PHP - Moved: Trouble With Phptestmanager
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Their unique code corresponds to an FTP folder, e.g. Quote \FTP_Root\Customer A & Co\ So when the user logs in, they can see their FTP directory and contents, in this case an Inbox and an Outbox. This works as the FTP folder contains both an Inbox and an Outbox. The problem I am having is when I try browsing within this directory (say the Inbox), ftp-chdir fails. I have a hyperlink that sends a 'dir' parameter to the same ftp.php page, and if set, will attempt to change to the given directory. So even though the URL Hyperlink reads: Quote ....server.co.uk/ftp.php?dir=/FTP_Root/Customer A & Co/Inbox When you click on this link, the page tries to load: Quote ....server.co.uk/ftp.php?dir=/FTP_Root/Customer A So my question is what do I need to change, the Hyperlink to read something like: Quote ....server.co.uk/ftp.php?dir=/FTP_Root/Customer%20A%20&%20Co/Inbox Do something to the ftp-chdir function, where I encode/decode/whatever to make sure it tries to change to the correct FTP directory? Or exclude all ampersand entirely? Hello everyone, I am new to this forum and PHP world. Doing my first project, a pretty complicated one to start with. I will be needing your help a lot to accomplish it. Here is the first one. 1. I have a certain field called 'country' 2. I have small flag icons for every country. WHAT DO I WANT TO DO? Example - If the country is U.S.A., the U.S. flag shows up and is a link to www.domain.com/usa If the country is Germany, the German flags shows up and is a link to www.domain.com/germany If the country is not set, no flag shows up. END. How do I execute this? This is what I am doing to get the image <img src="images/flags/<?php echo $row_rsPilots['country']; ?>.gif" alt="" name="Flag" width="20" height="20" id="Flag" /> How do make it a link to www.domain.com/'country' Thanks in advance Building a website for work. I am struggling with the login for some reason. I`m using a lot of the same code as I did for my personal site and a few other websites I`ve programmed which has always worked. But for some reason, it isn`t working now. I`ve already told it to display to me the information that`s being processed and that is all correct (it even updates the database like it`s supposed to). It just won`t show the person being logged in, which defeats the purpose of logging in, yanno? Here are all the files in question. login.php <?php include "file_calls.php"; $title = "Business Name (Beta): Log In"; include "functions.php"; session_start(); echo "$title"; echo "<p>"; echo "Log into the Business Name website. Only authorized members of the Business Name Staff can log into the website."; echo "<p>"; include "login_form.php"; ?> login_form.php <?php echo "<form action='logging.php' method='post'>"; echo "E-Mail Address:"; echo "<br><input type='text' name='email' size=60 maxlength=100>"; echo "<p>"; echo "Password:"; echo "<br><input type='password' name='pass' size=60 maxlength=25>"; echo "<p>"; $buttonlabel = "Log In"; include "formbutton_format.php"; echo "</form>"; ?> logging.php <?php include "file_calls.php"; $title = "Business Name (Beta): Logging In"; include "functions.php"; session_start(); echo "$title"; echo "<p>"; echo "Logging into the Business Name website. Only authorized members of the Business Name Staff can log into the website."; echo "<p>"; $email = $_POST['email']; $pass = $_POST['pass']; $entry_date = strftime("%B\ %e\,\ %Y %I:%M:%S %p", time()); $res = mysql_query("SELECT id, memlev, pwd1, pwd2, email, name FROM user_data WHERE email='$email'"); $by = mysql_fetch_row($res); mysql_free_result($res); $log = $by[4]; $pas = $by[2]; $pas2 = $by[3]; if ($email && $pass) { if ($by[0]) { if ($by[1] == 2) { $passwd = crypt($_REQUEST['pass'],$by[5]); if ($pass == $pas2) { mysql_query("UPDATE user_data SET lastlogin='$entry_date' WHERE email='$email'"); mysql_close($con); header("Location: index.php"); } elseif ($passwd != $pas) { header("Location: nolog.php?logout=1&m=4"); } } elseif ($by[1] == 1) { header("Location: nolog.php?logout=1&m=2"); } elseif ($by[1] == 0) { header("Location: nolog.php?logout=1&m=3"); } } elseif (!$by[0]) { header("Location: nolog.php?logout=1&m=1"); } } elseif (!$email || !$pass) { echo "<b>Error:</b> Both username and password must be entered in order to log in."; echo "<p>"; include "login_form.php"; } ?>[/php index.php [php]<?php include "file_calls.php"; $title = "Business Name (Beta)"; include "functions.php"; session_start(); echo "$title"; echo "<p>"; echo "This website is currently under construction. Thank you for your patience."; echo "<p>"; if ($lev > 1) { echo "Hello, $loggeduser !"; } elseif ($lev < 2) { echo "Not logged in."; } echo "<p>"; echo "$lev"; echo "<br>$loggeduser<br>$email"; ?> auth.php <?php // Defines DEFINE('SESSION_MAGIC','sadhjasklsad2342'); // Initialization @session_start(); @ob_start(); /* Redirects to another page */ function Redirect($to) { @session_write_close(); @ob_end_clean(); @header("Location: $to"); } /* Deletes existing session */ function RemoveSession() { $_SESSION = array(); if (isset($_COOKIE[session_name()])) { @setcookie(session_name(), '', time()+(60*60*24*365), '/'); } } /* Checks if user is logged in */ function isLoggedIn() { return(isset($_SESSION['magic']) && ($_SESSION['magic']==SESSION_MAGIC)); } /* read message count */ function CountMessages($id) { if ($res=mysql_query("SELECT * FROM user_data WHERE email='$email'")) { $count=mysql_num_rows($res); mysql_free_result($res); return($count); } return 0; } /* Go login go! */ function Login($email,$pass) { global $nmsg, $rows; $ok=false; if ($res=mysql_query("SELECT id, email, name, pwd1, pwd2, memlev FROM user_data WHERE email='$email' AND pwd2='$pass'")) { if ($rows=mysql_fetch_row($res)) { $_SESSION['sess_name'] = $rows[2]; $_SESSION['pass'] = $pass; $_SESSION['gal'] = $rows[0]; $_SESSION['level2'] = $rows[5]; $_SESSION['email'] = $rows[1]; $_SESSION['magic'] = SESSION_MAGIC; $nmsg = CountMessages($rows[0]); $ok=true; } else { include('login_failed.php'); } mysql_free_result($res); } return($ok); } /* Terminates an existing session */ function Logout() { @RemoveSession(); @session_destroy(); } /* Escape array using mysql */ function Escape(&$arr) { if (Count($arr)>0) { foreach($arr as $k => $v) { if (is_array($v)) { Escape($arr[$k]); } else { if (function_exists('get_magic_quotes')) { if(!get_magic_quotes_gpc()) { $arr[$k] = stripslashes($v); } } $arr[$k] = mysql_real_escape_string($v); } } } } // ----------------------------------------------- // Main // ----------------------------------------------- Escape($_POST); Escape($_GET); Escape($_COOKIE); Escape($_REQUEST); Escape($_GLOBALS); Escape($_SERVER); ?> file_calls.php <?php include "info_con.php"; include "auth.php"; ?> functions.php <?php echo "<title>$title</title>"; $lev=isset($_SESSION['level2'])?$_SESSION['level2']:0; $logged=isset($_SESSION['gal'])?$_SESSION['gal']:0; $loggeduser=$_SESSION['sess_name']; $nmsg = 0; $rows = isset($_SESSION['rows'])?$_SESSION['rows']:array(); $email = isset($_SESSION['email'])?$_SESSION['email']:''; $pass = isset($_SESSION['pass'])?$_SESSION['pass']:''; function rand_chars($c, $l, $u = FALSE) { if (!$u) for ($s = '', $i = 0, $z = strlen($c)-1; $i < $l; $x = rand(0,$z), $s .= $c{$x}, $i++); else for ($i = 0, $z = strlen($c)-1, $s = $c{rand(0,$z)}, $i = 1; $i != $l; $x = rand(0,$z), $s .= $c{$x}, $s = ($s{$i} == $s{$i-1} ? substr($s,0,-1) : $s), $i=strlen($s)); return $s; } function ShowLoggedInBar() { global $email,$pass,$rows,$logid; $nmes=""; if($nmsg){ $nmes="($nmsg New)"; } echo "Hello, $loggeduser !"; } /* check if we are logging out */ if (isset($_REQUEST['logout'])) { Logout(); } /* check if already logged in */ if (isset($_SESSION['magic']) && ($_SESSION['magic']==SESSION_MAGIC)) { ShowLoggedInBar(); } else { /* not logged in, is it a form post? */ if (isset($_REQUEST['email']) && isset($_REQUEST['pass'])) { $email = $_REQUEST['email']; $pass = crypt($_REQUEST['pass'],$email); Login($email,$pass); } else { } } ?> Can anyone see why it works on everything but getting the person logged in? Hi all, the last 15 minutes i wasted my time pulling my hair while looking at my php code. Of course I used mysqli_error() & mysqli_errno() to find out what was happening. I got something like this: Quote warning: mysqli_error() expects exactly 1 parameter, 0 given in /wicked/fatmonkeyseatbananas/zoo/index.php on line 12 That didnt really help me. I also echoed out my query. Until I thought let's double check the field names I have in the database. They were also correct. And that's when I found out that it was in fact the property of a my ID field. It was set as primary key, but not set to auto increment. Apparently each time a new row was inserted there was a conflict since the next row also had an id of 0. After I add auto increment it was all fixed. So if anyone ever has this problem, hope this helps now it's time for a beer btw. if anyone has a faster way of solving problems like this I love to hear it. I am trying to put the tree of an XML file into an array. sample.xml Code: [Select] <?xml version='1.0' encoding='UTF-8'?> <getHistogramsResponse xmlns="http://www.ebay.com/marketplace/search/v1/services"> <ack>Success</ack> <version>1.7.0</version> <timestamp>2010-09-01T12:13:31.480Z</timestamp> <aspectHistogramContainer> <domainName>UK_CamerasPhoto_DigitalCameras_DigitalCameras_JN</domainName> <domainDisplayName>Digital Cameras</domainDisplayName> <aspect name="Brand"> <valueHistogram valueName="Canon"> <count>4323</count> </valueHistogram> <valueHistogram valueName="Sony"> <count>2210</count> </valueHistogram> </aspect> </aspectHistogramContainer> </getHistogramsResponse> $this->cXmlToArray->m_arr = GetXMLTree('sample.xml') if (isset($this->cXmlToArray->m_arr['getHistogramsResponse']['aspectHistogramContainer'][0])) { $this->m_arrResult = $this->cXmlToArray->m_arr['getHistogramsResponse']['aspectHistogramContainer']; } else if (count($this->cXmlToArray->m_arr['getHistogramsResponse']['aspectHistogramContainer']) > 0) { $this->m_arrResult[] = $this->cXmlToArray->m_arr['getHistogramsResponse']['aspectHistogramContainer']; } However my output is showing as: Array ( [0] => Array ( [domainName] => UK_CamerasPhoto_DigitalCameras_DigitalCameras_JN [domainDisplayName] => Digital Cameras [aspect] => Array ( [0] => Array ( [valueHistogram] => Array ( [0] => Array ( [count] => 4311 ) [1] => Array ( [count] => 2209 ) ) ) ) So how do I get it to return the attributes name and valueName? Any help will be greatly appreciated. I am working on a debug script to see if users had setup my script correctly. The very last part I have is a little debug section where it checks for some common errors. I am having problems with one last part, which is checking to see if they correctly setup the chmod permissions to 0777. This is my coding below: // Directory Permissions $path = $_SERVER['DOCUMENT_ROOT']; $customavatar = $path."/customavatars"; $smilies = $path."/images/smilies"; $store = $path."/images/pointmarket/store"; $text4 = $vbphrase['market_maintenance_directory_good']; $icon4 = "$site_url/checkmark.png"; if (!is_writable(dirname($customavatar))) { $badfolder = $customavatar; $text4 = $vbphrase['market_maintenance_directory_bad']; $icon4 = "$site_url/warning.gif"; } if (!is_writable(dirname($smilies))) { $badfolder = $smilies; $text4 = $vbphrase['market_maintenance_directory_bad']; $icon4 = "$site_url/warning.gif"; } if (!is_writable(dirname($store))) { $badfolder = $store; $text4 = $vbphrase['market_maintenance_directory_bad']; $icon4 = "$site_url/warning.gif"; } Now I've physically tested it to see if I have all directories chmodded to 0777 and they all are. However, whenever I output each time I am being shown that the directories are showing me that they aren't writeable because $badfolder, $text4, and $icon4 are all being reassigned within the if statement. I've tried each one of the three above to see and all of them are going within the if statement even though they shouldn't be. Can anyone tell me what I might be doing wrong? could someone help me with this. here is what I have: <?php require_once('dbinfo.php'); ?> <?php mysql_connect($dbaddress,$username,$password); mysql_select_db($db) or die("Cannot find database!"); /* $query = "SELECT * FROM contacts"; $result = mysql_query($query); $numRows = mysql_numrows($result); $i = 0; while ($i < $numRows) { echo mysql_result($result, $i, "first") . " " . mysql_result($result, $i, "last"); $i++; } */ $fname = 'WrdScramble()'; $ftype = 'DAO'; $fdesc = 'Randomly rearranges the characters in a string or word.'; $fcode = ''; $query = "INSERT INTO functions VALUES ('','$fname','$fcode','$ftype','$fdesc')"; mysql_query($query); print($query); //$query = "SELECT fcode FROM contacts WHERE first = 'code'"; //$result = mysql_query($query); //$numRows = mysql_numrows($result); //$fcode = mysql_result($result, 0, "fcode"); ?> <? //close connection mysql_close(); ?> It is not inserting the data into the database, and if I use: mysql_error() and print it out, I get nothing with that either. I am not sure where to go! I also have used this: (fid, fname, fcode, ftype, fdesc) preceeding the 'VALUES' word in the statement just to see if it would take. No luck! Do I not see the obvious? Any help appreciated! thanks! (I have checked my field types and max length in PHPmyadmin and they are not an issue.) Trying to get my output to to show 2 decimal places but keep getting syntax errors please help. Here's the code i have. Code: [Select] <?php // Input Data From Form $Cost = $_POST['fielda']; //$Cost2 = number_format(,2); //Set Current Date & Shipping Date (15 days) $Date = date('l, F d, Y'); $ShipDate = date('F d, Y',strtotime('+ 15 days')); $Tax = ($Cost * .06); $Ship = 0; //$Sub = ($Cost + $Tax); //Calculate Shipping if ($Cost <= 25.99 ){ $Ship = 3.00; } if ($Cost >= 26.00 && $Cost <= 50.99 ){ $Ship = 4.00;} if ($Cost >= 51.00 && $Cost <= 75.00) { $Ship = 5.00;} if ($Cost > 75.00) { $Ship = 6.00;} //Calculate Order Total $Total = ($Cost + $Tax + $Ship); print " Date: $Date<br><br>"; print " Cost : $$Cost<br><br> Tax: $$Tax<br><br> Shipping: $$Ship<br><br> Total: $$Total<br><br>"; print "Estimated Ship Date is $ShipDate<br> " ; ?> MOD EDIT: code tags added. I am using PHP Version 5.3.1 and MYSQL 5.1.41 and I have my select statement here however i was not sure with the php that im using to filter the results if I should have or could have posted it here or in the PHP section, as part of my question pertains to it as well...if i am posting in the wrong place please let me know and i will move it. ok this is killin me its not working and when it works it works backwards...whats happening is when I use $DArea anywhere in the Query it doesnt work, If I choose a different Area than one in the database It executes the Else Statement(not supposed to do that) and it post data from the same Month, Year and Area to the database...once again what im trying to do is not allow any information from an Area "$DArea" in the database unless the Month or the Year is different. In other words If I fill out information from area "Gym" on November("DMonth") of 2010("DYear") the next time i will be allowed to enter data on it is next month. And if you were wondering Dmonth and DYear are variables tied to the time() function abd it does pass the correct month and time to the database. $DArea = $areaReported; $query = mysql_query("SELECT * FROM $table_name WHERE Month ='$DMonth' AND Year= '$DYear' AND Area = '$DArea'") or die(mysql_error()); $numrows = mysql_num_rows($query); if ($numrows!=0) { while ($row = mysql_fetch_assoc($query)) { $dbMonth = $row['Month']; $dbYear = $row['Year']; $dbArea = $row['Area']; } if ($DMonth==$dbMonth&&$DYear==$dbYear&&$DArea==$dbArea) { mysql_query("INSERT INTO $table_name (user_id, Date, Month, Year, Area, percent1, stat1) values ('{$user_id}', '{$realdate}', '{$DMonth}', '{$DYear}', '{$DArea}', '{$percent1}', '{$stat1}')"); // defining the output include("output.inc"); if ($total == "1"){ $output = $output1; } else if ($total == "2"){ $output = $output2; } else if ($total == "3"){ $output = $output3; } else if ($total == "4"){ $output = $output4; } else if ($total == "5"){ $output = $output5; } else if ($total == "6"){ $output = $output6; } else if ($total == "7"){ $output = $output7; } else if ($total == "8"){ $output = $output8; } else if ($total == "9"){ $output = $output9; } else if ($total == "10"){ $output = $output10; } } else echo "<p style=\"color:red;\">A Report for this $areaReported has already been submitted for this Month</p>"; } I have also tried the statement this way. ("SELECT 'Month', 'Year', 'Area' FROM 'data' WHERE 'Month' ='$DMonth' AND 'Year'='$DYear' AND 'Area' = '$DArea'") If it stops any info from hitting the database it stops it all and when the error message I want to appear when someone tries to enter a duplicate record from a specific area for the same month, it does not show unless its a month that does not exist in the database and to get that to even show up I have to take the Area Clause out of the Select statement. I am out of ideas on this one. Can anyone tell me if im even going in the right direction here? Please help Hello everyone, I am having major issues trying to figure this one out. So here is the code.... $useranswer= ( $_POST["Text1"]); $query = "SELECT Answer FROM images_guess WHERE id=16"; $result = mysql_query($query); while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { echo "Answer :{$row["Answer"]} <br>"; echo "User Answer:$useranswer <br>"; } $result2 = $result. " "; $useranswer2 = $useranswer . " "; if ($result2==$useranswer2) echo "right"; else echo "wrong"; ?> Basically, I am taking whatever was put into "Text1" and finding out if it matches what was in the Answer row in my table. The two variables I used, $result and $useranswer are both working because they do print out on the next page. The problem is the If/else statement is not working with them. They are both text if that helps. Any advice would help a ton, i'm at a loss. Thanks! Hi: I'm having trouble getting my data to display using SELECT. Maybe someone can point out what I'm missing. I want to display the City and Zip for all 50 states, depending on which abbr_state name is clicked. This my code: Code: [Select] <a href="z.php?abbr_state=NY">New York</a> <br /> <a href="z.php?abbr_state=PA">Pennsylvania</a> <?php include('include/myConn.php'); $result = mysql_query("SELECT city,abbr_state,zip FROM `zip_codes` WHERE `abbr_state` = " . mysql_real_escape_string ( $_GET['abbr_state'] );") while($row = mysql_fetch_array($result)){ echo $row['city']. " - ". $row['zip']; echo "<br />"; } ?> Getting this error: Code: [Select] Parse error: syntax error, unexpected ';' in /var/www/domains/sppcon.com/docs/z.php on line 29 What am I missing? The only ';' I see is at the end but removing it doesn't fix the error. Also - now that I think of it - how would I list the Cities in alphabetical order, and perhaps do pagination to I only show a certain amount each page? There are a lot of cities and I think displaying them all on one page would be a bit much. Thanks! |