Search Results

Search found 1913 results on 77 pages for 'die in sente'.

Page 48/77 | < Previous Page | 44 45 46 47 48 49 50 51 52 53 54 55  | Next Page >

  • PHP/MYSQL Trouble Selecting by Primary Key

    - by djs22
    Hi all, So I have a primary key column called key. I'm trying to select the row with key = 1 via this code: $query ="SELECT * FROM Bowlers WHERE key = '1'"; $result = mysql_query($query) or die(mysql_error()); For some reason, I'm getting this result: 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 'key = '1'' at line 1 The mysql statement works for using other keys, ie WHERE name = 'djs22'. Any ideas?

    Read the article

  • INSERT INTO error MYSQL/PHP

    - by bat
    I get this error: 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 'order (total, addy, cc) VALUES ('798' , '123 sadf' , '12124123')' at line 1 $total = addslashes(($_SESSION['total'])); $addy = addslashes(($_POST['addy'])); $cc = addslashes(($_POST['cc'])); echo "$total"; echo "$addy"; echo "$cc"; mysql_query("INSERT INTO order (total, addy, cc) VALUES ('$total' , '$addy' , '$cc')") or die(mysql_error()); help plz =]

    Read the article

  • PHP making input from web form case insensitive?

    - by Haskella
    So I have some code here that takes user input from a standard web form: $searchsport = $_POST['sport']; $sportarray = array( "Football" => "Fb01", "Cricket" => "ck32", "Tennis" => "Tn43", ); if(isset($sportarray[$searchsport])){ header("Location: ".$sportarray[$searchsport].".html"); die; } How would I go about modifying this (I think the word is parsing?) to make it case *in*sensitive? For example, I type in "fOoTbAlL" and php will direct me to Fb01.html normally.

    Read the article

  • Why is is giving me an SQL syntax error?

    - by Tibo
    Do you have any idea why i get this: 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 '``, `title` varchar(255) collate latin1_general_ci NOT NULL default ``,' at line 3 The code is like this (the part im having problem with...) $sql = 'CREATE TABLE `forum` ( `postid` bigint(20) NOT NULL auto_increment, `author` varchar(255) collate latin1_general_ci NOT NULL default ``, `title` varchar(255) collate latin1_general_ci NOT NULL default ``, `post` mediumtext collate latin1_general_ci NOT NULL, `showtime` varchar(255) collate latin1_general_ci NOT NULL default ``, `realtime` bigint(20) NOT NULL default `0`, `lastposter` varchar(255) collate latin1_general_ci NOT NULL default ``, `numreplies` bigint(20) NOT NULL default `0`, `parentid` bigint(20) NOT NULL default `0`, `lastrepliedto` bigint(20) NOT NULL default `0`, `author_avatar` varchar(30) collate latin1_general_ci NOT NULL default `default`, `type` varchar(2) collate latin1_general_ci NOT NULL default `1`, `stick` varchar(6) collate latin1_general_ci NOT NULL default `0`, `numtopics` bigint(20) NOT NULL default `0`, `cat` bigint(20) NOT NULL, PRIMARY KEY (`postid`) );'; mysql_query($sql,$con) or die(mysql_error()); Help would be greatly appreciated!

    Read the article

  • How can I start a TCP server in the background during a Perl unit test?

    - by John
    I am trying to write a unit test for a client server application. To test the client, in my unit test, I want to first start my tcp server (which itself is another perl file). I tried to start the TCP server by forking: if (! fork()) { system ("$^X server.pl") == 0 or die "couldn't start server" } So when I call make test after perl Makefile.PL, this test starts and I can see the server starting but after that the unit test just hangs there. So I guess I need to start this server in background and I tried the & at the end to force it to start in background and then test to continue. But, I still couldn't succeed. What am I doing wrong? Thanks.

    Read the article

  • What is wrong with my SQL syntax here?

    - by CT
    New to SQL. I'm looking to create a IT asset database. Here is one of the tables created with php: mysql_query("CREATE TABLE software( id VARCHAR(30), PRIMARY KEY(id), software VARCHAR(30), key VARCHAR(30))") or die(mysql_error()); echo "Software Table Created.</br />"; This is the output from the browser when I run the script: 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 'VARCHAR(30))' at line 5 I am running a standard LAMP stack on Ubuntu Server 10.04. Thank you.

    Read the article

  • we are getting .txt file but not getting proper alignment

    - by pmms
    we are getting the following texfile_screenshot1.JPG when we are exporting data to .txt file we need output which is shown in texfile_screenshot2.JPG following is the code $myFile = "user_password.txt"; $fh = fopen($myFile, 'a') or die("can't open file"); $newline ="\r\n"; fwrite ($fh,$newline); $stringData1 = $_POST['uname1']." "." "." " ; fwrite($fh, $stringData1); $stringData1 =$_POST['password1']." "." "." "; fwrite($fh,$stringData1); $stringData1 = $_POST['email1']." "." "." "; fwrite($fh, $stringData1); fclose($fh);

    Read the article

  • Is it possible to group validation?

    - by lambdabutz
    I am using a lot of my own validation methods to compare the data from one association to the other. I've noticed that I'm constantly checking that my associations aren't nil before trying to call anything on them, but I am also validating their presence, and so I feel that my nil checks are redundant. Here's an example: class House < ActiveRecord::Base has_one :enterance, :class => Door has_one :exit, :class => Door validates_presence_of :enterance, :exit validate :not_a_fire_hazard def not_a_fire_hazard if enterance && exit && enterance.location != exit.location errors.add_to_base('If there is a fire you will most likely die') return false end end end I feel like I am repeating myself by checking the existence of enterance and exit within my own validation. Is there a more "The Rails Way" to do this?

    Read the article

  • PHP Connect to 4D Database

    - by Matt Reid
    Trying to connect to 4D Database. PHPINFO says PDO is installed etc etc... Testing on localhost MAMP system. However when I run my code I get: Fatal error: Uncaught exception 'PDOException' with message 'could not find driver' in /Applications/MAMP/htdocs/4d/index.php:12 Stack trace: #0 /Applications/MAMP/htdocs/4d/index.php(12): PDO->__construct('4D:host=127.0.0...', 'test', 'test') #1 {main} thrown in /Applications/MAMP/htdocs/4d/index.php on line 12 My code is: $dsn = '4D:host=127.0.0.1;charset=UTF-8'; $user = 'test'; $pass = 'test'; // Connection to the 4D SQL server $db = new PDO($dsn, $user, $pass); try { echo "OK"; } catch (PDOException $e) { die("Error 4D : " . $e->getMessage()); } Can't put my finger on the error, i'm using the settings under the PHP tab... Thank you.

    Read the article

  • Nested mysql select statements

    - by Jimmy Kamau
    I have a query as below: $sult = mysql_query("select * from stories where `categ` = 'businessnews' and `stryid`='".mysql_query("SELECT * FROM comments WHERE `comto`='".mysql_query("select * from stories where `categ` ='businessnews'")." ORDER BY COUNT(comto) DESC")."' LIMIT 3") or die(mysql_error()); while($ow=mysql_fetch_array($sult)){ The code above should return the top 3 'stories' with the most comments {count(comto)}. The comments are stored in a different table from the stories. The code above does not return any values and doesn't show any errors. Could someone please help?

    Read the article

  • Using hash to check if page with $_POST values was refreshed

    - by Dieseltime
    When posting a form to the same PHP page, what is the correct method to find if the page was accidentally refreshed instead of submitted again? Here's what I'm using right now: $tmp = implode('',$_POST); $myHash = md5($tmp); if(isset($_SESSION["myHash"]) && $_SESSION["myHash"] == $myHash) { header("Location: index.php"); // page refreshed, send user somewhere else die(); } else { $_SESSION["myHash"] = $myHash; } // continue processing... Is there anything wrong with this solution?

    Read the article

  • M2Crypto: Is PKey a reference to a Public or a Private key?

    - by Andrea Zilio
    In the PKey class documentation of the M2Crypto python package (an OpenSSL wrapper for Python) it is said that PKey is a reference to a Public key. My opinion is instead that it's a reference to a Private Key because the init method of the PKey class calls the evp_pkey_new openssl function that, from this link: http://linux.die.net/man/3/evp_pkey_new , should allocate a new reference to a private key structure! There are two only possible explaination: The M2Crypto documentation is wrong or the link I've reported has wrong informations. Can someone help me to find the truth?

    Read the article

  • Perl to output processed XML file encoded as UTF-8 with UNIX line endings (in Win32 environment)?

    - by Umber Ferrule
    Running ActiveState Perl 5.8.8 on WinXP. As the title suggests, I'd like to output an XML file as UTF-8 with UNIX line endings. I've looked at the PerlDoc for binmode, but am unsure of the exact syntax (if I'm not barking up the wrong tree). The following doesn't do it (forgive my Perl - it's a learning process!): sub SaveFile { my($FileName, $Contents) = @_; my $File = "SAVE"; unless( open($File, ">:utf-8 :unix", $FileName) ) { die("Cannot open $FileName"); } print $File @$Contents; close($File); } Any suggestions? Thanks.

    Read the article

  • An appropriate C API for inspecting attribute values

    - by uk82
    There are two obvious ways in C to provide outside access to internal attribute values (A) provide a generic interface that accepts a list of attributes that changes over time (some added / some die) or (B) a specific interface for each and every attribute. Example A: int x_get_attribute_value(ATT att) { if (a) return a_val; if (b) return b_val; } Example B: A_Enum x_get_a_type_attribute() {} B_Enum x_get_b_type_attribute() {} I recall that Eclipse's API is very much like A (I could be wrong). What I can't do is come up with a compelling argument against either. A is clean - any user will no where to go to find out a property value. It can evolve cleanly without leaving dead interfaces around. B has type checking to a degree - this is C enums! Is there a big hitter argument that pushes the balance away from opinion?

    Read the article

  • draw line with php using coordinates from txt file

    - by netmajor
    I have file A2.txt with coordinate x1,y1,x2,y2 in every line like below : 204 13 225 59 225 59 226 84 226 84 219 111 219 111 244 192 244 192 236 209 236 209 254 223 254 223 276 258 276 258 237 337 in my php file i have that code. This code should take every line and draw line with coordinate from line. But something was wrong cause nothing was draw :/: <?php $plik = fopen("A2.txt", 'r') or die("blad otarcia"); while(!feof($plik)) { $l = fgets($plik,20); $k = explode(' ',$l); imageline ( $mapa , $k[0] , $k[1] , $k[2] , $k[3] , $kolor ); } imagejpeg($mapa); imagedestroy($mapa); fclose($plik) ; ?> If I use imagejpeg and imagedestroy in while its only first line draw. What to do to draw every line ?? Please help :)

    Read the article

  • MySQL parameter resource error

    - by Derek
    Here is my error: Warning: mysql_query() expects parameter 2 to be resource, null given... This refers to line 23 of my code which is: $result = mysql_query($sql, $connection) My entire query code looks like this: $query = "SELECT * from users WHERE userid='".intval( $_SESSION['SESS_USERID'] )."'"; $result = mysql_query($query, $connection) or die ("Couldn't perform query $query <br />".mysql_error()); $row = mysql_fetch_array($result); I don't have a clue what has happpened here. All I wanted to do was to have the value of the users 'fullname' displayed in the header section of my web page. So I am outputting this code immediately after to try and achieve this: echo 'Hello '; echo $row['fullname']; Before this change, I had it working perfectly, where the session variable of fullname was echoed $_SESSION['SESS_NAME']. However, because my user can update their information (including their name), I wanted the name displayed in the header to be updated accordingly, and not displaying the session value.

    Read the article

  • Deleting a database row with mysql - getting an error!?

    - by Nike
    Here's the code: mysql_query("DELETE " . $_GET['id'] . " FROM forum_favorites WHERE thread_id='" . $_GET['id'] . "' AND user='" . $userinfo['username'] . "'") or die(mysql_error()); And the error message: 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 '77 FROM forum_favorites WHERE thread_id='77' AND user='nike1'' at line 1 Anyone knows what's up here? I've been stuck here for hours now and i just can't figure out what the heck's wrong? The database name and the column names are correct. Thanks. -Nike

    Read the article

  • Moving to OOP hopefully!

    - by Luke
    So I am trying to understand OOP more and use it. The following code was written before i started using OOP. //loop through all the users $game = "$_POST[Game]_teams"; $result = mysql_query("SELECT username FROM `users`") or die(mysql_error()); while( $row=mysql_fetch_assoc($result) ) { $u[] = $row['username']; } I have put the query into my database page like following: function selectAllUsers() { $q = "SELECT username FROM ".TBL_USERS.""; mysql_query($q, $this->connection); } I'm a little confused about how the rest could be different? Would it be? Is it possible for anyone to help me without more code or understanding of my structure?

    Read the article

  • a query is inserted from PHPMYAdmin but not from PHP

    - by iyad al aqel
    i'm writing a php code to insert form values in a forum values $dbServer = mysql_connect("localhost" , "root", "") ; if(!$dbServer) die ("Unable to connect"); mysql_select_db("kfumWonder"); $name= $_POST['name'] ; $password= md5($_POST['password']); $email= $_POST['email'] ; $major= $_POST['major'] ; $dateOfBirth=$_POST['dateOfBirth'] ; $webSite = $_POST['website']; $joinDate= date("Y m d") ; $query = "INSERT INTO user (name, password, email, major, dob, website, join_date) Values ('$name', '$password', '$email', '$major', '$dateOfBirth', '$webSite' , '$joinDate')" ; //echo $query ; $result = mysql_query($query) ; if (! $result ) echo " no results " ; this works perfectly fine when i took the printed query and run it in PHPMyAdmin but when i run this code nothing happens , any ideas !?

    Read the article

  • Have threads run indefinitely in a java application

    - by TP
    I am trying to program a game in which I have a Table class and each person sitting at the table is a separate thread. The game involves the people passing tokens around and then stopping when the party chime sounds. how do i program the run() method so that once I start the person threads, they do not die and are alive until the end of the game One solution that I tried was having a while (true) {} loop in the run() method but that increases my CPU utilization to around 60-70 percent. Is there a better method?

    Read the article

  • Jquery Ajaxform with input type=file and multifile

    - by crunchingnumbers
    Im using ajax form with multifile. However ajaxform just seems to quietly die and does not do anything if using multifile. Multifile is just a jquery plugin that builds a list of input type=file so that you can upload multiple files at once, before which you can add/delete your file upload list. I've looked at multifile and made one change to ensure that it generated a unique name as well as id which it didn't do before and i've checked the form elements multifile is building which all appear to be correct. From looking at ajaxform, it seems that it shoulld just generate the iframe as normal and post the multiple input type=file but it does nothing. Has anyone else experienced problems with multiple input type=file uploads

    Read the article

  • How to show / debug PEAR::DB errors in webpage?

    - by Markus Ossi
    I am connecting to MySQL database on my webpage and have this copy-pasted code for errors: if(DB::isError($db)) die($db->getMessage()); I have the connection code in an outside file called connection.inc that I include at the beginning of my page before the DOCTYPE and html tags. For debugging purposes, how can I print the database errors on my webpage? I thought I could do something like this: echo 'Could not connect to database. The error was:' . $db->getMessage(); but this returns: Fatal error: Call to undefined method DB_mysql::getMessage()

    Read the article

  • Does retrieving an object from Doctrine2 cause __construct() of the model class to run?

    - by jiewmeng
    When I retrieve an object say by $em->find('Application\Models\User', 1); or other methods like DQL, findBy*() cause the __construct() of the model class to run? I am having a problem where I set variables there like reference to EntityManager and I find that its not set. I tried putting a die() in __construct() and it doesn't halt the application. Can I say that if I want to set other properties/fields like EntityManager $em I have to do it some other way? Perhaps something like protected function getEm() { if (!isset($this->em)) { $this->em = \Zend_Registry::get('em'); } return $this->em; }

    Read the article

  • Echo mysql results in a loop?

    - by Roy D. Porter
    I am using turn.js to make a book. Every div within the 'deathnote' div becomes a new page. <div id="deathnote"> //starts book <div style="background-image:url(images/coverpage.jpg);"></div> //creates new page <div style="background-image:url(images/paper.jpg);"></div> //creates new page <div style="background-image:url(images/paper.jpg);"></div> //creates new page </div> //ends book What I am doing is trying to get 3 'content' (content being a name and cause of death) divs onto 1 page, and then generate a new page. So here is what i want: <div id="deathnote"> //starts book <div style="background-image:url(images/coverpage.jpg);"></div> //creates new page <div style="background-image:url(images/paper.jpg);"></div> //creates new page <div style="background-image:url(images/paper.jpg);"> //creates new page but leaves it open <div> CONTENT </div> <div> CONTENT </div> <div> CONTENT </div> </div> //ends the page </div> //ends book Seems simple enough, however the content is data from a MySQL DB, so i have to echo it in using PHP. Here is what i have so far <div id="deathnote"> <div style="background-image:url(images/coverpage.jpg);"></div> <div style="background-image:url(images/paper.jpg);"></div> <div style="background-image:url(images/paper.jpg);"></div> <div style="background-image:url(images/paper.jpg);"></div> <div style="background-image:url(images/paper.jpg);"></div> <div style="background-image:url(images/paper.jpg);"></div> <?php $pagecount = 0; $db = new mysqli('localhost', 'username', 'passw', 'DB'); if($db->connect_errno > 0){ die('Unable to connect to database [' . $db->connect_error . ']'); } $sql = <<<SQL SELECT * FROM `TABLE` SQL; if(!$result = $db->query($sql)){ die('There was an error running the query [' . $db->error . ']'); } //IGNORE ALL OF THE GARBAGE ABOVE. IT IS SIMPLE CONNECTING SCRIPT THAT I KNOW WORKS //THE METHOD I AM HAVING TROUBLE WITH IS BELOW $pagecount = 0; while($row = $result->fetch_assoc()){ //GETS THE VALUE (and makes sure it isn't nothing echo '<div style="background-image:url(images/paper.jpg);">'; //THIS OPENS A NEW PAGE while ($pagecount !== 3) { //KEEPS COUNT OF HOW MUCH CONTENT DIVS IS ON THE PAGE while($row = $result->fetch_assoc()){ //START A CONTENT DIV echo '<div class="content"><div class="name">' . $row['victim'] . '</div><div class="cod">' . $row['cod'] . '</div></div>'; //END A CONTENT DIV $pagecount++; //UP THE PAGE COUNT } } $pagecount=0; //PUT IT BACK TO 0 echo '</div>'; //END PAGE } $db->close(); ?> <div style="background-image:url(images/backpage.jpg);"></div> //BACK PAGE </div> At the moment i seem to be causing and infinite loop so the page won't load. The problem resides within the while loops. Any help is greatly appreciated. Thanks in advance guys. :)

    Read the article

  • I want to keep the values on textbox after onchange function

    - by user1908045
    hello i have problem with this code..I want to keep the values of the textboxes when the pageload and didnt write again the values. I have two drop down list. the First one is the country when the country selected then the page load and appear the city to select but afte the pageload the values on textbox is empty. I want to keep the values of textbox when the page load. This is the code <head> <script type="text/javascript"> function Load_id() { var Count = document.getElementById("Count").value; var Count_txt = "?Count=" location = Count_txt + Count } </script> <meta charset="UTF-8"> </head> <body> <div class="main"> <div class="headers"> <table> <tr><td rowspan="2"><img alt="unipi" src="/Images/logo.jpeg" height="75" width="52"></td> <td>University</td></tr> <tr><td>Data</td></tr> </table> </div> <div class="form"> <h3>Personal</h3><br/><br/><br/> <form id="Page1" name="Page1" action="Form1Sub.php" method="Post"> <table style="width:520px;text-align:left;"> <tr><td><label>Number:</label></td> <td><input type="text" required="required" id="AM" name="AM" value=""/></td> </tr> <tr><td><label>Name:</label></td> <td><input type="text" required="required" name="Name"/></td> </tr> <?php $host="localhost"; $username=""; $password=""; $dbName="Database"; $connection = mysql_connect($host, $username, $password) or die("Couldn't Connect to the Server"); $db = mysql_select_db($dbName, $connection) or die("cannot select DataBase"); $Count = $_GET['Count']; echo "<tr><td><label>Country</label></td>\n"; $country = mysql_query("select DISTINCT Country FROM lut_country_city "); echo " <td><select id=\"Count\" name=\"cat\" onChange=\"Load_id(this)\">\n"; echo " `<option>Select Country</option>\n"; while($nt=mysql_fetch_array($country)){ $selected = ($nt["Country"] == $Count)? "SELECTED":""; echo"<option value=\"".$nt['Country']."\"". $selected." >".$nt['Country']."</option>"; } echo " </select></td></tr>\n"; echo"<tr><td><label>City:</label></td>\n"; $q2 = mysql_query("Select id,City,Country FROM lut_country_city WHERE Country = '$Count'"); echo"<td><select name=\"SelectCity\">\n"; while($row = mysql_fetch_array($q2)) { echo"<option value=\"".$row['id']."\">".$row['City']."</option>"; } echo " </select></td></tr>\n"; ?> </table> <p> <button type="submit" id="Next">Next</button> </form> <form id="form1" action="index.php"> <button id="Back" type="submit">Back</button> </form> </p> </div> </div> </body> </html>

    Read the article

< Previous Page | 44 45 46 47 48 49 50 51 52 53 54 55  | Next Page >