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PHP - If Variable Not Working
Hi guys,
I have a problem with the code on below. When I input the value into $name and $email method, the page come into blank page. <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'mydbusername'); define('DB_PASSWORD', 'mydbpassword'); define('DB_DATABASE', 'mydbname'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $name = clean($_GET['name']); $email = clean($_GET['email']); $comments = clean($_GET['comments']); $type = clean($_GET['type']); if($name == '') { $errmsg_arr[] = 'name or member ID missing'; $errflag = true; } if($email == '') { $errmsg_arr[] = 'email address ID missing'; $errflag = true; } else { } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $insert = array(); if(isset($_GET['name'])) { $insert[] = 'name = \'' . clean($_GET['name']) .'\''; // echo "tested"; } if(isset($_GET['email'])) { $insert[] = 'email = \'' . clean($_GET['email']) . '\''; } if(isset($_GET['type'])) { $insert[] = 'type = \'' . clean($_GET['type']) . '\''; } if(isset($_GET['comments'])) { $insert[] = 'comments = \'' . clean($_GET['comments']) . '\''; } if (count($insert)>0) { $names = implode(',',$insert); if(isset($name)) { $name = $_GET['name']; $headers = "From: "-f .$name."@myemail.com"; $to = "myname@mymail.com"; $subject = $type; $message = $comments . '; $header .= "MIME-Version: 1.0\l\n"; $add = "-f". $name ."@myemail.com"; mail($to, $subject, $message, $header, $add); echo "Thank you for sent us your feedback"; } else { if(isset($email)) { $email = $_GET['email']; $headers = "From: "$email"; $to = "myname@mymail.com"; $subject = $type; $message = $comments . '; $header .= "MIME-Version: 1.0\l\n"; $add = "-f". $email ."; mail($to, $subject, $message, $headers, $add); } echo "Thank you for sent us your email"; } } } ?> There must be the problem coming from this: if(isset($name)) { $name = $_GET['name']; $headers = "From: "-f .$name."@myemail.com"; $to = "myname@mymail.com"; $subject = $type; $message = $comments . '; $header .= "MIME-Version: 1.0\l\n"; $add = "-f". $name ."@myemail.com"; mail($to, $subject, $message, $header, $add); echo "Thank you for sent us your feedback"; } else { if(isset($email)) { $email = $_GET['email']; $headers = "From: "$email"; $to = "myname@mymail.com"; $subject = $type; $message = $comments . '; $header .= "MIME-Version: 1.0\l\n"; $add = "-f". $email ."; mail($to, $subject, $message, $headers, $add); } I am not sure where the problem is, so please could you help me?? Similar TutorialsI have a script that adds points together based upon the placing. This is the actual script: Code: [Select] <? $points = 0; if($place === '1st') {$points = $points + 50;} elseif($place === '2nd') {$points = $points + 45;} elseif($place === '3rd') {$points = $points + 40;} elseif($place === '4th') {$points = $points + 35;} elseif($place === '5th') {$points = $points + 30;} elseif($place === '6th') {$points = $points + 25;} elseif($place === '7th') {$points = $points + 20;} elseif($place === '8th') {$points = $points + 10;} elseif($place === '9th') {$points = $points + 10;} elseif($place === '10th') {$points = $points + 10;} elseif($place === 'CH') {$points = $points + 50;} elseif($place === 'RCH') {$points = $points + 40;} elseif($place === 'TT') {$points = $points + 30;} elseif($place === 'T5') {$points = $points + 30;} elseif($place === 'Champion') {$points = $points + 50;} elseif($place === 'Reserve Champion') {$points = $points + 40;} echo "Total HF Points: $points"; ?>What it *should* do (my friend's script works the same way and it works) it starts at points = 0, than if there is a first place, it adds 50, and so forth until it reaches the end. It is included into a file, in this area: Code: [Select] <div class="tabbertab"> <h2>Records</h2> <? $query92 = "SELECT * FROM THISTABLE WHERE VARIABLE='$id' OR VARIABLE = '$name' ORDER BY ABS(VARIABLE), VARIABLE"; $result92 = mysql_query($query92) or die (mysql_error()); echo "<table class='record'> <tr><th>Show</th> <th>Class</th> <th>Place</th></tr> "; while($row92 = mysql_fetch_array($result92)) { $class = $row92['class']; $place = $row92['place']; $entries = $row92['entries']; $race = $row92['show']; $purse = number_format($row92['purse'],2); echo "<tr><td>$race</td> <td>$class</td> <td>$place</td></tr>"; } ?> <tr><td colspan='3'><div align='right'><? include('includes/points.php'); ?></div></td></tr> </table> </div> This is the code that is relevant. When ended here, it echoes the last place that appears in the results (such as a 5th place echoing 30 points). When I move it to be included in the while loop, it shows Total Points: 50 Total Points: 25 Total Points: 10 (depending on the results displayed on that page). What am I doing wrong? Hello everyone, I can get Test 2 to successfully operate the if statement using a variable variable. But when I try the same method using a session variable (Test 1) the if statement is not executed. Please could you tell me why the if statement in Test 1 is not being executed? Code: [Select] <?php # TEST 1 $_SESSION[test_variable] = "abcd"; $session_variable_name = "_SESSION[test_variable]"; if ($$session_variable_name == "abcd") { echo "<br>line 373, abcd<br>"; } # TEST 2 $test_variable = "efgh"; $test_variable_name = "test_variable"; if ($$test_variable_name == "efgh") { echo "<br>line 379, efgh<br>"; } ?> Many thanks, Stu Hi, I have a variable named $siteName stored in the server.php file in the data directory data/server.php Now in the header.php page I want to retrieve the $siteName variable so it can be used in the header information. This is what I have but it isnt working. server.php Code: [Select] <?php $siteName = 'My+Test+Site'; ?><?php $adminEmail = 'pwithers2009@hotmail.co.uk'; ?><?php $sendmailLoc = '/usr/sbin/sendmail'; ?><?php $imgdir = 'http://www.tropicsbay.co.uk/images/'; ?><?php $imgdirbase = '/home/tropicsb/public_html/images/'; ?> header.php Code: [Select] <?php include("data/server.php"); $siteName = $_GET['siteName']; echo ' <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>$siteName Admin Panel</title> <link rel="stylesheet" href="admin.css" type="text/css" /> </head> <div id="pagehead"> <div id="navigation"> <img alt="" src="images/leftNav.gif" height="32" width="4" id="leftNav" /> <img alt="" src="images/rightNav.gif" height="32" width="4" id="rightNav" /> <div id="globalLink"> <a href="admin.php?cmd=manage&username=admin&password=$adminpw" id="gl1" class="glink" onmouseover="ehandler(event,menuitem1);">Users</a><a href="admin.php?cmd=editTemplate&username=admin&password=$adminpw" id="gl2" class="glink" onmouseover="ehandler(event,menuitem2);">Templates</a><a href="admin.php?cmd=mysqlBackup&username=admin&password=$adminpw" id="gl3" class="glink" onmouseover="ehandler(event,menuitem3);">Database</a><a href="admin.php?cmd=paymentLog&username=admin&password=$adminpw" id="gl4" class="glink" onmouseover="ehandler(event,menuitem4);">Payment</a><a href="admin.php?cmd=profileFields&username=admin&password=$adminpw" id="gl5" class="glink" onmouseover="ehandler(event,menuitem5);">Setup</a><a href="admin.php?cmd=changeAdminpw&username=admin&password=$adminpw" id="gl6" class="glink" onmouseover="ehandler(event,menuitem6);">Preferences</a><a href="admin.php?cmd=logout&username=admin&password=$adminpw" id="gl7" class="glink" onmouseover="ehandler(event,menuitem6);">Logout</a></div> </div></div> <div id="pagelayout"> <img alt="" src="images/leftCurve.gif" height="6" width="6" id="left" /> <img alt="" src="images/rightCurve.gif" height="6" width="6" id="right" /> <div id="pageName"> <h2>$siteName Admin Panel<h2> </div>'; ?> Any ideas on how to do this, it seems to work ok in one of my other scripts. Thanks i've been programming in PHP for years, and have done a substantial amount of work on applications of this nature. this problem has me stumped, not because i can't fix it (i did), but because i have no idea what the problem is. there are hundreds of lines of code involved here, so i'll break it down into a post-friendly format. take this for example, and forgive any typos. it's late, and i've been beating my head against this for over two hours... =\ this is from my form: Code: [Select] /* ... numerous form fields being passed as $_REQUEST arrays */ <input type="hidden" name="option_id[]" value="<?php print $query_result->option_id; ?>" /> /* a couple hundred more lines */ here's the DB update handler: Code: [Select] if (!empty($_REQUEST['option_name'])) { foreach ($_REQUEST['option_name'] as $k => $v) { if ($v != '') { $option_id = $_REQUEST['option_id'][$k]; $option_name = $_REQUEST['option_name'][$k]; $option_price = $_REQUEST['option_price'][$k]; $option_desc = htmlentities($_REQUEST['option_desc'][$k], ENT_QUOTES); if (!$option_id = '') { $sql_options = "UPDATE table SET" . " option_name = '" . $option_name . "', option_price = '" . $option_price . "', option_desc = '" . $option_desc . "' WHERE option_id = '" . $option_id "'"; if (!$query_function($sql_options)) { $error = true; } } else { $sql_options = "INSERT INTO table (option_name, option_price, option_desc)" . " VALUES ('" . $option_name . "', '" . $option_price . "', '" . $option_desc . "')"; if (!$query_function($sql_options)) { $error = true; } } } } } the above code doesn't post to the database because the $option_id variable returns a null value. however, if i replace the $option_id variable where i build the query string with $_REQUEST['option_id'], it works just fine. Code: [Select] /* in relevant part */ $sql_options = "UPDATE table SET" . " option_name = '" . $option_name . "', option_price = '" . $option_price . "', option_desc = '" . $option_desc . "' WHERE option_id = '" . $_REQUEST['option_id'] . "'"; needless to say i was infuriated by having spent a couple of hours to come to this conclusion. i only used the variables in the first place because i need to expand the function that this lives inside and i don't want to have to type $_REQUESTs over and over. the only thing i can think is that it might be a type issue. the data is coming out of the mysql table from an INT field and being placed into the value for the hidden field straight from the row collection. would forcing a variant data type by not strongly typing my variable have caused this problem? i haven't tested the theory because i'm still too ticked off to open my code editor. i'm bouncing this off the community and posting my experience in the hope that it might help someone who comes after. I've been baffled by this for 2 days now and cannot figure it out after exhaustive searches. I'd like to think I'm doing this correctly, but cannot get it to work.
I'm trying to use a variable within a query WHERE statement, and it shows 0 results. If I directly hardcode the text instead of using the variable, it works. The variable is pulling from a $_GET, and if I echo that variable, it is showing the correct text.
Here's my code:
$Domain = $_GET['Domain']; $result = mysql_query(SELECT Code, Title, Domain, Status FROM tablename WHERE Domain="$Domain" ORDER BY Code');If I swap out "$Domain" for direct text, like "ABC", it works. I have tried swapping out the quotes and single quotes throughout the statement, removing the quotes around $Domain, concatenating the statement separately....all yield erros or the same result. And as stated, if I echo $Domain, it shows "ABC" (or whatever it's supposed to show), so i know it's pulling correctly from the $_GET. Anyone know what I'm doing wrong? Hello Everyone, I am new to forum and could use some help with some php code that isn't working. I am very new to php/html/javascript and all of what I have learned, I learned from forums like this one so first....thank you! I am trying to assign a value from a php variable to the value of my form element. I'm sure there must be a way to do this but I can't seem to get the syntax right. here is my code... first I set the value of $loginname elsewhere in the script like so... <?php $loginname =strtolower(htmlspecialchars(strip_tags($_GET["loginname"]))); ?> This part works fine.. Then I try to set the value of my hidden text field inside the form to the value of $loginname to be passed to a javascript program. Everything works except that the value passed ends up being <?echo and not the expected user name inside of $loginname. <?php echo '<form name ="currentactivity" Id="currentactivity" action="<?php'.htmlspecialchars($_SERVER['PHP_SELF']).'?>" method="post">'; echo '<fieldset><legend><b>Your Current Activity Information</b></legend>'; echo '<input type="text" name="loginnm" style="visibility: hidden" value="<?php echo $loginname;?>">'; echo "<label for='myactivities'>Activity Name:</label>"; echo "<select name='myactivities' Id='myactivities' onchange=\"showdetails(this.form)\" value=''>"; echo "<option value = 'Select an activity'>Select an activity</option>"; for ($i = 1; $i <=$rowcount; $i++) { echo"<option value=$row[activity_name]>$row[activity_name] </option>"; $row = mysql_fetch_array($result); } echo "</select>"; echo '</fieldset>'; echo '</form>'; ?> Please note..the rest of the code is working perfectly, it is just this one value I can't seem to get. Any help you can give will be greatly appreciated. Hi Guys, I'm a PHP newbie and I'm having some trouble creating a function that I can call where I can perform a MYSQL SELECT querey which returns an associative array, and be able to pull a value from that array and return it to the call. My code is below. Everything seems to work up until where I try to set the $average variable. If I return $row instead of $round_val, I see me array as "Array ( [user_id] => 64 [AVG(twos_made)] => 5.0000 )" It seems as though my my functions arugment (twos_made - which is my $column_name var) is not getting passed through or something. Any help would be greatly appreciated! Thanks! Code: [Select] function avg_of($column_name) { global $connect_db; $query = "SELECT user_id, AVG($column_name) FROM table GROUP BY user_id"; $result = mysql_query($query, $connect_db); $row = mysql_fetch_assoc($result); $average = $row['AVG($column_name)']; $round_val = round($average, 1); return $round_val; } so my situation is something like this , i'm trying to fetch user details based on `id` that isset is getting, but some how the `variable that contains the $_GET value doesn't work` in query but when i put an static value to pdo query then it works and show the result. i have checked by doing `var_dump` of variable `$user` before query and it shows the correct value but not working in query. Below is the code i'm working with: hi guys. i am trying to set a variable to the session global variable. I've initialized the session with session_start(); at the very beginning of my website, and then i try finding if a session variable has been set like so if(isset($_SESSION['user'])) { print 'user is logged in'; } but this works in reverse. i have not set any session variables, but asking if it is set, results in an affirmative answer. so i am thinking this is because of register_globals. because i went to read about register_globals, and it says that if register_globals is turned off, i cannot use any other varibale except members of the session array. i dont know what that is yet, but a question befo in the php manual it says register_globals is deprecated, and it appears i need to enable it to set other variables as session variables besides it's associative array. if this is true, how to enable register_globals and allow other variables to be used as session variables? I am trying to create a script that will choose from 3 different Variables. This is much like a Chance game. Variable A has a 20% chance to be chosen Variable A has a 35% chance to be chosen Variable A has a 45% chance to be chosen How would I go about setting each of these variables chance to be chosen? This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=321230.0 This works: Code: [Select] form action="http://localhost/mysite/css/screen.css" method="post"> This does not work: Code: [Select] <?php $link_1 = "http://localhost/mysite/css/screen.css" ; $link_2 = '"' . "http://localhost/mysite/css/screen.css" .'"' ; ?> <form action=<?php echo $link_1;?> method="post"> <form action=<?php echo $link_2;?> method="post"> Why do neither of the PHP variables work? Thanks! Hi, I am trying to make some adjustments to uploadify.php which comes with the latest version of uploadify (3.0 beta), so that it works with a session variable that stores the login username and adds it to the path for uploads. Here is uploadify.php as it currently looks: Code: [Select] <?php session_name("MyLogin"); session_start(); $targetFolder = '/songs/' . $_SESSION['name']; // Relative to the root if (!empty($_FILES)) { $tempFile = $_FILES['Filedata']['tmp_name']; $targetPath = $_SERVER['DOCUMENT_ROOT'] . $targetFolder; $targetFile = rtrim($targetPath,'/') .'/'. $_FILES['Filedata']['name']; // Validate the file type $fileTypes = array('m4a','mp3','flac','ogg'); // File extensions $fileParts = pathinfo($_FILES['Filedata']['name']); if (in_array($fileParts['extension'],$fileTypes)) { move_uploaded_file($tempFile,$targetFile); echo '1'; } else { echo 'Invalid file type.'; } } echo $targetFolder; ?> I added Code: [Select] echo $targetFolder; at the bottom so that I could make sure that the string returned was correct, and it is, i.e. '/songs/nick'. For some reason though, uploads are not going to the correct folder, i.e. the username folder, but instead are going to the parent folder 'songs'. The folder for username exists, with correct permissions, and when I manually enter Code: [Select] $targetFolder = '/songs/nick';all works fine. Which strikes me as rather strange. I have limited experience of using php, but wonder how if the correct string is returned by the session variable, the upload works differently than with the manually entered string. Any help would be much appreciated. It's the last issue with a website that was due to go live 2 days ago! Thanks, Nick Hello all. I am very new to PHP, and I am not sure where to look or what I'm looking for in my current assignment. My task is to take in two numbers between 0-100. Once I take in that number, it should state beside it "The __ was accepted." The program should not accept any numbers greater than 100 or any characters. Once I do this, I must take a second number and do a similar thing. Finally, I must have a statement show up at the bottom stating which number is greater. Essentially, I need help in determining what I should use to place parameters, and how I can keep the program from echo ing any statement until input has been taken and tested for parameters. Any help you can provide will be greatly appreciated! I am trying to allow the user to update a variable he chooses by radio buttons, which they will then input text into a box, and submit, to change some attributes. I really need some help here. It works just fine until I add the second layer of variables on top of it, and I can't find the answer to this question anywhere. <?PHP require('connect.php'); ?> <form action ='' method='post'> <select name="id"> <?php $extract = mysql_query("SELECT * FROM cars"); while($row=mysql_fetch_assoc($extract)){ $id = $row['id']; $make= $row['make']; $model= $row['model']; $year= $row['year']; $color= $row['color']; echo "<option value=$id>$color $year $make $model</option> ";}?> </select> Which attribute would you like to change?<br /> <input type="radio" name="getchanged" value="make"/>Make<br /> <input type="radio" name="getchanged" value="model"/>Model<br /> <input type="radio" name="getchanged" value="year" />Year<br /> <input type="radio" name="getchanged" value="color" />Color<br /><br /> <br /><input type='text' value='' name='tochange'> <input type='submit' value='Change' name='submit'> </form> //This is where I need help... <?PHP if(isset($_POST['submit'])&&($_POST['tochange'])){ mysql_query(" UPDATE cars SET '$_POST[getchanged]'='$_POST[tochange]' where id = '$_POST[id]' ");}?> My login script stores the user's login name as $_SESSION[ 'name'] on login. For some unapparent reason, i'm getting errors stating that $user and $priv are undefined variables, though I've attempted to define $user as being equal to $_SESSION['name'], using $user to look up the the user's privilege level (stored as the su column ) in the SQL table, and then where the result of the sql query is $priv which is then evaluated in an if statement. I can't seem to figure out why this might not be working. The code I'm using:
<?php
session_start();
function verify() {
//verify that the user is logged in via the login page. Session_start has already been called.
if (!isset($_SESSION['loggedin'])) {
header('Location: /index.html');
exit;
}
//if user is logged in, we then lookup necessary privleges. $_SESSION['name'] was written with the login name upon login. Privleges // are written in db as a single-digit integer of of 0 for users, 1 for administrators, and 2 for special users.
$user === $_SESSION['name'];
//Connect to Databse
$link = mysqli_connect("127.0.0.1", "database user", "password", "database");
if (!$link) {
echo "Error: Unable to connect to MySQL." . PHP_EOL;
echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
exit;
}
//SQL Statement to lookup privlege information.
if ($result = mysqli_query($link, "SELECT su FROM accounts WHERE username = $user", MYSQLI_STORE_RESULT)) {
//LOOP TO CYCLE THROUGH SQL RESULTS AND STORE Privlege information as vairable $priv.
while ($row = $result->fetch_assoc()) {
$priv === $row["su"];
}
}
// close SQL connection.
mysqli_close($link);
// Verify privleges and take action. Only a privlege of "1" is allowed to view this page. A privlege of "2" indicates special
//accounts used in other scripts that have certain indermediate additional functions, but are not trusted administrators.
if ($priv !== 1) {
echo $_SESSION['name'];
echo "you have privlege level of $priv";
echo "<br>";
echo 'Your account does not have the privleges necessary to view this page';
exit;
}
}
verify();
?>
I have just re-installed Xampp and suddenly my sites are now displaying lots of: Notice: Use of undefined constant name - assumed 'name' in ... Notice: Use of undefined constant price - assumed 'price' in ... this is an example of the line its refering too: $defineProducts[1001] = array(name=>'This is a product', price=>123); I have a form that creates rows of data input textboxes depending on a user input number of things. I have a naming convention for all these textboxes that basically just keeps incrementing a number suffix for each row. All this is working fine. My problem is I need to get the data inserted into this table of textboxes into an array. Here's my code where I attempt to to this (it does not work): Code: [Select] $temp = $_SESSION['Num_Part']; $count = 1; while ($count <= $temp){ $temp2[$count] = "'Participant_P".$count."'"; //echo $temp2[$count]."<br/>"; $temp3[$count]=$_POST[$temp2[$count]]; //here's the problem $temp4[$count] = "'Result_P".$count."'"; $temp5[$count]=$_POST[$temp4[$count]]; //here's the problem //echo $temp4[$count]."<br/>"; $count++; } The problem is that the $_POST does not work with the variable in the argument position - even though the argument is formatted with single quotes. Can a variable be used in a POST argument and if so what is the correct syntax? If not, is there some other simple solution to harvest the data into an array. I understand I can harvest by explicitly accessing each key in the post assoc array. But this could be dozens of rows of input fields. Thanks in advance for your help here. I couldn't find anything online re this topic. Probably something simple but I have searched high and low and can't figure this one out. I have a variable that is of the datetime format. I have another variable that is of the time format. I need to add them together. Example: $var1 = 2012-02-24 06:38:22 $var2 = 02:00:00 $var3 = $var1 + $var2 = 2012-02-24 08:38:22 Thanks for the help! |
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