yogibear

Hi I have been trying to backup one of my tables using this code and it works however when i tried restoring the table using phpmyadmin i got an error on line 1. i believe the problem is lines terminated is ; and so is fields terminated when i export using phpmyadmin i can set what i what them to be how can i change lines terminated in the code to be : rather than ; without changing field terminated. $out = ''; // Get all fields names in table "name_list" in database "tutorial". $fields = mysql_list_fields($database,$table); // Count the table fields and put the value into $columns. $columns = mysql_num_fields($fields); // Add all values in the table to $out. while ($l = mysql_fetch_array($result)) { for ($i = 0; $i < $columns; $i++) { $out .='"'.$l["$i"].'";'; } $out .="\n"; } this is the export uing phpmyadmin "test";"test";"test";"test";"test";"test";"test";"2007-10-07";"2007-11-07";"0";"0";"2007-12-09";"0";"testt";"t";"test": this is the export using the code on my website "test";"test";"test";"test";"test";"test";"test";"2007-10-07";"2007-11-07";"0";"0";"2007-12-09";"0";"testt";"t";"test";

Hi thanks for your help works perfectly now many thanks yogi

Hi everyone quick question about adding the date to a file name this is what i have so far $date = date("Y/m/d"); // Open file 'backup.csv'. $f = fopen ('backup.csv','w'); // Put all values from $out to backup.csv. fputs($f, $out); fclose($f); header('Content-type: application/csv'); header('Content-Disposition: attachment; filename="backup.csv"'); readfile('backup.csv'); ?> I have tried a few things but got so many errors I thought I would remove them and ask for help before the whole thing stopped working. Many thanks yogi

Hi thank you so much, its so simple when you know how

Hi all I’m not total new to php but I still don’t understand a lot of php things so please go easy. I have 3 fields from a database going into a table. username, car and credits and a rank column the table is ordered by the total of 3 different fields in the database. I want the rank column to start at 1 for row 1, and 2 for row 2 and so on. I have tried some things that i found in another post but couldnt get them to work so I removed it. Rank Username Car Credits 1 qwert ford 100 2 test renault 50 this is the code i have <?php $con = mysql_connect("localhost","***","***"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("***", $con); $result = mysql_query("SELECT * FROM userinformation ORDER by credits + original + totalspent DESC"); echo "<table border='1'> <tr> <th>Rank</th> <th>Username</th> <th>Car</th> <th>Credits</th> </tr>";while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>". $row['rank'] . "</td>"; echo "<td>" . $row['username'] . "</td>"; echo "<td>" . $row['car'] . "</td>"; echo "<td>" . $row['credits'] . "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> I hope someone can help thanks in advanced yogi

Hi everyone I wasn’t sure were to put this post so please move if necessary. I have just finished the first half of my site when I thought if a user could change cookies or the contents cookies this would be a huge hack the cookies don’t contain any passwords or anything like that but a user being able to change cookies could give this user and advantage over other users. I know that cookies are client side but is the client able to change cookies some how and would using sessions solve the problem if there is one. Many thanks Yogi

works great thanks for your help

Hi now im getting an error Parse error: syntax error, unexpected $end in /home/sites/bf2c.co.uk/public_html/orderexhaust.php on line 27 This is all the code <? // You may copy this PHP section to the top of file which needs to access after login. session_start(); // Use session variable on this page. This function must put on the top of page. if(!session_is_registered("username")){ // if session variable "username" does not exist. header("location:index.php"); // Re-direct to index.php } $con = mysql_connect('localhost','***','***'); if (!$con) { die('Could not connect: ' . mysql_error()); }mysql_select_db("***", $con); $result = mysql_query("SELECT * FROM userinformation WHERE username='yogibear'"); while($row = mysql_fetch_array($result)) $a = $row['credits']; $b = 5000; $c = $a + $b; mysql_query("UPDATE userinformation SET credits = $c WHERE username = '". $_SESSION['username']."'); mysql_close($con); ?> Thanks for your help

Hi thanks for your fast reply, I thought everything was working when I got no error but then I noticed that it has stopped the calculation from working. This is all my work in progress code <? // You may copy this PHP section to the top of file which needs to access after login. session_start(); // Use session variable on this page. This function must put on the top of page. if(!session_is_registered("username")){ // if session variable "username" does not exist. header("location:index.php"); // Re-direct to index.php } $con = mysql_connect('localhost','***','***'); if (!$con) { die('Could not connect: ' . mysql_error()); }mysql_select_db("***", $con); $result = mysql_query("SELECT * FROM userinformation WHERE username='yogibear'"); while($row = mysql_fetch_array($result)) $a = $row['credits']; $b = 5000; $c = $a + $b; mysql_query("UPDATE userinformation SET credits = $c WHERE username = ". $_SESSION['username']"); mysql_close($con); ?> [code] The code is meant to take credits and add 5000 to it and then replace the original amount of credits with the total, but it doesn’t, it does when ". $_SESSION['username']"); is replaced with 'yogibear'"); Any ideas Thanks [/code]

Hi all I am quite new to php and am having a little problem with sessions. when a user logs in there username becomes a session and every page checks this session this all works fine, which is great but I would like it to search the database for this username this is part of the code that i have. this is the error I am getting sorry if i have not been clear, I hope you can help, thanks in advanced yogi

I was hoping it would be something different, cause im getting the same problem

do i just replace addslashes with mysql_real_escape_string

Hi i cant get your code to work it keeps saying "Unknown column 'test' in 'where clause'" the test is what i typed into the username textbox this is my code <?php if(isset($_POST['submit'])){ $dbHost = "localhost"; //Location Of Database usually its localhost $dbUser = "***"; //Database User Name $dbPass = "***"; //Database Password $dbDatabase = "***"; //Database Name $db = mysql_connect("$dbHost","$dbUser","$dbPass")or die("Error connecting to database."); //Connect to the databasse mysql_select_db("$dbDatabase", $db)or die("Couldn't select the database."); //Selects the database /* The Above code can be in a different file, then you can place include'filename.php'; instead. */ //Lets search the databse for the user name and password $sql = mysql_query("SELECT * FROM userinformation WHERE username=".addslashes($_POST['username'])." AND password=".addslashes($_POST['password'])." LIMIT 1")or die(mysql_error()); //Search for a row $row = mysql_fetch_array($sql); if($row){ //If there is a row start a session with values, then transfer to the users page session_start(); $_SESSION['username'] = $row['username']; $_SESSION['fname'] = $row['first_name']; $_SESSION['lname'] = $row['last_name']; $_SESSION['logged'] = 1; header("Location: users_page.php"); }else{ //If there isn't a row return the user to a login page header("Location: login_page.php"); } }else{ //If the form button wasn't submitted go to the index page header("Location: index.php"); } ?> any ideas whats wrong many thanks yogi

Hi still getting the same <?php // Connects to your Database mysql_connect("localhost", "***", "***") or die(mysql_error()); mysql_select_db("***") or die(mysql_error()); //This code runs if the form has been submitted if (isset($_POST['submit'])) { //This makes sure they did not leave any fields blank if (!$_POST['username'] | !$_POST['pass'] | !$_POST['pass2'] ) { die('You did not complete all of the required fields'); } // checks if the username is in use if (!get_magic_quotes_gpc()) { $_POST['username'] = addslashes($_POST['username']); } $usercheck = $_POST['username']; $check = mysql_query("SELECT username FROM userinformation WHERE username = '$usercheck'") or die(mysql_error()); $check2 = mysql_num_rows($check); //if the name exists it gives an error if ($check2 != 0) { die('Sorry, the username '.$_POST['username'].' is already in use.'); } // this makes sure both passwords entered match if ($_POST['pass'] != $_POST['pass2']) { die('Your passwords did not match. '); } // here we encrypt the password and add slashes if needed $_POST['pass'] = md5($_POST['pass']); if (!get_magic_quotes_gpc()) { $_POST['pass'] = addslashes($_POST['pass']); $_POST['username'] = addslashes($_POST['username']); } // now we insert it into the database $insert="INSERT INTO userinformation (username, password) VALUES ('$_POST[username]','$_POST[pass]')"; $add_member = mysql_query($insert); ?> <!-- Now we let them know if their registration was successful --> <h1>Registered</h1> <p>Thank you, you have registered - you may now login</a>.</p> <?php } else { ?> <!-- This is what they see before they have registered --> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <table border="0"> <tr><td>Username:</td><td> <input type="text" name="username" maxlength="60"> </td></tr> <tr><td>Password:</td><td> <input type="password" name="pass" maxlength="10"> </td></tr> <tr><td>Confirm Password:</td><td> <input type="password" name="pass2" maxlength="10"> </td></tr> <tr><th colspan=2><input type="submit" name="submit" value="Register"></th></tr> </table> </form> <?php } ?> This all seems to work fine but i am still get the Incorrect password, please try again. when i try to login thanks for your help yogi

well that sounds very complicated I think this login may be more than I need. Can any one recommend more simple version. many thanks

Hi all I found this login and thought it would be perfect for my site http://php.about.com/od/finishedphp1/ss/php_login_code_5.htm however its a little more advanced than i am used to and its not working right it keeps saying Incorrect password, please try again. when the password is correct all the validation works and it checks the username fine. <?php $host="localhost"; // Host name $username="***"; // Mysql username $password="***"; // Mysql password $db_name="***"; // Database name $tbl_name="userinformation"; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); //Checks if there is a login cookie if(isset($_COOKIE['ID_my_site'])) //if there is, it logs you in and directes you to the members page { $username = $_COOKIE['ID_my_site']; $pass = $_COOKIE['Key_my_site']; $check = mysql_query("SELECT * FROM userinformation WHERE username = '$username'")or die(mysql_error()); while($info = mysql_fetch_array( $check )) { if ($pass != $info['password']) { } else { header("Location: members.php"); } } } //if the login form is submitted if (isset($_POST['submit'])) { // if form has been submitted // makes sure they filled it in if(!$_POST['username'] | !$_POST['pass']) { die('You did not fill in a required field.'); } // checks it against the database if (!get_magic_quotes_gpc()) { $_POST['email'] = addslashes($_POST['email']); } $check = mysql_query("SELECT * FROM userinformation WHERE username = '".$_POST['username']."'")or die(mysql_error()); //Gives error if user dosen't exist $check2 = mysql_num_rows($check); if ($check2 == 0) { die('That user does not exist in our database. <a href=add.php>Click Here to Register</a>'); } while($info = mysql_fetch_array( $check )) { $_POST['pass'] = stripslashes($_POST['pass']); $info['password'] = stripslashes($info['password']); $_POST['pass'] = md5($_POST['pass']); //gives error if the password is wrong if ($_POST['pass'] != $info['password']) { die('Incorrect password, please try again.'); } else { // if login is ok then we add a cookie $_POST['username'] = stripslashes($_POST['username']); $hour = time() + 3600; setcookie(ID_my_site, $_POST['username'], $hour); setcookie(Key_my_site, $_POST['pass'], $hour); //then redirect them to the members area header("Location: members.php"); } } } else { // if they are not logged in ?> <form action="<?php echo $_SERVER['PHP_SELF']?>" method="post"> <table border="0"> <tr><td colspan=2><h1>Login</h1></td></tr> <tr><td>Username:</td><td> <input type="text" name="username" maxlength="40"> </td></tr> <tr><td>Password:</td><td> <input type="password" name="pass" maxlength="50"> </td></tr> <tr><td colspan="2" align="right"> <input type="submit" name="submit" value="Login"> </td></tr> </table> </form> <?php } ?> any ideas what the problem many thanks yogi

[quote author=jesirose link=topic=124160.msg514101#msg514101 date=1169830799] instead of echo, use header('Location: '.$newURL); [/quote] Works perfectly Many thanks

Hi my php is all working and everything is going well thanks to everyone that has helped me. ;D Just to make things look better how would i change an if statement so rather than displaying text it goes to a new page on the site. this is the current if statement [code] if ($num_rows > 0) {     echo "ALREADY IN DATABASE"; } else {     echo "Working!"; } [/code] many thanks yogi

[quote author=jesirose link=topic=123684.msg511617#msg511617 date=1169580331] [quote author=yogibear link=topic=123684.msg511611#msg511611 date=1169579981] [quote author=jesirose link=topic=123684.msg511609#msg511609 date=1169579762] if ($num_rows => 0) { => is syntax for assigning array values. You want >= [/quote] I'm still getting the same error thanks yogi [/quote] Actually, I think you'd just want > - you don't want it to go if it's 0, just if it's > 0 [/quote] IT WORKS THANK YOU ;D simcoweb got it going but it always said it was in the database. jesirose fixed the last problem many thanks everyone

[quote author=jesirose link=topic=123684.msg511609#msg511609 date=1169579762] if ($num_rows => 0) { => is syntax for assigning array values. You want >= [/quote] I'm still getting the same error thanks yogi

Hi this is my code i have no idea about it i have changed removed and added bits and now im lost with it. [code]<?php $con = mysql_connect('localhost','***','***'); mysql_select_db("web22-james"); if (!$con) {   die("Connection failed."); }else{ $username = $_POST['username']; $email = $_POST['email']; $pword1 = $_POST['pword1']; $pword2 = $_POST['pword2']; $query = "SELECT * FROM usersinformation WHERE username='$username' LIMIT 1"; $result = mysql_query($query) or die("Problem with the query:<pre>$query</pre><br>" . mysql_error()); $num_rows = mysql_num_rows($result); if ($num_rows => 0) {     echo "Our database shows this username has already been registered. Please choose another username."; } else {     echo "$results"; }     mysql_close(); ?>[/code] this is the html form [code]<form action="registerconfirm.php" method="POST"> Username: <input type="text" name="username" maxlength="16" /><br> E-mail: <input type="text" name="email"  /><br> Password: <input type="text" name="pword1"><br> Password (again): <input type="text" name="pword2"><br> <input type="submit" value="Register"> </form>[/code] when i click submit it says Parse error: syntax error, unexpected $end in /home/sites/***/public_html/registerconfirm.php on line 25

This is the code [code]<?php $con = mysql_connect('localhost','***','***'); if (!$con) {    die("Connection failed."); }else{ $username = $_POST['username']; $email = $_POST['email']; $pword1 = $_POST['pword1']; $pword2 = $_POST['pword2']; $query = "SELECT * FROM usersinformation WHERE username='$username' LIMIT 1";    // check if username already      $result = mysql_query($query); if(0 != mysql_num_rows($result)) {         echo "Our database shows this username has already been registered. Please choose another username.";     } else {             echo "WORKING!";     }          mysql_close(); } ?> [/code] Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/sites/***/public_html/registerconfirm.php on line 14 Many thanks yogi

[quote author=jesirose link=topic=123684.msg511547#msg511547 date=1169576103] This is why you don't just copy and paste code without understanding it. [b] isn't php, that's a bold tag. Take that stuff out, and in the future POST the error, not just that you got one. [/quote] hi i did post the error it was in my second post and i also tried it with and without the bold tag i didnt just Copy and Paste but that is why people get help because they dont understand thanks for help

hi im getting a error on line 14 now [code] <?php $con = mysql_connect('localhost','***','***'); if (!$con) {   die("Connection failed."); }else{ $username = $_POST['username']; $email = $_POST['email']; $pword1 = $_POST['pword1']; $pword2 = $_POST['pword2']; $query = "SELECT * FROM usersinformation WHERE username='$username' LIMIT 1";    // check if username already      $result = mysql_query($query); if(0 != mysql_num_rows($result)) [b]{[/b]         echo "Our database shows this username has already been registered. Please choose another username.";     } else {             echo "WORKING!";     }        mysql_close(); } ?> [/code]

Hi this is the error Parse error: syntax error, unexpected T_ELSE in /home/sites/***/public_html/registerconfirm.php on line 16 many thanks yogibear