up? Thank you!

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  • How to insert scraping data to mysql

    - by user1887288
    i am fetching data from other websites can any one tell me how to insert fetch data to mysql database Below code i am using to fetch results coming $urls = $_POST["urls"]; require_once('simple_html_dom.php'); $useragent = 'Googlebot/2.1 (http://www.googlebot.com/bot.html)'; foreach ($urls as $url) { $curl = curl_init(); curl_setopt($curl, CURLOPT_URL, $url); curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1); curl_setopt($curl, CURLOPT_CONNECTTIMEOUT, 20); curl_setopt($curl, CURLOPT_USERAGENT, $useragent); $str = curl_exec($curl); curl_close($curl); $html= str_get_html($str); foreach($html->find('span.price') as $e) echo $e->innertext . '<br>'; }

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  • Can't Connect To Local Mysql Using IP Address, but CAN connect from remote server

    - by user1782041
    Here's an interesting one that does not seem to fall into any of the mysql connection issues I've read about or searched for: On an Ubuntu 12.04 box I had some system updates waiting to install, and I took care of that this evening. After the install, I started seeing some errors in my syslog complaining about a particular php script that could no longer connect to the mysql instance on the box. Here is the specific error: PHP Warning: mysql_connect(): Can't connect to MySQL server on '192.168.0.40' (4) Now, the server's IP address is 192.168.0.40, and I've checked to make sure that I have mysql listening on 0.0.0.0 so that I can connect using either "localhost" or "192.168.0.40". Here's where things get odd: From the local machine, if I try the following: mysql -uroot -p -h192.168.0.40 I get this error: ERROR 2003 (HY000): Can't connect to MySQL server on '192.168.0.40' (110) I've checked, and error 110 indicates an OS timeout, and error 2003 is the mysql generic "can't connect" error. This indicates that it is not permissions with the user. However, if I do the same thing from a remote machine (say, from 192.168.0.30), I log right in with no problems. Futher, other scripts on the local machine that connect to mysql using "localhost" for the host rather than "192.168.0.40" connect with no problems. Also, I can connect via the mysql socket with no problems both from the command line and php scripts. So, this feels like a networking issue of some kind on the local box, but there are no iptables rules on this box (it is firewalled externally) and I can't figure out what else may be causing this. This problematic script worked perfectly prior to the latest system update. For now, I'll simply change the script to connect via localhost, but I'd really like to know why it broke for 2 reasons: There may be other scripts that connect using 192.168.0.40 that don't run very often which are now broken. Auditing them all will take more time than I feel like devoting at the moment. I'm curious, and want to know why it broke so I can fix it correctly. Any help?

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  • mysql command line not working

    - by Sandeepan Nath
    I have mysql running in my fedora system. I have xampp setup on the system and php projects present in the webspace are working fine. PhpMyAdmin is working fine. echoing phpinfo() in a PHP script also shows mysql enabled. But running mysql connect command mysql -u[username] -p[password] Gives this - bash: mysql: command not found How do I fix that? Any pointers? I guess I need to do some pointing (define some path in some file) so that my system knows that mysql is installed. What exactly do I have to do? Additional Details This system was someone else's and he is not available here. May be PHP/Mysql was setup already in the system. I just freshly extracted xampp for linux into /opt/lampp/ and have put all the above mentioned things (PHP projects and PhpMyAdmin) there. After doing that I had a socket problem (PhpMyAdmin was not working and showing this)- #2002 - The server is not responding (or the local MySQL server's socket is not correctly configured) I restarted lampp using ./lampp restart but problem remained. Then after turning on system today, I started lampp and everything worked just fine. No project issues anymore only command line Mysql not working

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  • Set up linux box for hosting a-z

    - by microchasm
    I am in the process of reinstalling the OS on a machine that will be used to host a couple of apps for our business. The apps will be local only; access from external clients will be via vpn only. The prior setup used a hosting control panel (Plesk) for most of the admin, and I was looking at using another similar piece of software for the reinstall - but I figured I should finally learn how it all works. I can do most of the things the software would do for me, but am unclear on the symbiosis of it all. This is all an attempt to further distance myself from the land of Configuration Programmer/Programmer, if at all possible. I can't find a full walkthrough anywhere for what I'm looking for, so I thought I'd put up this question, and if people can help me on the way I will edit this with the answers, and document my progress/pitfalls. Hopefully someday this will help someone down the line. The details: CentOS 5.5 x86_64 httpd: Apache/2.2.3 mysql: 5.0.77 (to be upgraded) php: 5.1 (to be upgraded) The requirements: SECURITY!! Secure file transfer Secure client access (SSL Certs and CA) Secure data storage Virtualhosts/multiple subdomains Local email would be nice, but not critical The Steps: Download latest CentOS DVD-iso (torrent worked great for me). Install CentOS: While going through the install, I checked the Server Components option thinking I was going to be using another Plesk-like admin. In hindsight, considering I've decided to try to go my own way, this probably wasn't the best idea. Basic config: Setup users, networking/ip address etc. Yum update/upgrade. Upgrade PHP/MySQL: To upgrade PHP and MySQL to the latest versions, I had to look to another repo outside CentOS. IUS looks great and I'm happy I found it! Add IUS repository to our package manager cd /tmp wget http://dl.iuscommunity.org/pub/ius/stable/Redhat/5/x86_64/epel-release-1-1.ius.el5.noarch.rpm rpm -Uvh epel-release-1-1.ius.el5.noarch.rpm wget http://dl.iuscommunity.org/pub/ius/stable/Redhat/5/x86_64/ius-release-1-4.ius.el5.noarch.rpm rpm -Uvh ius-release-1-4.ius.el5.noarch.rpm yum list | grep -w \.ius\. # list all the packages in the IUS repository; use this to find PHP/MySQL version and libraries you want to install Remove old version of PHP and install newer version from IUS rpm -qa | grep php # to list all of the installed php packages we want to remove yum shell # open an interactive yum shell remove php-common php-mysql php-cli #remove installed PHP components install php53 php53-mysql php53-cli php53-common #add packages you want transaction solve #important!! checks for dependencies transaction run #important!! does the actual installation of packages. [control+d] #exit yum shell php -v PHP 5.3.2 (cli) (built: Apr 6 2010 18:13:45) Upgrade MySQL from IUS repository /etc/init.d/mysqld stop rpm -qa | grep mysql # to see installed mysql packages yum shell remove mysql mysql-server #remove installed MySQL components install mysql51 mysql51-server mysql51-devel transaction solve #important!! checks for dependencies transaction run #important!! does the actual installation of packages. [control+d] #exit yum shell service mysqld start mysql -v Server version: 5.1.42-ius Distributed by The IUS Community Project Upgrade instructions courtesy of IUS wiki: http://wiki.iuscommunity.org/Doc/ClientUsageGuide Install rssh (restricted shell) to provide scp and sftp access, without allowing ssh login cd /tmp wget http://dag.wieers.com/rpm/packages/rssh/rssh-2.3.2-1.2.el5.rf.x86_64.rpm rpm -ivh rssh-2.3.2-1.2.el5.rf.x86_64.rpm useradd -m -d /home/dev -s /usr/bin/rssh dev passwd dev Edit /etc/rssh.conf to grant access to SFTP to rssh users. vi /etc/rssh.conf Uncomment or add: allowscp allowsftp This allows me to connect to the machine via SFTP protocol in Transmit (my FTP program of choice; I'm sure it's similar with other FTP apps). rssh instructions appropriated (with appreciation!) from http://www.cyberciti.biz/tips/linux-unix-restrict-shell-access-with-rssh.html Set up virtual interfaces ifconfig eth1:1 192.168.1.3 up #start up the virtual interface cd /etc/sysconfig/network-scripts/ cp ifcfg-eth1 ifcfg-eth1:1 #copy default script and match name to our virtual interface vi ifcfg-eth1:1 #modify eth1:1 script #ifcfg-eth1:1 | modify so it looks like this: DEVICE=eth1:1 IPADDR=192.168.1.3 NETMASK=255.255.255.0 NETWORK=192.168.1.0 ONBOOT=yes NAME=eth1:1 Add more Virtual interfaces as needed by repeating. Because of the ONBOOT=yes line in the ifcfg-eth1:1 file, this interface will be brought up when the system boots, or the network starts/restarts. service network restart Shutting down interface eth0: [ OK ] Shutting down interface eth1: [ OK ] Shutting down loopback interface: [ OK ] Bringing up loopback interface: [ OK ] Bringing up interface eth0: [ OK ] Bringing up interface eth1: [ OK ] ping 192.168.1.3 64 bytes from 192.168.1.3: icmp_seq=1 ttl=64 time=0.105 ms And this is where I'm at. I will keep editing this as I make progress. Any tips on how to Configure virtual interfaces/ip based virtual hosts for SSL, setting up a CA, or anything else would be appreciated.

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  • MySQL partition "full"?

    - by gdea73
    I have a server that runs Debian 6.2, with Apache, PHP5, and MySQL. Well, I hadn't done anything with MySQL at all so far, just Apache and PHP; I must have installed it (mysql-server) at some point along the line, and I decided to login to the database for the first time a couple days ago as I was considering using the database for a future website project. I noticed that the "root" user had a password, and I didn't recall having set one. My usual root password was incorrect. So I attempted to reset the password. sudo service mysql stop (stopped successfully) sudo /usr/bin/mysqld_safe --skip-grant-tables --skip-networking & started successfully, from what I can tell. However, mysql itself returns "Can't connect to local MySQL server through socket '/var/run/mysqld/mysqld,sock' (2)", and additionally sudo service mysql start returns "/etc/init.d/mysql: ERROR: The partition with /var/lib/mysql is too full! ... failed!" df -h tells me that / is 26% used, a 20GB partition, and /home, roughly 900GB, has only 5% usage. On a potentially related note, I've been experiencing random hangs since I noticed this problem, my tty2 randomly froze several times while idle, and the entire system is suddenly unstable. gnome-terminal also does not open. (Gnome-terminal apparently works now, disregard that part, but the server is still being somewhat unstable, I randomly lost connection when I was SSHed into it from my laptop, twice now.)

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  • Spring-mvc project can't select from a particular mysql table

    - by Dan Ray
    I'm building a Spring-mvc project (using JPA and Hibernate for DB access) that is running just great locally, on my dev box, with a local MySQL database. Now I'm trying to put a snapshot up on a staging server for my client to play with, and I'm having trouble. Tomcat (after some wrestling) deploys my war file without complaint, and I can get some response from the application over the browser. When I hit my main page, which is behind Spring Security authentication, it redirects me to the login page, which works perfectly. I have Security configured to query the database for user details, and that works fine. In fact, a change to a password in the database is reflected in the behavior of the login form, so I'm confident it IS reaching the database and querying the user table. Once authenticated, we go to the first "real" page of the app, and I get a "data access failure" error. The server's console log gets this line (redacted): ERROR org.hibernate.util.JDBCExceptionReporter - SELECT command denied to user 'myDbUser'@'localhost' for table 'asset' However, if I go to MySQL from the shell using exactly the same creds, I have no problem at all selecting from the asset table: [development@tomcat01stg]$ mysql -u myDbUser -pmyDbPwd dbName ... mysql> \s -------------- mysql Ver 14.12 Distrib 5.0.77, for redhat-linux-gnu (i686) using readline 5.1 Connection id: 199 Current database: dbName Current user: myDbUser@localhost ... UNIX socket: /var/lib/mysql/mysql.sock -------------- mysql> select count(*) from asset; +----------+ | count(*) | +----------+ | 19 | +----------+ 1 row in set (0.00 sec) I've broken down my MySQL access settings, cleaned out the user and re-run the grant commands, set up a version of the user from 'localhost' and another from '%', making sure to flush permissions.... Nothing is changing the behavior of this thing. What gives?

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  • Upgrading PHP, MySQL old-passwords issue

    - by Rushyo
    I've inherited a Windows 2k3 server running an XAMPP-installation from the stone age. I needed to upgrade PHP to facilitate an upgrade to MediaWiki to facilitate a new MediaWiki extension (to facilitate some documentation to facilitate doing my job to facilitate getting paid to facilit... you get the idea). However... installing a new version of PHP resulted in PHP's MySQL libraries refusing to communicate using MySQL's 'old style' 152-bit passwords. Not a problem in theory. The MySQL installation is post-4.1, so it should have the functionality to upgrade the user's passwords from 152-bit to 328-bit (what a weird hashing algorithm...). I ran the following: SET PASSWORD = PASSWORD('foo'); on MySQL but querying: SELECT user, password FROM mysql.user; returned just the same password I started out with - 152-bit. Now... I suspect you're thinking 'AHA! old-passwords is on!'. Unfortunately it's not - I've disabled it in the configuration (explicitly set it to 0), made doubly sure I have an absolute reference to that configuration file and ensured the service isn't using the --old-passwords flag. The service was reset after each and every operation. So I went onto another system and generated the 328-bit hash on there, copying the hash over to the first MySQL instance. Unfortunately, that didn't work either (I did remember to FLUSH PRIVILEGES). The application error is: "'mysqlnd cannot connect to MySQL 4.1+ using the old insecure authentication. Please use an administration tool [...snip...] Is there anything else I can try to get PHP to recognise MySQL as not using the 'old insecure authentication'? MySQL seems to be stuck in 'old-passwords' mode and I can't get it out of it.

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  • Mysql query help - Alter this mysql query to get these results?

    - by sandeepan-nath
    Please execute the following queries first to set up so that you can help me:- CREATE TABLE IF NOT EXISTS `Tutor_Details` ( `id_tutor` int(10) NOT NULL auto_increment, `firstname` varchar(100) NOT NULL default '', `surname` varchar(155) NOT NULL default '', PRIMARY KEY (`id_tutor`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=41 ; INSERT INTO `Tutor_Details` (`id_tutor`,`firstname`, `surname`) VALUES (1, 'Sandeepan', 'Nath'), (2, 'Bob', 'Cratchit'); CREATE TABLE IF NOT EXISTS `Classes` ( `id_class` int(10) unsigned NOT NULL auto_increment, `id_tutor` int(10) unsigned NOT NULL default '0', `class_name` varchar(255) default NULL, PRIMARY KEY (`id_class`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=229 ; INSERT INTO `Classes` (`id_class`,`class_name`, `id_tutor`) VALUES (1, 'My Class', 1), (2, 'Sandeepan Class', 2); CREATE TABLE IF NOT EXISTS `Tags` ( `id_tag` int(10) unsigned NOT NULL auto_increment, `tag` varchar(255) default NULL, PRIMARY KEY (`id_tag`), UNIQUE KEY `tag` (`tag`), KEY `id_tag` (`id_tag`), KEY `tag_2` (`tag`), KEY `tag_3` (`tag`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=18 ; INSERT INTO `Tags` (`id_tag`, `tag`) VALUES (1, 'Bob'), (6, 'Class'), (2, 'Cratchit'), (4, 'Nath'), (3, 'Sandeepan'), (5, 'My'); CREATE TABLE IF NOT EXISTS `Tutors_Tag_Relations` ( `id_tag` int(10) unsigned NOT NULL default '0', `id_tutor` int(10) default NULL, KEY `Tutors_Tag_Relations` (`id_tag`), KEY `id_tutor` (`id_tutor`), KEY `id_tag` (`id_tag`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1; INSERT INTO `Tutors_Tag_Relations` (`id_tag`, `id_tutor`) VALUES (3, 1), (4, 1), (1, 2), (2, 2); CREATE TABLE IF NOT EXISTS `Class_Tag_Relations` ( `id_tag` int(10) unsigned NOT NULL default '0', `id_class` int(10) default NULL, `id_tutor` int(10) NOT NULL, KEY `Class_Tag_Relations` (`id_tag`), KEY `id_class` (`id_class`), KEY `id_tag` (`id_tag`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1; INSERT INTO `Class_Tag_Relations` (`id_tag`, `id_class`, `id_tutor`) VALUES (5, 1, 1), (6, 1, 1), (3, 2, 2), (6, 2, 2); In the present system data which I have given , tutor named "Sandeepan Nath" has has created class named "My Class" and tutor named "Bob Cratchit" has created class named "Sandeepan Class". Requirement - To execute a single query with limit on the results to show search results as per AND logic on the search keywords like this:- If "Sandeepan Class" is searched , Tutor Sandeepan Nath's record from Tutor Details table is returned(because "Sandeepan" is the firstname of Sandeepan Nath and Class is present in class name of Sandeepan's class) If "Class" is searched Both the tutors from the Tutor_details table are fetched because Class is present in the name of the class created by both the tutors. Following is what I have so far achieved (PHP Mysql):- <?php $searchTerm1 = "Sandeepan"; $searchTerm2 = "Class"; mysql_select_db("test"); $sql = "SELECT td.* FROM Tutor_Details AS td LEFT JOIN Tutors_Tag_Relations AS ttagrels ON td.id_tutor = ttagrels.id_tutor LEFT JOIN Classes AS wc ON td.id_tutor = wc.id_tutor LEFT JOIN Class_Tag_Relations AS wtagrels ON td.id_tutor = wtagrels.id_tutor LEFT JOIN Tags as t1 on ((t1.id_tag = ttagrels.id_tag) OR (t1.id_tag = wtagrels.id_tag)) LEFT JOIN Tags as t2 on ((t2.id_tag = ttagrels.id_tag) OR (t2.id_tag = wtagrels.id_tag)) where t1.tag LIKE '%".$searchTerm1."%' AND t2.tag LIKE '%".$searchTerm2."%' GROUP BY td.id_tutor LIMIT 10 "; $result = mysql_query($sql); echo $sql; if($result) { while($rec = mysql_fetch_object($result)) $recs[] = $rec; //$rec = mysql_fetch_object($result); echo "<br><br>"; if(is_array($recs)) { foreach($recs as $each) { print_r($each); echo "<br>"; } } } ?> But the results are :- If "Sandeepan Nath" is searched, it does not return any tutor (instead of only Sandeepan's row) If "Sandeepan Class" is searched, it returns Sandeepan's row (instead of Both tutors ) If "Bob Class" is searched, it correctly returns Bob's row If "Bob Cratchit" is searched, it does not return any tutor (instead of only

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  • mySQL to XLS using PHP?

    - by kielie
    Hi guys, how can I create a .XLS document from a mySQL table using PHP? I have tried just about everything, with no success. Basically, I need to take form data, and input it into a database, which I have done, and then I need to retrieve that table data and parse it into a microsoft excel file, which needs to be saved automatically onto the web server. <?php // DB TABLE Exporter // // How to use: // // Place this file in a safe place, edit the info just below here // browse to the file, enjoy! // CHANGE THIS STUFF FOR WHAT YOU NEED TO DO $dbhost = "-"; $dbuser = "-"; $dbpass = "-"; $dbname = "-"; $dbtable = "-"; // END CHANGING STUFF $cdate = date("Y-m-d"); // get current date // first thing that we are going to do is make some functions for writing out // and excel file. These functions do some hex writing and to be honest I got // them from some where else but hey it works so I am not going to question it // just reuse // This one makes the beginning of the xls file function xlsBOF() { echo pack("ssssss", 0x809, 0x8, 0x0, 0x10, 0x0, 0x0); return; } // This one makes the end of the xls file function xlsEOF() { echo pack("ss", 0x0A, 0x00); return; } // this will write text in the cell you specify function xlsWriteLabel($Row, $Col, $Value ) { $L = strlen($Value); echo pack("ssssss", 0x204, 8 + $L, $Row, $Col, 0x0, $L); echo $Value; return; } // make the connection an DB query $dbc = mysql_connect( $dbhost , $dbuser , $dbpass ) or die( mysql_error() ); mysql_select_db( $dbname ); $q = "SELECT * FROM ".$dbtable." WHERE date ='$cdate'"; $qr = mysql_query( $q ) or die( mysql_error() ); // start the file xlsBOF(); // these will be used for keeping things in order. $col = 0; $row = 0; // This tells us that we are on the first row $first = true; while( $qrow = mysql_fetch_assoc( $qr ) ) { // Ok we are on the first row // lets make some headers of sorts if( $first ) { foreach( $qrow as $k => $v ) { // take the key and make label // make it uppper case and replace _ with ' ' xlsWriteLabel( $row, $col, strtoupper( ereg_replace( "_" , " " , $k ) ) ); $col++; } // prepare for the first real data row $col = 0; $row++; $first = false; } // go through the data foreach( $qrow as $k => $v ) { // write it out xlsWriteLabel( $row, $col, $v ); $col++; } // reset col and goto next row $col = 0; $row++; } xlsEOF(); exit(); ?> I just can't seem to figure out how to integrate fwrite into all that to write the generated data into a .xls file, how would I go about doing that? I need to get this working quite urgently, so any help would be greatly appreciated. Thanx guys.

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  • MySQL query works in PHPMyAdmin but not PHP

    - by Su4p
    I do not understand what's happening. I have a query in PHP who crashes -with a strange error-. When I copy/paste the exact same request in PHPMyAdmin it works as expected. What am I doing wrong here ? SELECT oms_patient.id, oms_patient.date, oms_patient.date_modif, date_modif, AES_DECRYPT(nom,"xxxxx") AS "Nom", AES_DECRYPT(prenom,"xxxxx") AS "Prénom usuel", DATE_FORMAT(ddn, "%d/%m/%Y") AS "Date de naissance", villeNaissance AS "Lieu de naissance (ville)", CONCAT(oms_departement.libelle,"(",id_departement,")") AS "Lieu de vie", CONCAT(oms_pays.libelle,"(",id_pays,")") AS "Pays", CONCAT(patientsexe.libelle,"(",id_sexe,")") AS "Sexe", CONCAT(patientprofession.libelle,"(",id_profession,")") AS "Profession", IF(asthme>0,"Oui","Non") AS "Asthme", IF(rhinite>0,"Oui","Non") AS "Rhinite", IF(bcpo>0,"Oui","Non") AS "BPCO", IF(insuffisanceResp>0,"Oui","Non") AS "Insuffisance respiratoire chronique", IF(chirurgieOrl>0,"Oui","Non") AS "Chirurgie ORL du ronflement", IF(autreChirurgie>0,"Oui","Non") AS "Autre chirurgie ORL", IF(allergies>0,"Oui","Non") AS "Allergies", IF(OLD>0,"Oui","Non") AS "OLD", IF(hypertensionArterielle>0,"Oui","Non") AS "Hypertension artérielle", IF(infarctusMyocarde>0,"Oui","Non") AS "Infarctus du myocarde", IF(insuffisanceCoronaire>0,"Oui","Non") AS "Insuffisance coronaire", IF(troubleRythme>0,"Oui","Non") AS "Trouble du rythme", IF(accidentVasculaireCerebral>0,"Oui","Non") AS "Accident vasculaire cérébral", IF(insuffisanceCardiaque>0,"Oui","Non") AS "Insuffisance cardiaque", IF(arteriopathie>0,"Oui","Non") AS "Artériopathie", IF(tabagismeActuel>0,"Oui","Non") AS "Tabagisme actuel", CONCAT(nbPaquetsActuel," ","PA") AS "", IF(tabagismeAncien>0,"Oui","Non") AS "Tabagisme ancien", CONCAT(nbPaquetsAncien," ","PA") AS "", IF(alcool>0,"Oui","Non") AS "Alcool (conso régulière)", IF(refluxGastro>0,"Oui","Non") AS "Reflux gastro-oesophagien", IF(glaucome>0,"Oui","Non") AS "Glaucome", IF(diabete>0,"Oui","Non") AS "Diabète", CONCAT(patienttypeDiabete.libelle,"(",id_typeDiabete,")") AS "", IF(hypercholesterolemie>0,"Oui","Non") AS "Hypercholestérolémie", IF(hypertriglyceridemie>0,"Oui","Non") AS "Hypertriglycéridémie", IF(dysthyroidie>0,"Oui","Non") AS "Dysthyroïdie", IF(depression>0,"Oui","Non") AS "Dépression", IF(sedentarite>0,"Oui","Non") AS "Sédentarité", IF(syndromeDApneesSommeil>0,"Oui","Non") AS "SAS", IF(obesite>0,"Oui","Non") AS "Obésité", IF(dysmorphieFaciale>0,"Oui","Non") AS "Dysmorphie faciale", TextObservations AS "", id_user FROM oms_patient LEFT JOIN oms_departement ON oms_departement.id = id_departement LEFT JOIN oms_pays ON oms_pays.id = id_pays LEFT JOIN patientsexe ON patientsexe.id = id_sexe LEFT JOIN patientprofession ON patientprofession.id = id_profession LEFT JOIN patienttypeDiabete ON patienttypeDiabete.id = id_typeDiabete WHERE oms_patient.id=1 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 'small"(conso régulière)", IF(refluxGastro0,"Oui","Non") as "Reflux ga' at line 1 "near 'small" <-- where is small o_O The PHP code isn't really relevant cause you won't see a lot. $db = mysql_connect(); mysql_select_db();//TODO SWITCH TO PDO mysql_query("SET NAMES UTF8"); $fields = $form->getFields($form); $settingsForm = $form->getSettings(); $sql = 'SELECT oms_patient.id,oms_patient.date,oms_patient.date_modif,'; foreach ($fields as $field) { if (!$field->isMultiSelect()) { $field->select_full(&$sql, 'oms_patient', null); } } if (isset($settingsForm['linkTo'])) { $idLinkTo = 'id_' . str_replace('oms_', '', $settingsForm['linkTo']); $sql .= $idLinkTo; } $sql.=' FROM oms_patient'; foreach ($fields as $field) { if (!$field->isMultiSelect() && $field->getTable('oms_patient')) { $sql .=' LEFT JOIN ' . $field->getTable('oms_patient') . ' ON ' . $field->getTable('oms_patient') . '.id = '.$field->getFieldName().' '; } } $sql.=' where oms_patient.id=' . $this->m_settings['e']; $result = mysql_query($sql) or die('Erreur SQL !<br>' . $sql . '<br>' . mysql_error()); $data = mysql_fetch_assoc($result); var_dump of $sql string(2663) "SELECT oms_patient.id,oms_patient.date,oms_patient.date_modif,date_modif,AES_DECRYPT(nom,"xxxxx") as "Nom",AES_DECRYPT("prenom","xxxxx") as "Prénom usuel",DATE_FORMAT(ddn, "%d/%m/%Y") as "Date de naissance",villeNaissance as "Lieu de naissance (ville)",CONCAT(oms_departement.libelle,"(",id_departement,")") as "Lieu de vie",CONCAT(oms_pays.libelle,"(",id_pays,")") as "Pays",CONCAT(patientsexe.libelle,"(",id_sexe,")") as "Sexe",CONCAT(patientprofession.libelle,"(",id_profession,")") as "Profession", IF"... can't go further to see what is in the output after the "..." <-- if you have an idea

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  • php form-input validation

    - by fusion
    i have a html page in which i enter data which then submits and inserts in a database on a php page. how would i validate in php that the data received is not a duplicate of the data in the database? any help appreciated.

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  • Polling a running php cli script

    - by B_
    I want to run a php script from the command line that is always running and constantly updating a variable. I then want any php script that is run in the meantime (probably but not necessarily from the web) to be able to read that variable at any time. Anyone know how I can do this? Thanks.

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  • PHP SASL(PECL) sasl_server_init(app) works with CLI but not with ApacheModule

    - by ZokRadonh
    I have written a simple auth script so that Webusers can type in their username and password and my PHP script verifies them by SASL. The SASL Library is initialized by php function sasl_server_init("phpfoo"). So phpfoo.conf in /etc/sasl2/ is used. phpfoo.conf: pwcheck_method: saslauthd mech_list: PLAIN LOGIN log_level: 9 So the SASL library now tries to connect to saslauthd process by socket. saslauthd command line looks like this: /usr/sbin/saslauthd -r -V -a pam -n 5 So saslauthd uses PAM to authenticate. In the php script I have created sasl connection by sasl_server_new("php", null, "myRealm"); The first argument is the servicename. So PAM uses the file /etc/pam.d/php to see for further authentication information. /etc/pam.d/php: auth required pam_mysql.so try_first_pass=0 config_file=/etc/pam.d/mysqlconf.nss account required pam_permit.so session required pam_permit.so mysqlconf.nss has all information that is needed for a useful MySQL Query to user table. All of this works perfectly when I run the script by command line. php ssasl.php But when I call the same script via webbrowser(php apache module) I get an -20 return code (SASL_NOUSER). In /var/log/messages there is May 18 15:27:12 hostname httpd2-prefork: unable to open Berkeley db /etc/sasldb2: No such file or directory I do not have anything with a Berkeley db for authentication with SASL. I think authentication using /etc/sasldb2 is the default setting. In my opinion it does not read my phpfoo.conf file. For some reason the php-apache-module ignores the parameter in sasl_server_init("phpfoo"). My first thought was that there is a permission issue. So back in shell: su -s /bin/bash wwwrun php ssasl.php "Authentication successful". - No file-permission issue. In the source of the sasl-php-extension we can find: PHP_FUNCTION(sasl_server_init) { char *name; int name_len; if (zend_parse_parameters(1 TSRMLS_CC, "s", &name, &name_len) == FAILURE) { return; } if (sasl_server_init(NULL, name) != SASL_OK) { RETURN_FALSE; } RETURN_TRUE; } This is a simple pass through of the string. Are there any differences between the PHP CLI and PHP ApacheModule version that I am not aware of? Anyway, there are some interesting log entries when I run PHP in CLI mode: May 18 15:44:48 hostname php: SQL engine 'mysql' not supported May 18 15:44:48 hostname php: auxpropfunc error no mechanism available May 18 15:44:48 hostname php: _sasl_plugin_load failed on sasl_auxprop_plug_init for plugin: sqlite May 18 15:44:48 hostname php: sql_select option missing May 18 15:44:48 hostname php: auxpropfunc error no mechanism available May 18 15:44:48 hostname php: _sasl_plugin_load failed on sasl_auxprop_plug_init for plugin: sql Those lines are followed by lines of saslauthd and PAM which results in authentication success.(I do not get any of them in ApacheModule mode) Looks like that he is trying auxprop pwcheck before saslauthd. I have no other .conf file in /etc/sasl2. When I change the parameter of sasl_server_init to something other then I get the same error in CLI mode as in ApacheModule mode.

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  • Does PHP have job control like bash does?

    - by Andrew
    Hello, does PHP support something like ampersand in bash (forking)? Let's say I wanted to use cURL on 2 web pages concurrently, so script doesn't have to wait before first cURL command finnishes, how could one achieve that in PHP? Something like this in bash: curl www.google.com & curl www.yahoo.com & wait

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  • Problems with CGI wrapper for PHP

    - by user205878
    I'm having a bugger of a time with a CGI wrapper for PHP. I know very little about CGI and PHP as CGI. Here's what I know about the system: Solaris 10 on a 386 Suhosin PHP normally running as CGI, with cgiwrap (http://cgiwrap.sourceforge.net/). I am not able to find an example wrapper.cgi on the server to look at. Shared hosting (virtual host), so I don't have access to Apache config. But the admins are not helpful. Switching hosts is not an option. Options directive cannot be overridden in .htaccess (ExecCGI, for example). .htaccess: AddHandler php-handler .php Action php-handler "/bin/test.cgi" ~/public_html/bin/test.cgi: #!/usr/bin/sh # Without these 2 lines, I get an Internal Server Error echo "Content-type: text/html" echo "" exec "/path/to/php-cgi" 'foo.php'; /bin/foo.php: <?php echo "this is foo.php!"; Output of http://mysite.com/bin/test.cgi: X-Powered-By: PHP/5.2.11 Content-type: text/html echo "Content-type: text/html" echo "" exec "/path/to//php-cgi" 'foo.php'; Output of http:/ /mysite.com/anypage.php: X-Powered-By: PHP/5.2.11 Content-type: text/html echo "Content-type: text/html" echo "" exec "/path/to//php-cgi" 'foo.php'; The things I note are: PHP is being executed, as noted by the X-Powered-By ... header. The source of /bin/test.cgi is output in the results. No matter what I put as the second argument of exec, it isn't passed to the php binary. I've tried '-i' to get phpinfo, '-v' to get the version... When I execute test.cgi via the shell, I get the expected results (the argument is passed to php, and it is reflected in the output). Any ideas about how to get this to work? UPDATE It appears that the reason the source of the test.cgi was appearing was due to errors. Anytime fatal error occurred, either within the cgi itself or with the command being executed by exec, it would cause the source of the cgi to appear. Within test.cgi, I can get the proper output with exec "/path/to/php-cgi" -h (I get the same thing as I would from CLI).

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  • logic for a php function

    - by danish hashmi
    i need to make a php code for checking hotel room avaibility where user from the present day can book rooms upto 90 days or less and there are total 30 rooms available in the hotel,so if once i store the data for a user like his booking from one date till another next time if i want to check the avaibility how should i do it in php,what would be the logic. obviously i simple query like this isn't correct for eg $this->db->select('*') ->from('default_bookings') ->where('booking_from <',$input['fromdate']) ->where('booking_till >',$input['tilldate']);

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  • Php date() giving the wrong time after parsing

    - by Kirill
    This is confusing as hell, here's the php I'm using: <?php echo date('H:i D j, F',$j->date); ?> This is what it gives me: 01:33 Thu 1, January Which seems fine, until you look at the actual time that is being given ($j-date provides): 2010-06-12 21:12:23 Why is it giving me a January and what am I doing wrong?

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  • PHP URL Security Question

    - by TaG
    I want to have users store the url in my database I'm using php mysql and htmlpurifier I was wondering if the following code was good way to filter out bad data? Here is the Partial PHP code. $url = mysqli_real_escape_string($mysqli, $purifier->purify(htmlspecialchars(strip_tags($_POST['url'])));

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