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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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • searching for multiple columns mysql and php

    - by addi
    i'm trying to search for multiple columns using this code: <?php // Connection Database $search = $_POST ['Search']; mysql_connect("xxxxxx", "xxxxxx", "xxxxx") or die ("Error Connecting to Database"); mysql_select_db("xxxxx") or die('Error'); $data = mysql_query("SELECT CourseName, CourseDescription, CourseLeader FROM course MATCH (CourseName, CourseDescription, CourseLeader) AGAINST ('". $search ."') or die('Error'); Print "<table border cellpadding=3>"; while($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>"; ?> i'm getting the following error: Parse error: syntax error, unexpected T_STRING in /home/a7105766/public_html/website/scripts/coursesearchdb.php on line 30 what am I doing wrong?? cheers

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  • database connection OK,result not appear

    - by klox
    hi..all.for now i'm already connected to database but the result not appear at Tuner range is"+res+"this is my code: var str=data[0]; var matches=str.match(/[EE|EJU].*D/i); $.ajax({ type:"post", url:"process1.php", data:"tversion="+matches+"&action=tunermatches", cache:false, async:false, success: function(res){ $('#value').replacewith("<div id='value'><h6>Tuner range is"+res+".</h6></div>"); } }); }); and this is my process file: //connect to database $dbc=mysql_connect(_SRV,_ACCID,_PWD) or die(_ERROR15.": ".mysql_error()); $db=mysql_select_db("qdbase",$dbc) or die(_ERROR17.": ".mysql_error()); switch(postVar('action')) { case 'tunermatches' : tunermatches(postVar('tversion')); break; function tunermatches($tversion)){ $Tuner=mysql_real_escape_string($tversion); $sql= "SELECT remark FROM settingdata WHERE itemname='Tuner_range' AND itemdata='".$Tunermatches."'"; $res=mysql_query($sql) or die (_ERROR26.":".mysql_error()); $dat=mysql_fetch_array($res,MYSQL_NUM); if($dat[0]>0) { echo $dat[0]; } mysql_close($dbc); }

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  • Most efficient way to write over file after reading

    - by Ryan McClure
    I'm reading in some data from a file, manipulating it, and then overwriting it to the same file. Until now, I've been doing it like so: open (my $inFile, $file) or die "Could not open $file: $!"; $retString .= join ('', <$inFile>); ... close ($inFile); open (my $outFile, $file) or die "Could not open $file: $!"; print $outFile, $retString; close ($inFile); However I realized I can just use the truncate function and open the file for read/write: open (my $inFile, '+<', $file) or die "Could not open $file: $!"; $retString .= join ('', <$inFile>); ... truncate $inFile, 0; print $inFile $retString; close ($inFile); I don't see any examples of this anywhere. It seems to work well, but am I doing it correctly? Is there a better way to do this?

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  • authorise user from mysql database

    - by Jacksta
    I suck at php, and cant find the error here. The script gets 2 variables "username" and "password" from a html from then check them against a MySQL databse. When I run this I get the follow error "Query was empty" <? if ((!$_POST[username]) || (!$_POST[password])) { header("Location: show_login.html"); exit; } $db_name = "testDB"; $table_name = "auth_users"; $connection = @mysql_connect("localhost", "admin", "pass") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); $slq = "SELECT * FROM $table_name WHERE username ='$_POST[username]' AND password = password('$_POST[password]')"; $result = @mysql_query($sql, $connection) or die(mysql_error()); $num = mysql_num_rows($result); if ($num != 0) { $msg = "<p>Congratulations, you're authorised!</p>"; } else { header("Location: show_login.html"); exit; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Secret Area</title> </head> <body> <? echo "$msg"; ?> </body> </html>

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  • php + MySQL editing table data.

    - by Jacksta
    This question is relating to 2 php scripts. The first script is called pick_modcontact.php where I choose a contact (from a contact book like phone book), then posts to the script show_modcontact.php When I click the submit button on the form on pick.modcontact.php. As a result of submitting the form I am then taken to show_modcontact.php. As the variables are not present the user is directed back to pick_modcontact.php I can not work out how to correct the code so that it will show the results of the script show_modcontact.php 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", "pass") 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="text" name="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> note for site admin, I am re posting this question with the hope of someone else reading over it. older questions seem to go dead after a while.

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  • Error with MySQL Query

    - by Ken
    Okay, I must be an idiot, because this is my 3rd question for today. Here's my code: date_default_timezone_set("America/Los_Angeles"); include("mainmenu.php"); $con = mysql_connect("localhost", "root", "********"); if(!$con){ die(mysql_error()); } $usrname = $_POST['usrname']; $fname = $_POST['fname']; $lname = $_POST['lname']; $password = $_POST['password']; $email = $_POST['email']; mysql_select_db("`users`, $con) or die(mysql_error()"); $query = ("INSERT INTO `users`.`data` (`id`, `usrname`, `fname`, `lname`, `email`, `password`) VALUES (NULL, '$usrname', '$fname', '$lname', '$email', 'password'))"); mysql_query('$query') or die(mysql_error()); mysql_close($con); echo("Thank you for registering!"); I always get the error returned as: "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '$query' at line 1. Help a newbie. I'm about to stab my monitor.

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  • I can't insert data into my database

    - by Ken
    I don't know why, but my data doesn't go into my database 'users' with the table 'data'. <html> <body> <?php date_default_timezone_set("America/Los_Angeles"); include("mainmenu.php"); $con = mysql_connect("localhost", "root", "g00dfor@boy"); if(!$con){ die(mysql_error()); } $usrname = $_POST['usrname']; $fname = $_POST['fname']; $lname = $_POST['lname']; $password = $_POST['password']; $email = $_POST['email']; mysql_select_db(`users`, $con) or die(mysql_error()); mysql_query(INSERT INTO `users`.`data` (`id`, `usrname`, `fname`, `lname`, `email`, `password`) VALUES (NULL, '$usrname', '$fname', '$lname', '$email', 'password')) or die(mysql_error()); mysql_close($con) echo("Thank you for registering!"); ?> </body> </html> All i get is a blank page.

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  • Why doesn't this PHP execute?

    - by cam
    I copied the code from this site exactly: http://davidwalsh.name/web-service-php-mysql-xml-json as follows, /* require the user as the parameter */ if(isset($_GET['user']) && intval($_GET['user'])) { /* soak in the passed variable or set our own */ $number_of_posts = isset($_GET['num']) ? intval($_GET['num']) : 10; //10 is the default $format = strtolower($_GET['format']) == 'json' ? 'json' : 'xml'; //xml is the default $user_id = intval($_GET['user']); //no default /* connect to the db */ $link = mysql_connect('localhost','username','password') or die('Cannot connect to the DB'); mysql_select_db('db_name',$link) or die('Cannot select the DB'); /* grab the posts from the db */ $query = "SELECT post_title, guid FROM wp_posts WHERE post_author = $user_id AND post_status = 'publish' ORDER BY ID DESC LIMIT $number_of_posts"; $result = mysql_query($query,$link) or die('Errant query: '.$query); /* create one master array of the records */ $posts = array(); if(mysql_num_rows($result)) { while($post = mysql_fetch_assoc($result)) { $posts[] = array('post'=>$post); } } /* output in necessary format */ if($format == 'json') { header('Content-type: application/json'); echo json_encode(array('posts'=>$posts)); } else { header('Content-type: text/xml'); echo '<posts>'; foreach($posts as $index => $post) { if(is_array($post)) { foreach($post as $key => $value) { echo '<',$key,'>'; if(is_array($value)) { foreach($value as $tag => $val) { echo '<',$tag,'>',htmlentities($val),'</',$tag,'>'; } } echo '</',$key,'>'; } } } echo '</posts>'; } /* disconnect from the db */ @mysql_close($link); } And the php doesn't execute, it just displays as plain text. What's the dealio? The host supports PHP, I use it to run a Wordpress blog and other things.

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  • Javascript function with PHP throwing a "Illegally Formed XML Syntax" error

    - by Joe
    I'm trying to learn some javascript and i'm having trouble figuring out why my code is incorrect (i'm sure i'm doing something wrong lol), but anyways I am trying to create a login page so that when the form is submitted javascript will call a function that checks if the login is in a mysql database and then checks the validity of the password for the user if they exist. however I am getting an error (Illegally Formed XML Syntax) i cannot resolve. I'm really confused, mostly because netbeans is saying it is a xml syntax error and i'm not using xml. here is the code in question: function validateLogin(login){ login.addEventListener("input", function() { $value = login.value; if (<?php //connect to mysql mysql_connect(host, user, pass) or die(mysql_error()); echo("<script type='text/javascript'>"); echo("alert('MYSQL Connected.');"); echo("</script>"); //select db mysql_select_db() or die(mysql_error()); echo("<script type='text/javascript'>"); echo("alert('MYSQL Database Selected.');"); echo("</script>"); //query $result = mysql_query("SELECT * FROM logins") or die(mysql_error()); //check results against given login while($row = mysql_fetch_array($result)){ if($row[login] == $value){ echo("true"); exit(0); } } echo("false"); exit(0); ?>) { login.setCustomValidity("Invalid Login. Please Click 'Register' Below.") } else { login.setCustomValidity("") } }); } the code is in an external js file and the error throws on the last line. Also from reading i understand best practices is to not mix js and php so how would i got about separating them but maintaining the functionality i need? thanks!

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  • HELP A NEWB (AGAIN) PLZ

    - by Ken
    Okay, I must be an idiot, because this is my 3rd question for today. Here's my code: date_default_timezone_set("America/Los_Angeles"); include("mainmenu.php"); $con = mysql_connect("localhost", "root", "********"); if(!$con){ die(mysql_error()); } $usrname = $_POST['usrname']; $fname = $_POST['fname']; $lname = $_POST['lname']; $password = $_POST['password']; $email = $_POST['email']; mysql_select_db("`users`, $con) or die(mysql_error()"); $query = ("INSERT INTO `users`.`data` (`id`, `usrname`, `fname`, `lname`, `email`, `password`) VALUES (NULL, '$usrname', '$fname', '$lname', '$email', 'password'))"); mysql_query('$query') or die(mysql_error()); mysql_close($con); echo("Thank you for registering!"); I always get the error returned as: "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '$query' at line 1. Help a newbie. I'm about to stab my monitor.

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  • Error loading font file using Imager::Font module in perl

    - by user211808
    use strict; use Imager; use Imager::Font; my $img = Imager->new(); my $file = "D:\\table.png"; $img->open(file=>$file) or die $img->errstr(); # Create smaller version my $thumb = $img->scale(scalefactor=>1.2); my $black = Imager::Color->new( 0, 0, 0 ); my $format; # Autostretch individual channels $thumb->filter(type=>'autolevels'); my $font_filename = "D:\\courbd.ttf"; my $font = Imager::Font->new(file=>$font_filename) or die "Cannot load $font_filename: ", Imager->errstr; for $format ( qw( png gif jpg tiff ppm ) ) { # Check if given format is supported if ($Imager::formats{$format}) { $file.="_low.$format"; print "Storing image as: $file\n"; $thumb->string(x => 50, y => 70, font =>$font, string => "Hello, World!", color => 'red', size => 30, aa => 1); $thumb->write(file=>$file) or die $thumb->errstr; } }

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  • How to loop through an array return from the Query of Mysql

    - by Jerry
    This might be easy for you guys but i could't get it. I have a php class that query the database and return the query result. I assign the result to an array and wants to use it on my main.php script. I have tried to use echo $var[0] or echo $var[1] but the output are 'array' instead of my value. Anyone can help me about this issue? Thanks a lot! My php class <?php class teamQuery { function teamQuery(){ } function getAllTeam(){ $connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS); if (!$connection) { die("Database connection failed: " . mysql_error()); } $db_select = mysql_select_db(DB_NAME,$connection); if (!$db_select) { die("Database selection failed: " . mysql_error()); } $teamQuery=mysql_query("SELECT * FROM team", $connection); if (!$teamQuery){ die("database has errors: ".mysql_error()); } $ret = array(); while($row=mysql_fetch_array($teamQuery)){ $ret[]=$row; } mysql_free_result($teamQuery); return $ret; } } ?> My php on the main.php $getTeam=new teamQuery(); $team=$getTeam->getAllTeam(); //echo $team[0] or team[1] output 'array' string! // while($team){ // do something } can't work either // How to loop through the values?? Thanks!

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  • How to output multiple rows from an SQL query using the mysqli object

    - by Jonathan
    Assuming that the mysqli object is already instantiatied (and connected) with the global variable $mysql, here is the code I am trying to work with. class Listing { private $mysql; function getListingInfo($l_id = "", $category = "", $subcategory = "", $username = "", $status = "active") { $condition = "`status` = '$status'"; if (!empty($l_id)) $condition .= "AND `L_ID` = '$l_id'"; if (!empty($category)) $condition .= "AND `category` = '$category'"; if (!empty($subcategory)) $condition .= "AND `subcategory` = '$subcategory'"; if (!empty($username)) $condition .= "AND `username` = '$username'"; $result = $this->mysql->query("SELECT * FROM listing WHERE $condition") or die('Error fetching values'); $this->listing = $result->fetch_array() or die('could not create object'); foreach ($this->listing as $key => $value) : $info[$key] = stripslashes(html_entity_decode($value)); endforeach; return $info; } } there are several hundred listings in the db and when I call $result-fetch_array() it places in an array the first row in the db. however when I try to call the object, I can't seem to access more than the first row. for instance: $listing_row = new Listing; while ($listing = $listing_row-getListingInfo()) { echo $listing[0]; } this outputs an infinite loop of the same row in the db. Why does it not advance to the next row? if I move the code: $this->listing = $result->fetch_array() or die('could not create object'); foreach ($this->listing as $key => $value) : $info[$key] = stripslashes(html_entity_decode($value)); endforeach; if I move this outside the class, it works exactly as expected outputting a row at a time while looping through the while statement. Is there a way to write this so that I can keep the fetch_array() call in the class and still loop through the records?

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  • Is it best to make fewer calls to the database and output the results in an array?

    - by Jonathan
    I'm trying to create a more succinct way to make hundreds of db calls. Instead of writing the whole query out every time I wanted to output a single field, I tried to port the code into a class that did all the query work. This is the class I have so far: class Listing { /* Connect to the database */ private $mysql; function __construct() { $this->mysql = new mysqli(DB_LOC, DB_USER, DB_PASS, DB) or die('Could not connect'); } function getListingInfo($l_id = "", $category = "", $subcategory = "", $username = "", $status = "active") { $condition = "`status` = '$status'"; if (!empty($l_id)) $condition .= "AND `L_ID` = '$l_id'"; if (!empty($category)) $condition .= "AND `category` = '$category'"; if (!empty($subcategory)) $condition .= "AND `subcategory` = '$subcategory'"; if (!empty($username)) $condition .= "AND `username` = '$username'"; $result = $this->mysql->query("SELECT * FROM listing WHERE $condition") or die('Error fetching values'); $info = $result->fetch_object() or die('Could not create object'); return $info; } } This makes it easy to access any info I want from a single row. $listing = new Listing; echo $listing->getListingInfo('','Books')->title; This outputs the title of the first listing in the category "Books". But if I want to output the price of that listing, I have to make another call to getListingInfo(). This makes another query on the db and again returns only the first row. This is much more succinct than writing the entire query each time, but I feel like I may be calling the db too often. Is there a better way to output the data from my class and still be succinct in accessing it (maybe outputting all the rows to an array and returning the array)? If yes, How?

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  • Storing data from database [mysql_num_rows]

    - by user1717305
    So I have this code to pass items from database to my order table. When I'm echoing the session. The session variable contains something so there's no problem with that. But when I echo those variables under numrows, it only shows nothing. Is there something wrong? <?php error_reporting(E_ALL ^ E_NOTICE); session_start(); require("connect.php"); $UserID = $_SESSION['CustNum']; $UserN = $_SESSION['UserName']; $ProdGTotal = $_SESSION['ProdGTotal']; $queryord = mysql_query("SELECT * FROM customer WHERE UserName = '$UserN'"); $numrows = mysql_num_rows($queryord); if(numrows == 1){ $row = mysql_fetch_assoc($queryord)or die ('Unable to run query:'.mysql_error()); // fetch associated: get function from a query for a database $dbstreet = $row['Street']; $dhousenum = $row['HouseNum']; $dbcnum = $row['CelNum']; $dbarea = $row['Area']; $dbbuilding = $row['Building']; $dbcity = $row['City']; $dbpnum = $row['PhoneNum']; $dbfname = $row['FName']; $dblname = $row['LName']; } else die(mysql_error()); $query4=mysql_query("INSERT INTO orderdetails VALUES ('', '$UserID', Now(), '$dbhousenum', '$dbstreet', '$dbarea', '$dbbuilding', '$dbcity', '$dbfname', '$dblname', '$dbcnum', '$dbpnum', '$ProdGTotal')",$connect); if ($query4){ header("location:index.php"); } else die(mysql_error()); ?>

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  • how to install nginx after removed it manually

    - by april
    I have installed nginx using app sudo apt-get install software-properties-common sudo add-apt-repository ppa:nginx/stable sudo apt-get install software-properties-common sudo apt-get update apt -get install nginx Than I use whereis "nginx" and remove all files manually (rm) now i wanna re-install nginx but its not work it was return error awk: cannot open /etc/nginx/nginx.conf (No such file or directory) i create /etc/nginx/nginx.conf then use apt-get install nginx its complete install but not work

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