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  • Perl Lingua giving weird error on install

    - by user299306
    I am trying to install perl Lingua onto a unix system (ubuntu, latest version). Of course I am root. when I go into the package to install using 'perl Makefile.pl' I get this dumb error: [root@csisl27 Lingua-Lid-0.01]# perl Makefile.PL /opt/ls//lib does not exist at Makefile.PL line 48. I have tried playing with the path on line 48, nothing changes, here is what line 48-50 looks like: Line 48: die "$BASE/lib does not exist" unless -d "$BASE/lib"; Line 49: die "$BASE/include does not exist" unless -d "$BASE/include"; Line 50: die "lid.h is missing in $BASE/include" unless -e "$BASE/includ/lid.h"; The variable $BASE is declared as this: $BASE = "/opt/ls/" if ($^O eq "linux" or $^O eq "solaris"); $BASE = "/usr/local/" if ($^O eq "freebsd"); $BASE = $ENV{LID_BASE_DIR} if (defined $ENV{LID_BASE_DIR}); Now the perl program I am trying to write simply look like this (just my base): #!/usr/bin/perl use Lingua::LinkParser; use strict; print "Hello world!\n"; When I run this trying to use Lingua, here is my error: [root@csisl27 assign4]# ./perl_parser_1.pl Can't locate Lingua/LinkParser.pm in @INC (@INC contains: /usr/lib/perl5/site_perl/5.10.0/x86_64-linux-thread-multi /usr/lib/perl5/site_perl/5.10.0 /usr/lib/perl5/vendor_perl/5.10.0/x86_64-linux-thread-multi /usr/lib/perl5/vendor_perl/5.10.0 /usr/lib/perl5/5.10.0/x86_64-linux-thread-multi /usr/lib/perl5/5.10.0 /usr/lib/perl5/site_perl /usr/lib/perl5/vendor_perl .) at ./perl_parser_1.pl line 3. BEGIN failed--compilation aborted at ./perl_parser_1.pl line 3. Tried insalling this from cpan, still doesn't properly work.

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  • php edit database entery script not working.

    - by Jacksta
    I have two files which show all contats in a database, then edits them. The second script keeps directing the browser back due to this part. I cant work out how to make it work! :) <?php if (!$_POST[id]) { header( "Location: pick_modcontact.php"); exit; } else { session_start(); } if ($_SESSION[valid] != "yes") { header( "Location: pick_modcontact.php"); exit; } This script shows all contacts in a database which is an "address book" this part works fine. please see below. Name:pick_modcontact.php if ($_SESSION[valid] != "yes") { header( "Location: contact_menu.php"); exit; } $db_name = "testDB"; $table_name = "my_contacts"; $connection = @mysql_connect("localhost", "admin", "user") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); $sql = "SELECT id, f_name, l_name FROM $table_name ORDER BY f_name"; $result = @mysql_query($sql, $connection) or die(mysql_error()); $num = @mysql_num_rows($result); if ($num < 1) { $display_block = "<p><em>Sorry No Results!</em></p>"; } else { while ($row = mysql_fetch_array($result)) { $id = $row['id']; $f_name = $row['f_name']; $l_name = $row['l_name']; $option_block .= "<option value\"$id\">$f_name, $l_name</option>"; } $display_block = "<form method=\"POST\" action=\"show_modcontact.php\"> <p><strong>Contact:</strong> <select name=\"id\">$option_block</select> <input type=\"submit\" name=\"submit\" value=\"Select This Contact\"></p> </form>"; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Modify A Contact</title> </head> <body> <h1>My Contact Management System</h1> <h2><em>Modify a Contact</em></h2> <p>Select a contact from the list below, to modify the contact's record.</p> <? echo "$display_block"; ?> <br> <p><a href="contact_menu.php">Return to Main Menu</a></p> </body> </html> This script is for modifying the contact: named show_modcontact.php <?php if (!$_POST[id]) { header( "Location: pick_modcontact.php"); exit; } else { session_start(); } if ($_SESSION[valid] != "yes") { header( "Location: pick_modcontact.php"); exit; } $db_name = "testDB"; $table_name = "my_contacts"; $connection = @mysql_connect("localhost", "admin_cantsayno", "cantsayno") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); $sql = "SELECT f_name, l_name, address1, address2, address3, postcode, prim_tel, sec_tel, email, birthday FROM $table_name WHERE id = '$_POST[id]'"; $result = @mysql_query($sql, $connection) or die(mysql_error()); while ($row = mysql_fetch_array($result)) { $f_name = $row['f_name']; $l_name = $row['l_name']; $address1 = $row['address1']; $address2 = $row['address2']; $address3 = $row['address3']; $country = $row['country']; $prim_tel = $row['prim_tel']; $sec_tel = $row['sec_tel']; $email = $row['email']; $birthday = $row['birthday']; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Modify A Contact</title> </head> <body> <form action="do_modcontact.php" method="post"> <input type="id" value="<? echo "$_POST[id]"; ?>" /> <table cellpadding="5" cellspacing="3"> <tr> <th>Name & Address Information</th> <th> Other Contact / Personal Information</th> </tr> <tr> <td align="top"> <p><strong>First Name:</strong><br /> <input type="text" name="f_name" value="<? echo "$f_name" ?>" size="35" maxlength="75" /></p> <p><strong>Last Name:</strong><br /> <input type="text" name="l_name" value="<? echo "$l_name" ?>" size="35" maxlength="75" /></p> <p><strong>Address1:</strong><br /> <input type="text" name="f_name" value="<? echo "$address1" ?>" size="35" maxlength="75" /></p> <p><strong>Address2:</strong><br /> <input type="text" name="f_name" value="<? echo "$address2" ?>" size="35" maxlength="75" /></p> <p><strong>Address3:</strong><br /> <input type="text" name="f_name" value="<? echo "$address3" ?>" size="35" maxlength="75" /> </p> <p><strong>Postcode:</strong><br /> <input type="text" name="f_name" value="<? echo "$postcode" ?>" size="35" maxlength="75" /></p> <p><strong>Country:</strong><br /> <input type="text" name="f_name" value="<? echo "$country" ?>" size="35" maxlength="75" /> </p> <p><strong>First Name:</strong><br /> <input type="text" name="f_name" value="<? echo "$f_name" ?>" size="35" maxlength="75" /></p> </td> <td align="top"> <p><strong>Prim Tel:</strong><br /> <input type="text" name="f_name" value="<? echo "$prim_tel" ?>" size="35" maxlength="75" /></p> <p><strong>Sec Tel:</strong><br /> <input type="text" name="f_name" value="<? echo "$sec_tel" ?>" size="35" maxlength="75" /></p> <p><strong>Email:</strong><br /> <input type="text" name="f_name" value="<? echo "$email" ?>" size="35" maxlength="75" /> </p> <p><strong>Birthday:</strong><br /> <input type="text" name="f_name" value="<? echo "$birthday" ?>" size="35" maxlength="75" /> </p> </td> </tr> <tr> <td align="center"> <p><input type="submit" name="submit" value="Update Contact" /></p> <br /> <p><a href="contact_menu.php">Retuen To Menu</a></p> </td> </tr> </table> </form> </body> </html>

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  • Simple perl program failing to execute

    - by yves Baumes
    Here is a sample that fails: #!/usr/bin/perl -w # client.pl #---------------- use strict; use Socket; # initialize host and port my $host = shift || 'localhost'; my $port = shift || 55555; my $server = "10.126.142.22"; # create the socket, connect to the port socket(SOCKET,PF_INET,SOCK_STREAM,(getprotobyname('tcp'))[2]) or die "Can't create a socket $!\n"; connect( SOCKET, pack( 'Sn4x8', AF_INET, $port, $server )) or die "Can't connect to port $port! \n"; my $line; while ($line = <SOCKET>) { print "$line\n"; } close SOCKET or die "close: $!"; with the error: Argument "10.126.142.22" isn't numeric in pack at D:\send.pl line 16. Can't connect to port 55555! I am using this version of Perl: This is perl, v5.10.1 built for MSWin32-x86-multi-thread (with 2 registered patches, see perl -V for more detail) Copyright 1987-2009, Larry Wall Binary build 1006 [291086] provided by ActiveState http://www.ActiveState.com Built Aug 24 2009 13:48:26 Perl may be copied only under the terms of either the Artistic License or the GNU General Public License, which may be found in the Perl 5 source kit. Complete documentation for Perl, including FAQ lists, should be found on this system using "man perl" or "perldoc perl". If you have access to the Internet, point your browser at http://www.perl.org/, the Perl Home Page. While I am running the netcat command on the server side. Telnet does work.

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  • Java HttpURLConnection bekommt keine cookies

    - by TeNNoX
    ich versuche über eine HttpURLConnection einen Login auf einer Webseite durchzuführen, und davon dann die cookies zu erhalten... Bei meinen Testseiten auf einem eigenen Server geht es problemlos, ich sende "a=3&b=5" und als cookie erhalte ich "8", also die Summe. Wenn ich dies allerdings auf der gewollten Seite anwende, kommt einfach nur die Seite, als ob ich gar nichts per POST gesendet hätte... :( Generelle Verbesserungsvorschläge sind auch erwünscht! :) Mein Code: HttpURLConnection conn = (HttpURLConnection) new URL(url).openConnection(); conn.setDoInput(true); conn.setDoOutput(true); conn.setRequestMethod("POST"); conn.setRequestProperty("useragent", "Mozilla/5.0 (Windows NT 6.1; WOW64; rv:17.0) Gecko/20100101 Firefox/17.0"); conn.setRequestProperty("Connection", "keep-alive"); DataOutputStream out = new DataOutputStream(conn.getOutputStream()); out.writeBytes("USER=tennox&PASS=*****"); out.close(); BufferedReader in = new BufferedReader(new InputStreamReader(conn.getInputStream())); String line; String response = new String(); while ((line = in.readLine()) != null) { response = response + line + "\n"; } in.close(); System.out.println("headers:"); int i = 0; String header; while ((header = conn.getHeaderField(i)) != null) { String key = conn.getHeaderFieldKey(i); System.out.println(((key == null) ? "" : key + ": ") + header); i++; } String cookies = conn.getHeaderField("Set-Cookie"); System.out.println("\nCookies: \"" + cookies + "\"");

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  • MySQL Connection Error in PHP

    - by user309381
    I have set the password for root and grant all privileges for root. Why does it say it is denied? ****mysql_query() [function.mysql-query]: Access denied for user 'SYSTEM'@'localhost' (using password: NO) in C:\wamp\www\photo_gallery\includes\database.php on line 56 Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in C:\wamp\www\photo_gallery\includes\database.php on line 56 The Query has problemAccess denied for user 'SYSTEM'@'localhost' (using password: NO) Code as follows: <?php include("DB_Info.php"); class MySQLDatabase { public $connection; function _construct() { $this->open_connection(); } public function open_connection() { /* $DB_SERVER = "localhost"; $DB_USER = "root"; $DB_PASS = ""; $DB_NAME = "photo_gallery";*/ $this->connection = mysql_connect($DBSERVER,$DBUSER,$DBPASS); if(!$this->connection) { die("Database Connection Failed" . mysql_error()); } else { $db_select = mysql_select_db($DBNAME,$this->connection); if(!$db_select) { die("Database Selection Failed" . mysql_error()); } } } function mysql_prep($value) { if (get_magic_quotes_gpc()) { $value = stripslashes($value); } // Quote if not a number if (!is_numeric($value)) { $value = "'" . mysql_real_escape_string($value) . "'"; } return $value; } public function close_connection() { if(isset($this->connection)) { mysql_close($this->connection); unset($this->connection); } } public function query($sql) { //$sql = "SELECT*FROM users where id = 1"; $result = mysql_query($sql); $this->confirm_query($result); //$found_user = mysql_fetch_assoc($result); //echo $found_user; return $found_user; } private function confirm_query($result) { if(!$result) { die("The Query has problem" . mysql_error()); } } } $database = new MySQLDatabase(); ?>

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  • Shouldn't prepared statements be much faster?

    - by silversky
    $s = explode (" ", microtime()); $s = $s[0]+$s[1]; $con = mysqli_connect ('localhost', 'test', 'pass', 'db') or die('Err'); for ($i=0; $i<1000; $i++) { $stmt = $con -> prepare( " SELECT MAX(id) AS max_id , MIN(id) AS min_id FROM tb "); $stmt -> execute(); $stmt->bind_result($M,$m); $stmt->free_result(); $rand = mt_rand( $m , $M ).'<br/>'; $res = $con -> prepare( " SELECT * FROM tb WHERE id >= ? LIMIT 0,1 "); $res -> bind_param("s", $rand); $res -> execute(); $res->free_result(); } $e = explode (" ", microtime()); $e = $e[0]+$e[1]; echo number_format($e-$s, 4, '.', ''); // and: $link = mysql_connect ("localhost", "test", "pass") or die (); mysql_select_db ("db") or die ("Unable to select database".mysql_error()); for ($i=0; $i<1000; $i++) { $range_result = mysql_query( " SELECT MAX(`id`) AS max_id , MIN(`id`) AS min_id FROM tb "); $range_row = mysql_fetch_object( $range_result ); $random = mt_rand( $range_row->min_id , $range_row->max_id ); $result = mysql_query( " SELECT * FROM tb WHERE id >= $random LIMIT 0,1 "); } defenitly prepared statements are much more safer but also every where it says that they are much faster BUT in my test on the above code I have: - 2.45 sec for prepared statements - 5.05 sec for the secon example What do you think I'm doing wrong? Should I use the second solution or I should try to optimize the prep stmt?

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  • Reading a CSV with file_get_contents in PHP

    - by JPro
    I am reading a 'kind' of csv file and exploding it and storing it in array. The file I am reading has this structure Id,Log,Res,File mydb('Test1','log1','Pass','lo1.txt'). mydb('Test2','log2','Pass','lo2.txt'). mydb('Test3','log3','Pass','lo3.txt'). Now what I am trying to do is : reading the last record in my database, get the Name, lets say in this case 'Test1' and then I am searching through my file and where I can find the position of 'Test1' and get the next lines in the file, extract the ID,s and add it to database. I am getting the position of desired string in the file, but I am not sure how to get the next lines too. Here's my code so far. <?php mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db("testing") or die(mysql_error()); $result = mysql_query("select ID from table_1 order by S_no DESC limit 1") or die(mysql_error()); $row = mysql_fetch_array( $result ); $a = $row['ID']; echo 'Present Top Row is '.$a.'<br>'; $addresses = explode("\n", file_get_contents('\\\\fil1\\logs\\tes.pl')); foreach($addresses as $val) { $pos = strstr($val, $a); if ($pos === false) { } else { echo "The string <br> '$a' <br>was found in the string <br>'$val' <br>"; echo " and exists at position <br>$pos<br>"; } }

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  • displaying a group once in php mysql

    - by JPro
    I have some data like this : 1 TC1 PASS 2 TC2 FAIL 3 TC3 INCONC 4 TC1 FAIL 5 TC21 FAIL 6 TC4 PASS 7 TC3 PASS 8 TC2 FAIL 9 TC1 TIMEOUT 10 TC21 FAIL If I try the below code : <?php mysql_connect("localhost", "root", "pop") or die(mysql_error()); mysql_select_db("jpd") or die(mysql_error()); $oustanding_fails = mysql_query("select * from SELECT_PASS ") or die(mysql_error()); $resultSetArray = array(); $platform; while($row1 = mysql_fetch_array( $oustanding_fails )) { if(trim($row1['TESTCASE']) <> trim($platform)) { echo $row1['TESTCASE']."-"; $platform = $row1['TESTCASE']; } echo $row1['RESULT'] ."<br>"; } ?> to get a result like this : TC1 PASS FAIL TIMEOUT TC2 FAIL FAIL TC3 INCONC PASS TC4 PASS AND SO ON. I am unable to get the result I want. Any ideas where exactly I am making mistake? Thanks.

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  • How to display external database data in VBulletin 4 Forums Custom PHP Block?

    - by NJTechGuy
    Hi guys! I want to display data feed from an external database in a sidebar in the forums section. PHP Block Code : $host = 'db.123.net'; $dbUser = 'db49'; $dbPass = 'iReVbY'; $db = 'db6578h8'; mysql_connect("$host", "$dbUser", "$dbPass") or die(mysql_error()); mysql_select_db("$db") or die(mysql_error()); ob_start(); $result = mysql_query("SELECT id, title from abc") or die(mysql_error()); while($row = mysql_fetch_array( $result )) { print"<center>"; print "<a href=\"http://abc.com/?id=" . $row['id'] . "\"></a>"; print "</center>"; } $output .= ob_get_contents(); return $output; ob_end_clean(); How do I return an array to display in a PHP block in the sidebar (forums section)? Please help me out of this! Thank you..

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  • sending email in php unable to use HTML tags

    - by JPro
    I am trying to send an email through the result sets generated in MySQL in PHP This is the code. <?php $txtMsg = '<table><th>Name</th>'; $txtMsg = ''; mysql_connect("localhost", "root", "pop") or die(mysql_error()); mysql_select_db("jpd") or die(mysql_error()); $oustanding = mysql_query("select Name from results") or die(mysql_error()); $num=mysql_num_rows($oustanding); while($row1 = mysql_fetch_array( $oustanding )) { ?> <tr> <td><h3><?php echo $row1['Name']; ?></h3></td> </tr> <?php $txtMsg .= "<tr><td>".$row1['Name']."</td></tr>"; } ini_set ( "SMTP", "xy.domain.com" ); $mail_to= '[email protected]'; $mail_from='[email protected]'; $mail_sub='OutStanding Results'; $mail_mesg=$txtMsg; //Check for success/failure of delivery if(mail($mail_to,$mail_sub,$mail_mesg,"From: $mail_from")) echo "<br><br>Email Successfully Sent!"; else echo "<br><br>Error Sending Email!"; } ?> The problem is , I want the results to be displayed in table format. But instead of processing the html tags, they are getting printed as well. How to get the email with table format? Thanks.

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  • AJAX HTML PHP question

    - by Jordan Pagaduan
    This is my scripts.js function showHint(str) { if (str.length==0) { document.getElementById("txtHint").innerHTML=""; return; } if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); } else {// code for IE6, IE5 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById("txtHint").innerHTML=xmlhttp.responseText; } } xmlhttp.open("GET","inputProcess.php?q="+str,true); xmlhttp.send(); } This is my HTML <script type="text/javascript" src="scripts.js"></script> <form> Type your name here : <input type="text" onkeypress="showHint(this.value)" name="name" /> </form> This is my PHP file <?php $q = $_GET['q']; $dbc=mysql_connect("localhost","root","") or die (mysql_error()); mysql_select_db('input_oop') or die (mysql_error()); $sql = "INSERT INTO users set name = '".$q."'"; mysql_query($sql) or die (mysql_error()); ?> When the text is save in my database it will save 5 times or up.

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  • security deleting a mysql row with jQuery $.post

    - by FFish
    I want to delete a row in my database and found an example on how to do this with jQuery's $.post() Now I am wondering about security though.. Can someone send a POST request to my delete-row.php script from another website? JS function deleterow(id) { // alert(typeof(id)); // number if (confirm('Are you sure want to delete?')) { $.post('delete-row.php', {album_id:+id, ajax:'true'}, function() { $("#row_"+id).fadeOut("slow"); }); } } PHP: delete-row.php <?php require_once("../db.php"); mysql_connect(DB_SERVER, DB_USER, DB_PASSWORD) or die("could not connect to database " . mysql_error()); mysql_select_db(DB_NAME) or die("could not select database " . mysql_error()); if (isset($_POST['album_id'])) { $query = "DELETE FROM albums WHERE album_id = " . $_POST['album_id']; $result = mysql_query($query); if (!$result) die('Invalid query: ' . mysql_error()); echo "album deleted!"; } ?>

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  • ajax to php to curl and back..

    - by pfunc
    I am trying to make an ajax call to a php script. The php script calls an rss feed using curl, gets the data, and returns the data to the funciton. I keep getting an error "Warning: Wrong parameter count for curl_error() in" .... Here is my php code:1 $ch = curl_init() or die(curl_error()); curl_setopt($ch, CURLOPT_URL, $feed); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); $data1 = curl_exec($ch) or die(curl_error()); echo $data1; and the ajax call: $.ajax({ url: "getSingleFeed.php", type: "POST", data: "feedURL=" + window.feedURL, success: function(feed){ alert(feed); }}); I tested all the variables, they are being passed correctly, I can echo them out. But this line: $data1 = curl_exec($ch) or die(curl_error()); is what is giving me the error. I am doing the same thing with curl on other pages, just without ajax, and it is working fine. Is there anything special I need to do with ajax to do this?

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  • i am getting error like mysql_connect() acces denied for system@localhost(using password NO)

    - by user309381
    class MySQLDatabase { public $connection; function _construct() { $this->open_connection();} public function open_connection() {$this->connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS); if(!$this->connection){die("Database Connection Failed" . mysql_error());} else{$db_select = mysql_select_db(DB_NAME,$this->connection); if(!$db_select){die("Database Selection Failed" . mysql_error()); } }} public function close_connection({ if(isset($this->connection)){ mysql_close($this->connection); unset($this->connection);}} public function query(/*$sql*/){ $sql = "SELECT*FROM users where id = 1"; $result = mysql_query($sql); $this->confirm_query($result); //return $result;while( $found_user = mysql_fetch_assoc($result)) { echo $found_user ['username']; } } private function confirm_query($result) { if(!$result) { die("The Query has problem" . mysql_error()); } } } $database = new MySQLDatabase(); $database->open_connection(); $database->query(); $database->close_connection(); ?>

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  • save html-formatted text to database

    - by yozhik
    Hi all! I want to save html-formatted text to database, but when I do that it is don't save html-symbols like < / ' and others This is how I read article from database for editing: <p class="Title">??????????? ???????:</p> <textarea name="EN" cols="90" rows="20" value="<?php echo $articleArr['EN']; ?>" ></textarea> And this is how I save it to database: function UpdateArticle($ArticleID, $ParentName, $Title, $RU, $EN, $UKR) { //fetch data from database for dropdown lists //connect to db or die) $db = mysql_connect($GLOBALS["gl_dbName"], $GLOBALS["gl_UserName"], $GLOBALS["gl_Password"] ) or die ("Unable to connect"); //to prevenr ????? symbols in unicode - utf-8 coding mysql_query("SET NAMES 'UTF8'"); mysql_set_charset('utf8'); mysql_query("SET NAMES 'utf8' COLLATE 'utf8_general_ci'"); //select database mysql_select_db($GLOBALS["gl_adminDatabase"], $db); $sql = "UPDATE Articles SET AParentName='".$ParentName."', ATitle='".$Title."', RU='".$RU."', EN='".$EN."', UKR='".$UKR."' WHERE ArticleID='".$ArticleID."';"; //execute SQL-query $result = mysql_query($sql, $db); if (!$result) { die('Invalid query: ' . mysql_error()); } //close database = very inportant mysql_close($db); } Help me please, how can I save such texts properly, Thanx!

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  • Selecting a whole database over an individual table to output to file

    - by Daniel Wrigley
    :::::::: EDIT :::::::: New code for people to have a look at, one question I have with this is where do I set were the *.gz file is saved? $backupFile = $dbname . date("Y-m-d-H-i-s") . '.gz'; $command = "mysqldump --opt -h $dbhost -u $dbuser -p $dbpass $dbname | gzip > $backupFile"; system($command); Also why the hell can you not reply yo your own post with answering it? :( :::::::: EDIT :::::::: Ok Im having trouble finding out how to select a full database for backup as an *.sql file rather than only an individual table. On the localhost I have several databases with one named "foo" and it is that which I want to backup and not any of the individual tables inside the database "foo". The code to connect to the database; //Database Information $dbhost = "localhost"; $dbname = "foo"; $dbuser = "bar"; $dbpass = "rulz"; //Connect to database mysql_connect ($dbhost, $dbuser, $dbpass) or die("Could not connect: ".mysql_error()); mysql_select_db($dbname) or die(mysql_error()); The code to backup the database; // Grab the time to know when this post was submitted $time = date('Y-m-d-H-i-s'); $tableName = 'foo'; $backupFile = '/sql/backup/'. $time .'.sql'; $query = "SELECT * INTO OUTFILE '". $backupFile ."' FROM ". $tableName .""; $result = mysql_query($query)or die("Database query died: " . mysql_error()); My brain is hurting near to the end of the day so no doubts i've missed something out very obvious. Thanks in advance to anyone helping me out.

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  • php fopen function dies, though I have file permissions set to read and write

    - by Matthew Robert Keable
    I'm following a tutorial on php, and am having difficulty getting this to work. I set the appropriate directory permissions to read and write, but every time I run this, I get the die string. The code is: $ourFileName = "testFile.txt"; $ourFileHandle = fopen($ourFileName, 'w') or die("can't open file"); fclose($ourFileHandle); As far as my basic understanding goes, if "testFile.txt" does not exist, fopen should create that file (I have basic knowledge of Python, and remember this same principle in that language). But it...it doesn't. Even if I create the aforementioned file, and put it up, that line of code still returns a die string. My hosting account does not give me permission to execute. Is this a problem? My server runs on Windows. I am using Dreamweaver CS5, on OSX 10.5.8. I've done some searching on this, and see other people having similar issues - but none of them keyed to exactly my range of problems. Being that I'm a beginner, I feel that it might be something I'm overlooking. Thanks!!

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  • change column widths in mysql query

    - by addi
    <?php // Connection Database $search = $_POST ['Search']; mysql_connect("xxxxxx", "xxxxxxxxx", "xxxxxx") or die ("Error Connecting to Database"); mysql_select_db("xxxxxx") or die('Error'); $data = mysql_query("SELECT* FROM course WHERE MATCH (CourseName, CourseDescription, CourseLeader) AGAINST ('". $search ."')") or die (mysql_error()); Print "<table border cellpadding=3>"; while($info = mysql_fetch_array( $data )) { Print "<tr>"; Print "<th>Course Name:</th> <td>".$info['CourseName'] . "</td> "; Print "<th>Course Description:</th><td>".$info['CourseDescription'] . "</td> "; Print "<th>Course Leader:</th><td>".$info['CourseLeader'] . " </td></tr>"; } Print "</table>"; ?> In my php code I print the columns CourseName, CourseDescription, CourseLeader after a search, as a resultset. CourseDescription has a lot of text, how do I print it all? is there a way to change the column widths?

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  • How do I get PHP variables from this MySQL query?

    - by CT
    I am working on an Asset Database problem using PHP / MySQL. In this script I would like to search my assets by an asset id and have it return all related fields. First I query the database asset table and find the asset's type. Then depending on the type I run 1 of 3 queries. <?php //make database connect mysql_connect("localhost", "asset_db", "asset_db") or die(mysql_error()); mysql_select_db("asset_db") or die(mysql_error()); //get type of asset $type = mysql_query(" SELECT asset.type From asset WHERE asset.id = 93120 ") or die(mysql_error()); switch ($type){ case "Server": //do some stuff that involves a mysql query mysql_query(" SELECT asset.id ,asset.company ,asset.location ,asset.purchase_date ,asset.purchase_order ,asset.value ,asset.type ,asset.notes ,server.manufacturer ,server.model ,server.serial_number ,server.esc ,server.user ,server.prev_user ,server.warranty FROM asset LEFT JOIN server ON server.id = asset.id WHERE asset.id = 93120 "); break; case "Laptop": //do some stuff that involves a mysql query mysql_query(" SELECT asset.id ,asset.company ,asset.location ,asset.purchase_date ,asset.purchase_order ,asset.value ,asset.type ,asset.notes ,laptop.manufacturer ,laptop.model ,laptop.serial_number ,laptop.esc ,laptop.user ,laptop.prev_user ,laptop.warranty FROM asset LEFT JOIN laptop ON laptop.id = asset.id WHERE asset.id = 93120 "); break; case "Desktop": //do some stuff that involves a mysql query mysql_query(" SELECT asset.id ,asset.company ,asset.location ,asset.purchase_date ,asset.purchase_order ,asset.value ,asset.type ,asset.notes ,desktop.manufacturer ,desktop.model ,desktop.serial_number ,desktop.esc ,desktop.user ,desktop.prev_user ,desktop.warranty FROM asset LEFT JOIN desktop ON desktop.id = asset.id WHERE asset.id = 93120 "); break; } ?> So far I am able to get asset.type into $type. How would I go about getting the rest of the variables (laptop.model to $model, asset.notes to $notes and so on)? Thank you.

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  • using jquery in mysql php

    - by JPro
    I am new to using Jquery using mysql and PHP I am using the following code to pull the data. But there is not data or error displayed. JQUERY: <html> <head> <script> function doAjaxPost() { // get the form values var field_a = $("#field_a").val(); $("#loadthisimage").show(); $.ajax({ type: "POST", url: "serverscript.php", data: "ID="+field_a, success: function(resp){ $("#resposnse").html(resp); $("#loadthisimage").hide(); }, error: function(e){ alert('Error: ' + e); } }); } </script> </head> <body> <select id="field_a"> <option value="data_1">data_1</option> <option value="data_2">data_2</option> </select> <input type="button" value="Ajax Request" onClick="doAjaxPost()"> <a href="#" onClick="doAjaxPost()">Here</a> </form> <div id="resposnse"> <img src="ajax-loader.gif" style="display:none" id="loadthisimage"> </div> </body> and now serverscript.php <?php if(isset($_POST['ID'])) { $nm = $_POST['ID']; echo $nm; //insert your code here for the display. mysql_connect("localhost", "root", "pop") or die(mysql_error()); mysql_select_db("JPro") or die(mysql_error()); $result1 = mysql_query("select Name from results where ID = \"$nm\" ") or die(mysql_error()); // store the record of the "example" table into $row while($row1 = mysql_fetch_array( $result1 )) { $tc = $row1['Name']; echo $tc; } } ?>

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  • Php 5 and mysql connecting gives error ...code and error in there check out and plz..plz help....

    - by user309381
    **mysql_query() [function.mysql-query]: Access denied for user 'SYSTEM'@'localhost' (using password: NO) in C:\wamp\www\photo_gallery\includes\database.php on line 56 Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in C:\wamp\www\photo_gallery\includes\database.php on line 56 The Query has problemAccess denied for user 'SYSTEM'@'localhost' (using password: NO) i have set the password for root and grant privilege all for root.Why it soes show like SYSTEM@localhost i dont have SYSTEM .** class MySQLDatabase { public $connection; function _construct() { $this-open_connection(); } public function open_connection() { /* $DB_SERVER = "localhost"; $DB_USER = "root"; $DB_PASS = ""; $DB_NAME = "photo_gallery";*/ $this-connection = mysql_connect($DBSERVER,$DBUSER,$DBPASS); if(!$this-connection) { die("Database Connection Failed" . mysql_error()); } else { $db_select = mysql_select_db($DBNAME,$this-connection); if(!$db_select) { die("Database Selection Failed" . mysql_error()); } } } function mysql_prep($value) { if (get_magic_quotes_gpc()) { $value = stripslashes($value); } // Quote if not a number if (!is_numeric($value)) { $value = "'" . mysql_real_escape_string($value) . "'"; } return $value; } public function close_connection() { if(isset($this->connection)) { mysql_close($this->connection); unset($this->connection); } } public function query($sql) { //$sql = "SELECT*FROM users where id = 1"; $result = mysql_query($sql); $this->confirm_query($result); //$found_user = mysql_fetch_assoc($result); //echo $found_user; return $found_user; } private function confirm_query($result) { if(!$result) { die("The Query has problem" . mysql_error()); } } } $database = new MySQLDatabase(); ?

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  • My SQL query is only returning I need the parent aswell

    - by sico87
    My sql query is only returning the children of the parent I need it to return the parent as well, public function getNav($cat,$subcat){ //gets all sub categories for a specific category if(!$this->checkValue($cat)) return false; //checks data $query = false; if($cat=='NULL'){ $sql = "SELECT itemID, title, parent, url, description, image FROM p_cat WHERE deleted = 0 AND parent is NULL ORDER BY position;"; $query = $this->db->query($sql) or die($this->db->error); }else{ //die($cat); $sql = "SET @parent = (SELECT c.itemID FROM p_cat c WHERE url = '".$this->sql($cat)."' AND deleted = 0); SELECT c1.itemID, c1.title, c1.parent, c1.url, c1.description, c1.image, (SELECT c2.url FROM p_cat c2 WHERE c2.itemID = c1.parent LIMIT 1) as parentUrl FROM p_cat c1 WHERE c1.deleted = 0 AND c1.parent = @parent ORDER BY c1.position;"; $query = $this->db->multi_query($sql) or die($this->db->error); $this->db->store_result(); $this->db->next_result(); $query = $this->db->store_result(); } return $query; }

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  • mysql_connect() acces denied for system@localhost(using password NO)

    - by user309381
    class MySQLDatabase { public $connection; function _construct() { $this-open_connection(); } public function open_connection() { $this-connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS); if(!$this-connection) { die("Database Connection Failed" . mysql_error()); } else { $db_select = mysql_select_db(DB_NAME,$this-connection); if(!$db_select) { die("Database Selection Failed" . mysql_error()); } } } function mysql_prep($value) { if (get_magic_quotes_gpc()) { $value = stripslashes($value); } // Quote if not a number if (!is_numeric($value)) { $value = "'" . mysql_real_escape_string($value) . "'"; } return $value; } public function close_connection() { if(isset($this->connection)) { mysql_close($this->connection); unset($this->connection); } } public function query(/*$sql*/) { $sql = "SELECT*FROM users where id = 1"; $result = mysql_query($sql); $this->confirm_query($result); //return $result; while( $found_user = mysql_fetch_assoc($result)) { echo $found_user ['username']; } } private function confirm_query($result) { if(!$result) { die("The Query has problem" . mysql_error()); } } } $database = new MySQLDatabase(); $database-open_connection(); $database-query(); $database-close_connection(); ?

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  • i dont understand error while connecting php and mysql? user denied ? plz help me out to solve. ?

    - by user309381
    class MySQLDatabase { public $connection; function _construct() { $this->open_connection();} public function open_connection() {$this->connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS); if(!$this->connection){die("Database Connection Failed" . mysql_error());} else{$db_select = mysql_select_db(DB_NAME,$this->connection); if(!$db_select){die("Database Selection Failed" . mysql_error()); } }} public function close_connection({ if(isset($this->connection)){ mysql_close($this->connection); unset($this->connection);}} public function query(/*$sql*/){ $sql = "SELECT*FROM users where id = 1"; $result = mysql_query($sql); $this->confirm_query($result); //return $result;while( $found_user = mysql_fetch_assoc($result)) { echo $found_user ['username']; } } private function confirm_query($result) { if(!$result) { die("The Query has problem" . mysql_error()); } } } $database = new MySQLDatabase(); $database->open_connection(); $database->query(); $database->close_connection(); I am getting error like denied for user system@locahost(using password no).i have also other database but it runs fine and i dont also i have set the password after encountered the error what else can do to solve plz help ?

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  • php and mysql listing databases and looping through results

    - by Jacksta
    Beginner help needed :) I am doign an example form a php book which lists tables in databases. I am getting an error on line 36: $db_list .= "$table_list"; <?php //connect to database $connection = mysql_connect("localhost", "admin_cantsayno", "cantsayno") or die(mysql_error()); //list databases $dbs = @mysql_list_dbs($connection) or die(mysql_error()); //start first bullet list $db_list = "<ul>"; $db_num = 0; //loop through results of functions while ($db_num < mysql_num_rows($dbs)) { //get database names and make each a list point $db_names[$db_num] = mysql_tablename($dbs, $db_num); $db_list .= "<li>$db_names[$db_num]"; //get table names and make another list $tables = @mysql_list_tables($db_names[$db_num]) or die(mysql_error()); $table_list = "<ul>"; $table_num = 0; //loop through results of function while ($table_num < mysql_num_rows($tables)){ //get table names and make each bullet point $table_names[$table_num] = mysql_tablename($tables, $table_num); $table_list .= "<li>$table_names[$table_num]"; $table_num++; } //close inner bullet list and increment number to continue $table_list .= "</ul>" $db_list .= "$table_list"; $db_num++; } //close outer bullet list $db_list .= "</ul>"; ?> <html> <head> <title>MySQL Tables</title> </head> <body> <p><strong>Data bases and tables on local host</strong></p> <? echo "$db_list"; ?> </body>

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