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  • Why I cannot create DB in MySQL throw PHP?

    - by Roman
    I have this code: $link = mysql_connect("localhost", "ctman", "blablabla"); if ( ! $link ) die ("I cannot connect to MySQL.<br>\n"); else print "Connection is established.<br>\n"; // Create the "ct" database. mysql_query("create database ct", $link) or die("I cannot create the DB: ".mysql_error()."<br>\n"); And I get this error message: I cannot create the DB: Access denied for user 'ctmanager'@'%' to database 'ct' Does anybody have any idea why I cannot create a DB and why I have '@%' symbols in the error message?

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  • Can't select database table even though the code is right

    - by Lisa
    I am trying to display a list of my vbulliten threads on a non-vbulliten portion of my site. However I can't select the vbulliten database: <?php $host = "localhost"; $user = "my username"; $pass = "my password"; $dbname = "tableprefix_forum"; mysql_connect($host, $user, $pass) or die ("Could not connect to database server."); mysql_select_db($dbname) or die ("Could not select database."); ?> I am substituting some things here in this example but all my credentials are correct including my db server username, password and forum db name. So what is the problem? Is it due to some internal security feature in vbulliten, does this system not allow you to connect to it's db if the page trying to connect to it is a non-vbulliten page?

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  • Archive tar files to a different location inperl

    - by user314261
    Hi, I am a newbee in Perl. I am reading a directory having some archive files and uncompressing the archive files one by one. Everything seems well however the files are getting uncompressed in the folder which has the main perl code module which is running the sub modules. I want the archive to be generated in the folder I specify. This is my code: sub ExtractFile { #Read the folder which was copied in the output path recursively and extract if any file is compressed my $dirpath = $_[0]; opendir(D, "$dirpath") || die "Can't open dir $dirpath: $!\n"; my @list = readdir(D); closedir(D); foreach my $f (@list) { print " \$f = $f"; if(-f $dirpath."/$f") { #print " File in directory $dirpath \n ";#is \$f = $f\n"; my($file_name, $file_dirname,$filetype)= fileparse($f,qr{\..*}); #print " \nThe file extension is $filetype"; #print " \nThe file name is is $file_name"; # If compressed file then extract the file if($filetype eq ".tar" or $filetype eq ".tzr.gz") { my $arch_file = $dirpath."/$f"; print "\n file to be extracted is $arch_file"; my $tar = Archive::Tar->new($arch_file); #$tar->extract() or die ("Cannot extract file $arch_file"); #mkdir($dirpath."/$file_name"); $tar->extract_file($arch_file,$dirpath."/$file_name" ) or die ("Cannot extract file $arch_file"); } } if(-d $dirpath."/$f") { if($f eq "." or $f eq "..") { next; } print " Directory\n";# is $f"; ExtractFile($dirpath."/$f"); } } } The method ExtractFile is called recursively to loop all the archives. When using $tar-extract() it uncompresses in the folder which calls this metohd. when I use $tar-extract_file($arch_file,$dirpath."/$file_name" ) I get an error : No such file in archive: '/home/fsang/dante/workspace/output/s.tar' at /home/fsang/dante/lib/Extraction.pm line 80 Please help I have checked that path and input output there is no issue with it. Seems some usage problem I am not aware of for $tar-extract_file(). Many thanks for anyone resolving this issue. Regards, Sakshi

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  • PHP export to text file - Only saving first line.

    - by wertz8090
    I'm trying to export some extracted $_POST information into a text file, however my code is only capturing the first variable and ignoring the rest. I'm saving the information to the text file in this manner: $values = "First Name: $fName\r\n"; $values .= "Last Name: $lName\r\n"; $values .= "Address: $address\r\n"; etc. This is the code I use to write to the text file: $fp = @fopen("person.data", "w") or die("Couldn't open person.data for writing!"); $numBytes = @fwrite($fp, $values) or die("Couldn't write values to file!"); @fclose($fp); Any ideas on why it would only save the first $values ($fName) variable but not the rest of them? It actually saves the first part of the $values string for all of them (so I see Last Name:, Address:, etc. on separate lines in the text file) but the called variables $lName and $address do not appear.

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  • PHP New Line will not work

    - by user364333
    Hi I am trying to create some code that first reads the existing contents of the file in and then adds the new line of code on a new line but the code i am using just adds it on the new text on to the already existing line instead of the new line... Here is the code i am using: <?php $id = $_GET['id']; $userfile = "user1.txt"; $fo = fopen($userfile, 'r') or die("can't open favourites file"); $currentdata = fread($fo, filesize($userfile)); fclose($fo); $fw = fopen($userfile, 'w') or die("can't open favourites file"); $currentprocessed = "$currentdata\n"; fwrite($fw, $currentprocessed); fwrite($fw, $id); fclose($fw); ?> I have tried a whole range of different ideas but nothing has worked, any help would be appreciated.

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  • PHP MySQL Syntax Error 'You have an error in your SQL syntax'

    - by Alec
    I cannot figure out the issue with my code here. I am trying to take info from the table, then subtract 1 second from Current_Time which looks like '2:00'. The problem is, I get: "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 'Current_Time) VALUES('22')' at line 1" I don't even understand where it gets 22 from. Thanks, I really appreciate it. if (isset($_GET['id']) && isset($_GET['time'])) { mysql_select_db("aleckaza_pennyauction", $connection); $query = "SELECT Current_Time FROM Live_Auctions WHERE ID='1'"; $results = mysql_query($query) or die(mysql_error()); while ($row = mysql_fetch_array($results)) { $newTime = $row['Current_Time'] - 1; $query = "INSERT INTO Live_Auctions(Current_Time) VALUES('".$newTime."')"; $results = mysql_query($query) or die(mysql_error()); } }

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  • php class crash course

    - by rabidmachine9
    Hello people, I'm sorry for this question but I'm getting crazy trying to write my first php clash, it is supposed to connect me to server and to a database but I'm always getting this error: Parse error: syntax error, unexpected ',', expecting T_PAAMAYIM_NEKUDOTAYIM in /Applications/XAMPP/xamppfiles/htdocs/classTest/test.php on line 9 thanks in advance, here is the code: <?php // include 'DBConnect.php'; class DBConnect { function connection($hostname = "localhost", $myname = "root", $pass = ''){ mysql_connect(&hostname, &myname, &pass) or die("Could not connect!"); //Connect to mysql } function choose($dbnam){ mysql_select_db(&dbnam) or die("Couldn't find database!"); //fnd specific db } } $a = new DBConnect(); $connect = $a->connection("localhost", "root"); $a->choose('mydb'); ?>

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  • Using Java, can I have one JVM spawn another, and then have the original one exit?

    - by CarlG
    I have a need to have a running JVM start another JVM and then exit. I'm currently trying to do this via Runtime.getRuntime().exec(). The other JVM starts, but my original JVM won't exit until the "child" JVM process stops. It appears that using Runtime.getRuntime().exec() creates a parent-child relationship between the processes. Is there some way to de-couple the spawned process so that the parent can die, or some other mechanism to spawn a process without any relationship to the creating process? Note that this seems exactly like this question: http://stackoverflow.com/questions/2566502/using-java-to-spawn-a-process-and-keep-it-running-after-parent-quits but the accepted answer there doesn't actually work, at least not on my system (Windows 7, Java 5 and 6). It seems that maybe this is a platform-dependent behavior. I'm looking for a platform independent way to reliably invoke the other process and let my original process die.

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  • PHP loop position

    - by Jordan Pagaduan
    Can someone help me on this. I'm made an image uploader and i want the image to make another tr if it reach to 5 pics so it will not overflow. Here is my code: Can someone help me on this. I'm made an image uploader and i want the image to make another tr if it reach to 5 pics so it will not overflow. Here is my code: $dbc = mysql_connect("localhost" , "root" , "") or die (mysql_error()); mysql_select_db('blog_data') or die (mysql_error()); $sql = "SELECT * FROM img_uploaded"; $result = mysql_query($sql); while($rows=mysql_fetch_array($result)) { if ($rows) { echo "<tr><td><img src='user_images/".$rows['img_name'] . "' width='100' height='100'></td></tr>"; } else { echo "<td><img src='user_images/".$rows['img_name'] . "' width='100' height='100'></td>"; } }

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  • PHP MySQL Insert Help

    - by user364333
    Hey I am trying to make a page that inserts some strings into a MySQL table but it just dosn't seem to be working for me. Here is the code I am using at the moment. <?php mysql_connect($address, $username, $password); @mysql_select_db($database) or die("Unable to select database"); $query = "insert INTO user (movieid, moviename)('" . $id . "','" . $name . "') or die(mysql_error())"; mysql_query($query); mysql_close(); ?> Where am i going wrong?

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  • ajax how to read out $_post variable

    - by alex
    Hi, I am trying to filter/search a database with ajax $.ajax({ type: "POST", url: "filterSearch.php", queryString: qry, success: function(data){ alert( "Data Saved: " + data ); $('#searchResult').html(data); // Fill the search results box } }); Now in filterSearch.php i have the following test codes if(isset($_POST['queryString'])) { echo "TEST"; } if($_POST['runquery']==1) { $sql = "SELECT * FROM fs_vacatures WHERE here-the-like-query?"; $msg = $sql; echo $msg; die(); } die(); But nor TEST or the $sql is return in the alert??

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  • How can I insert a line at the beginning of a file with Perl's Tie::File?

    - by thebourneid
    I'm trying to insert/add a line 'COMMENT DUMMY' at the beginnig of a file as a first row if /PATTERN/ not found. I know how to do this with OPEN CLOSE function. Probably after reading the file it should look something like this: open F, ">", $fn or die "could not open file: $!"; ; print F "COMMENT DUMMY\n", @array; close F; But I have a need to implement this with the use of the Tie::File function and don't know how. use strict; use warnings; use Tie::File; my $fn = 'test.txt'; tie my @lines, 'Tie::File', $fn or die "could not tie file: $!"; untie @lines;

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  • Strange behaviour of mb_detect_order() in PHP

    - by termopro
    I would like to detect encoding of some text (using PHP). For that purpose i use mb_detect_encoding() function. The problem is that the function returns different results if i change the order of possible encodings with mb_detect_order() function. Consider the following example $html = <<< STR ????????????????????????????????????????????????????????????????????????????????????????????????????????? STR; mb_detect_order(array('UTF-8','EUC-JP', 'SJIS', 'eucJP-win', 'SJIS-win', 'JIS', 'ISO-2022-JP','ISO-8859-1','ISO-8859-2')); $originalEncoding = mb_detect_encoding($str); die($originalEncoding); // $originalEncoding = 'UTF-8' However if you change the order of encodings in mb_detect_order() the results will be different: mb_detect_order(array('EUC-JP','UTF-8', 'SJIS', 'eucJP-win', 'SJIS-win', 'JIS', 'ISO-2022-JP','ISO-8859-1','ISO-8859-2')); die($originalEncoding); // $originalEncoding = 'EUC-JP' So my questions are: Why is that happening ? Is there a way in PHP to correctly and unambiguously detect encoding of text ?

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  • Stop Submit With empty and error Input Values Using PHP

    - by user3615781
    I am using the following code for sending data to database, but it sends the data even the values of the fields are incorrect or empty. So can anyone help me solve this by using php? Here is my code: <?php //Connecting to sql db $connect = mysqli_connect("localhost","root","","form"); /* check connection */ if (!$connect) { die('Connect Error: ' . mysqli_connect_error()); } //Sending data to sql db $result = mysqli_query($connect,"INSERT INTO students(name,email,website,comment,gender) VALUES('$name','$email','$website','$comment','$gender')"); if (!$result) { die('Query Error:'. mysqli_error($connect)); } ?>

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  • Where I get wrong?

    - by Ole Jak
    I have PHP code like this require_once( "includes" . DIRECTORY_SEPARATOR . "constants.php"); $userID = validKey( $key ); $query = "select a.username, b.streamID from user a, streams b where a.id = b.userID;)"; $connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS); if (!$connection) { die("Database connection failed: " . mysql_error()); } $db_select = mysql_select_db(DB_NAME,$connection); if (!$db_select) {die( "Database selection failed: " . mysql_error()); } $generated_table = mysql_fetch_array(mysql_query($query, $connection)); // line (*) print_r($generated_table); But PHP gives me error/worning and does not print anything out Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given on line (*) 1) Why It does not printed me anything out 2) Where am I wrong?

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  • gauge chart is not displaying any thing

    - by Sandy
    i am trying to display the latest speed in mysql database on guage chart. i have tried so many things but gauge is not display plz any can help me...my code is attached and php part shows the correct value but dont know why guage is not display <?php $host="localhost"; // Host name $username="root"; // Mysql username $password=""; // Mysql password $db_name="mysql"; // Database name $tbl_name="gpsdb"; // Table name // Connect to server and select database. $con=mysql_connect("$host", "$username")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $data = mysql_query("SELECT speed FROM gpsdb WHERE DeviceId=1234 ORDER BY TIME DESC LIMIT 1") or die(mysql_error()); while ($nt = mysql_fetch_assoc($data)) { $speed = $nt['speed']; $jsonTable = json_encode($speed); echo $jsonTable; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="content-type" content="text/html; charset=utf-8"/> <title> Google Visualization API Sample </title> <script type="text/javascript" src="//www.google.com/jsapi"></script> <script type="text/javascript"> google.load('visualization', '1', {packages: ['gauge']}); </script> <script type="text/javascript"> function drawVisualization() { // Create and populate the data table. var data = new google.visualization.DataTable(<?=$speed?>); // Create and draw the visualization. new google.visualization.Gauge(document.getElementById('visualization')). draw(data); } google.setOnLoadCallback(drawVisualization); </script> </head> <body style="font-family: Arial;border: 0 none;"> <div id="visualization" style="width: 600px; height: 300px;"></div> </body> </html>

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  • SFTP not supported error with PHP & cURL

    - by Bad Programmer
    I followed the advice from this Stack Overflow question thread, but I keep hitting a snag. I am receiving the following error message: Unsupported protocol: sftp Here is my code: $ch = curl_init(); if(!$ch) { $error = curl_error($ch); die("cURL session could not be initiated. ERROR: $error.""); } $fp = fopen($docname, 'r'); if(!$fp) { $error = curl_error($ch); die("$docname could not be read."); } curl_setopt($ch, CURLOPT_URL, "sftp://$user_name:$user_pass@$server:22/$docname"); curl_setopt($ch, CURLOPT_UPLOAD, 1); curl_setopt($ch, CURLOPT_PROTOCOLS, CURLPROTO_SFTP); curl_setopt($ch, CURLOPT_INFILE, $fp); curl_setopt($ch, CURLOPT_INFILESIZE, filesize($docname)); //this is where I get the failure $exec = curl_exec ($ch); if(!$exec) { $error = curl_error($ch); die("File $docname could not be uploaded. ERROR: $error."); } curl_close ($ch); I used the curl_version() function to see my curl information, and found that sftp doesn't seem to be in the array of supported protocols: [version_number] => 462597 [age] => 2 [features] => 1597 [ssl_version_number] => 0 [version] => 7.15.5 [host] => x86_64-redhat-linux-gnu [ssl_version] => OpenSSL/0.9.8b [libz_version] => 1.2.3 [protocols] => Array ( [0] => tftp [1] => ftp [2] => telnet [3] => dict [4] => ldap [5] => http [6] => file [7] => https [8] => ftps ) Is this a matter of my version of cURL being outdated, or is the SFTP protocol simply not supported at all? Any advice is greatly appreciated.

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  • cant made mySQL connection

    - by Andika Evangelion Thirteenth S
    I've use following code $con = mysql_connect("localhost","root","");if (!$con) {die('Could not connect: ' . mysql_error());} It's works. but when $db_host='localhost';$db_id='root';$db_pass=''; $con = mysql_connect($db_host,$db_id,$db_pass);if (!$con) {die('Could not connect: ' . mysql_error());} it didn't works, trying to swap ("),('),and empty in mysql_connect() and in $var and vice versa. Any help would be appreciated. Thanks.

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  • ftp php file uploadin

    - by Issak
    <?php ini_set('max_execution_time', '0'); $host = '234.546.155.485'; $usr = 'fgfgfgdf'; $pwd = 'fghghh'; // file to move: $file = 'http://vsomesite.com/file.flv'; $ftp_path = '/public_html/video57242/test.flv'; // connect to FTP server (port 21) $conn_id = ftp_connect($host, 21) or die ("Cannot connect to host"); // send access parameters ftp_login($conn_id, $usr, $pwd) or die("Cannot login"); // turn on passive mode transfers (some servers need this) // ftp_pasv ($conn_id, true); // perform file upload $upload = ftp_put($conn_id, $ftp_path, $local_file, FTP_ASCII); // check upload status: print (!$upload) ? 'Cannot upload' : 'Upload complete'; ?> upload fails when the file is remote not local. what problem?

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  • Guice Problem with Tests

    - by D3orn
    Hey I wrote a LudoGame and now I like to test it with a little GuiceInjection^^ I have an Interface IDie for my die. Now for the game I only need an IDie instead of a realdie = in tests I simply give the LudoGame a MokeDie to set up the Numbers I like to roll. The IDie has only one method: roll() which returns a int. BUT the mokeDie now has another public method: sendNextNumber() (should be clear what this does^^) Now I like to @Inject a Die and if @UseMokeDie is before a Test I'll like to pass the MokeDie but I'm very new to Guice... Need some advices please! Thx for Answers

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  • MySQL Access in Php to a database created in C-Panel

    - by Rhys Drury
    Basically, i'm having trouble connecting to a mysql database using a php web page. I created the database in C-panel using the wizard i'm connecting like this $db_host = "localhost"; //your host Database address $db_username = "xxxx"; //your account username $db_pass = "xxxxx"; //your account password $db_name = "xxxxx"; //your database name @mysql_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql"); @mysql_select_db("$db_name") or die ("no database"); But all my page seems to do is trigger the " could not connect to mysql" my page does have wordpress installed at the minute but I'm planning to get rid of it as I'm creating my site myself. I'm just baffled why it can't connect, because in Phpmyadmin ( a feature on C-panel) it says the database is in localhost.

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  • PHP/MySQL Interview - How would you have answered?

    - by martincarlin87
    I was asked this interview question so thought I would post it here to see how other users would answer: Please write some code which connects to a MySQL database (any host/user/pass), retrieves the current date & time from the database, compares it to the current date & time on the local server (i.e. where the application is running), and reports on the difference. The reporting aspect should be a simple HTML page, so that in theory this script can be put on a web server, set to point to a particular database server, and it would tell us whether the two servers’ times are in sync (or close to being in sync). This is what I put: // Connect to database server $dbhost = 'localhost'; $dbuser = 'xxx'; $dbpass = 'xxx'; $dbname = 'xxx'; $conn = mysql_connect($dbhost, $dbuser, $dbpass) or die (mysql_error()); // Select database mysql_select_db($dbname) or die(mysql_error()); // Retrieve the current time from the database server $sql = 'SELECT NOW() AS db_server_time'; // Execute the query $result = mysql_query($sql) or die(mysql_error()); // Since query has now completed, get the time of the web server $php_server_time = date("Y-m-d h:m:s"); // Store query results in an array $row = mysql_fetch_array($result); // Retrieve time result from the array $db_server_time = $row['db_server_time']; echo $db_server_time . '<br />'; echo $php_server_time; if ($php_server_time != $db_server_time) { // Server times are not identical echo '<p>Database server and web server are not in sync!</p>'; // Convert the time stamps into seconds since 01/01/1970 $php_seconds = strtotime($php_server_time); $sql_seconds = strtotime($db_server_time); // Subtract smaller number from biggest number to avoid getting a negative result if ($php_seconds > $sql_seconds) { $time_difference = $php_seconds - $sql_seconds; } else { $time_difference = $sql_seconds - $php_seconds; } // convert the time difference in seconds to a formatted string displaying hours, minutes and seconds $nice_time_difference = gmdate("H:i:s", $time_difference); echo '<p>Time difference between the servers is ' . $nice_time_difference; } else { // Timestamps are exactly the same echo '<p>Database server and web server are in sync with each other!</p>'; } Yes, I know that I have used the deprecated mysql_* functions but that aside, how would you have answered, i.e. what changes would you make and why? Are there any factors I have omitted which I should take into consideration? The interesting thing is that my results always seem to be an exact number of minutes apart when executed on my hosting account: 2012-12-06 11:47:07 2012-12-06 11:12:07

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  • undefined GET id?

    - by Azzyh
    <?php $s = $_GET["s"]; if($s) { $hent_b = mysql_query("SELECT * FROM member_battles WHERE state = '1' ORDER BY id DESC LIMIT 0,200") or die(mysql_error()); }else{ $hent_b = mysql_query("SELECT * FROM member_battles WHERE state = '0' ORDER BY id DESC LIMIT 0,200") or die(mysql_error()); } while($vis = mysql_Fetch_array($hent_b)) { ?> I have this now i want when i enter my site (index.php) it should not come up undefined $_GET["s"]; how do i do this? but i want when you do index.php?s then it should change the query

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  • Won't connect to the database

    - by user1657958
    I'm confused...I'm using the same code in a different document and in there it's not a problem to get a connection to the database. But in the new document it's just not working...(password, username, database name...all is checked and correct) :-/ <?php define ("DB_HOST", "db1234567.db.hello.com"); // set database host define ("DB_USER", "db1234567"); // set database user define ("DB_PASS","password123"); // set database password define ("DB_NAME","db1234567"); // set database name $link = mysql_connect(DB_HOST, DB_USER, DB_PASS) or die("Couldn't make connection."); $db = mysql_select_db(DB_NAME, $link) or die("Couldn't select database"); ?> In the browser I get this: "Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'db1234567'@'123.123.12.12 (using password: YES) in /homepages/12/1234567/test/test.php on line 8 Couldn't make connection." Would be cool if someone could help me :) I'm not seeing any error... Thx!

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  • PHP Try and Catch for SQL Insert

    - by meme
    I have a page on my website (high traffic) that does an insert on every page load. I am curious of the fastest and safest way to (catch an error) and continue if the system is not able to do the insert into MySQL. Should I use try/catch or die or something else. I want to make sure the insert happens but if for some reason it can't I want the page to continue to load anyway. ... $db = mysql_select_db('mobile', $conn); mysql_query("INSERT INTO redirects SET ua_string = '$ua_string'") or die('Error #10'); mysql_close($conn); ...

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