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PHP - Preventing Errors From Being Displayed (which Function Is Better?)
I know that the following lines of code can be used to prevent errors from being displayed:
Code: [Select] <?php error_reporting(0); ini_set('display_errors', 0); ?> Is there a reason to use one over the other? Is it better to use both? Similar TutorialsHello, I was trying out some code to check out how some of the php settings work and when i tried allow_call_time_pass_reference to see if it works, it doesn't spew out any warning or error like the comments in the .ini file mention (or in the manual sites). Maybe i'm not using some other setting in order to see this? Here is my EXACT code. <?php ini_set('display_errors',1); error_reporting(-1); $variable = 'String'; function display($variable) { return $variable; } echo display(&$variable); ?> Outputs : String. No errors, no warnings. I'm running PHP Version 5.3.2-1ubuntu4.5 Been just trying to test out some basic code that utilizes gd. In Chrome i receive a blank page. In firefox I get "The image "image.php" cannot be displayed because it contains errors". There is no whitespace before or after the php tags, which seems to cause this error sometimes. $image = imagecreate(200,20); $background = imagecolorallocate($image, 0, 0, 0); $foreground = imagecolorallocate($image,255,255,255); imagestring($image, 5, 5, 1, "Test", $foreground); header("Content-type: image/jpeg"); $imagejpeg($image); I am just following a tutorial, and at this point in the tutorial i should get at least some garbled text, if not an image. I looked at other common causes. gd2 is un-commented in php.ini. phpinfo shows: GD Support enabled GD Version bundled (2.0.34 compatible) FreeType Support enabled FreeType Linkage with freetype FreeType Version 2.4.3 GIF Read Support enabled GIF Create Support enabled JPEG Support enabled libJPEG Version 6b PNG Support enabled libPNG Version 1.2.46 WBMP Support enabled XBM Support enabled gd.jpeg_ignore_warning 0 0 I have not found any other solutions to try, or maybe I have made an error is checking these solutions. Any advice with be great! Thank you for your time. Please help me fix these errors in my function. Warning: mysql_result(): Unable to jump to row 0 on MySQL result index 10 in Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in Code: [Select] function verify_user_fanned_company($uid, $cid) { $uid = (int)$uid; $sql = " SELECT `company_id` FROM `company_fans` WHERE `user_id` = {$uid} AND `company_id` = {$cid} "; $query = mysql_query($sql); $result = mysql_result($query, 0); $num = mysql_num_rows($result); if ($num === 1) { return true; } } Code: [Select] <?php $is_fanned = verify_user_fanned_company($user_info['uid'], $info["companyid"]); if ($is_fanned = true) { ?> do stuff <?php } ?> Hi, I'm trying to output some details from my database, along with a BLOB image. At the moment I'm having some warnings and notices and I don't know how to solve them. Here's the warnings and notices: Notice: Undefined index: id in G:\xampp\htdocs\xampp\dsa\search_func.php on line 38 Notice: Undefined variable: terms in G:\xampp\htdocs\xampp\dsa\search_func.php on line 13 Warning: mysql_query() expects parameter 2 to be resource, object given in G:\xampp\htdocs\xampp\dsa\search_func.php on line 16 Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in G:\xampp\htdocs\xampp\dsa\search_func.php on line 22 Notice: Trying to get property of non-object in G:\xampp\htdocs\xampp\dsa\search_func.php on line 27 Notice: Trying to get property of non-object in G:\xampp\htdocs\xampp\dsa\search_func.php on line 27 Notice: Trying to get property of non-object in G:\xampp\htdocs\xampp\dsa\search_func.php on line 27 Notice: Trying to get property of non-object in G:\xampp\htdocs\xampp\dsa\search_func.php on line 27 Notice: Trying to get property of non-object in G:\xampp\htdocs\xampp\dsa\search_func.php on line 28 Notice: Trying to get property of non-object in G:\xampp\htdocs\xampp\dsa\search_func.php on line 28 Notice: Trying to get property of non-object in G:\xampp\htdocs\xampp\dsa\search_func.php on line 28 Here's the function code: Code: [Select] <?php /* return the details of an employee parameter empno - employee number eg empHTML.php?empno=7521 This is very basic e.g no error checking */ function search($dbc, $id) { // define the query string $query = "SELECT * FROM stick, location, identification WHERE make LIKE '%$terms%' AND stick.sid = location.lid AND identification.stickID = stick.sid "; // execute the query and return a pointer to the resultant SQL relation $dbresult = mysql_query($query,$dbc); // return the result as a simple HTML table $output = Array(); // get each tuple in the table as an object $image = mysql_fetch_object($dbresult); // interpolate the required employee values -> is the equivalent of the java DOT notation $output[] = '<ul>'; $output[] = '<li> Type: '.$image->make .'<br />Size: '.$image->size .$image->type .'<br />Colour: '.$image->colour . '<br /> Street Missing in: ' .$image->street.'<br />Town missing in: '.$image->town .'<br />City missing in: '.$image->city.'</li>'; $output[] = '</ul>'; return join('',$output); } // connect $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); // get the department name from the URL $id = $_REQUEST['id']; // print the table of employees for this department // put it ina div HTML element with an emp class to support CSS styling print search($dbc, $id); print "<img src='getPhoto.php?id=$id'/>"; // close the database mysql_close($dbc); I have this code that has Code: [Select] $this->BeginTransaction();all over it. These are in classes where no such function is defined, and they don't have parents either. It's driving my crazy trying to figure out why they are there, why there's not so much as a warning being given and what, if anything, is being called. Now I do have a Code: [Select] set_error_handler("amfErrorHandler");and in amfErrorHandler I have Code: [Select] if( error_reporting() != 0 && ($amfphpErrorLevel | $level) == $amfphpErrorLevel ) { throw new VerboseException($string, $level, $file, $line); } I don't remember what $amfphpErrorLevel is, except that it should only be ignoring things like E_STRICT. Also, no exception is being thrown. So what do I do w/this code? HI guys I have vreated a site and am now trying to validate the ip address i have created a function but am failing to get it to work i need to add the errors to add to the errors array but instead it keeps resetting and leaving me with the last error that is returned any help would be greatly appreciated im a newbie so if i can understand the concept behind whats happening it will be of great benefit Thanks Sean heres the code // Function validateIpaddress($address,$record,$errors) // { // if ($address !=="" or $address !=="ip_address") // { // Print_r($address); // echo "<br><br>"; // $fullstoppos = strpos(($address), "."); // echo "strpos = $fullstoppos"; // $fullstoppos2 = strpos(($address), ".", $fullstoppos+1); // echo "strpos2 = $fullstoppos2"; // $fullstoppos3 = strpos(($address), ".", $fullstoppos2+1); // echo "strpos3 = $fullstoppos3"; // } // if ($fullstoppos !==4 && $fullstoppos2 !==9 && $fullstoppos3 !==14) // { // $errors[] = "<br><font color='red'>Please correct the following.Or contact your client manager for assitance.<br>* $record is a required field.An ip address is a 16 digit numerical value. </font>"; // } // } $errors = validateIpaddres($_POST['destinationPoint'],A/AAAA record, $errors) I am trying to write a php/mysql that will allow a church keep attendance on their members in bible study. I am also going to try to prevent doing a circular reference between tables and just can't figure it out how since I am just starting to learn mysql. Here are the tables:
Table 1: Members:
---------------
1:Name:
2:Address:
3:Bible Study Group it belongs to:
Table 2: Cells
--------------
1:Bible Study Lider:
2:Bible Group Name
Table 3: Attendance
--------------
1:Date
2:Member
3:Bible Group
As you all can see, Table 2:2 makes a lookup at Table 1 for the member(in this case, the leader). BUT Table 1:3 makes a lookup to Table 2:2
and is a circular lookup.
Anyone have an idea on how to properly do this without any circular problems?
Thanks in advance!
I've run into a little bit of a logistical nightmare on some registration pages I've taken over work on. On these pages, parents register their kids for classes. The pages have been coded as such: page one: the user enters name and personal info, and selects one or two classes to register for, on submit, they go to page two. page two: their info is entered into the mysql database's registration table on a unique id The user verifies the total, enters a discount and submits page three: credit card info is added to registration table, and sent to authorize.net. page four: payment processed, and they enter the data for their kids into the attendees table page five: confirmation and done. My issue is that, at first I saw people entering page one data, going to page two, then for some reason, hitting the back button. They could then enter the data again. I'd have two entries in the registration table for the same person. I was going to put some sort of unique key on the name and an email, but then I saw scenario two... Another person enters data and registers one kid.. then for whatever personal reason goes back and registers a second kid in a completely separate transaction. So I can't put that key on there, but is there a way to prevent them from going back and reentering twice. I don't want to have to blow up the code to do it at the end of everything. and that would take implementing sessions, wouldn't it? I'm not so versed at that. Any thoughts? I hope this is enough info, but I have this code (at the bottom) that grabs images from my S3 account (Amazon Web Services)... If there is no bucket for the name given, it gives this warning... Code: [Select] Warning: S3::getBucket(): [NoSuchBucket] The specified bucket does not exist in /home/xxxxxxx/public_html/xxxxxxx.com/S3.php on line 126 Assuming it's not specific syntax to S3 and is general PHP syntax, how could I prevent that warning from showing, by saying in plain English "If a bucket by that name doesn't exist, I'm gonna do something else and NOT spit out this warning?" Basically, I want to echo a message like "There is no bucket yet" instead of having that ugly warning. Code: [Select] // Get the contents of our bucket $contents = $s3->getBucket("mybucket"); foreach ($contents as $file){ $fname = $file['name']; $furl = "http://mybucket.s3.amazonaws.com/".$fname; //output a link to the file echo "<a href=\"$furl\">$fname</a><br />"; echo "<img src='$furl' width='50' /><br />"; echo "$furl<br />"; } Hello every1,
I'm trying to create a magazine wesite and I dont want anybody to access my images folder which is located in the root directory.
I want some thing like this...
I want to call the images in my website using <img> tag but I dont want any1 to access the directory directly.
It should also protect from software bots also like (HTTrack or website Copier)
It should show "Access forbidden".
Any help will be greatly appreciated.
Thank you...
Basically, I have the following code ($c2 is my connection variable): Code: [Select] $rid = $_GET['id']; $q = mysql_query("SELECT * FROM reports WHERE id = $rid", $c2) or die(mysql_error()); $report = mysql_fetch_array($q); $report is used later on to gather more information that is outputted to the user. However, if in the URL, someone were to put id=1', they would have an error message spit out to them (something along the lines of: 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 '\'' at line 1), indicating a SQL Injection exploit. How would I go about fixing this, and also preventing SQL Injection? Thanks a bunch, Mark Would this work to prevent remote file inclusion vulnerability? $file = "../include/links.php"; if ($file = '../include/links.php'){ include $file; } I imagine there are lots of ways to answer this question, so I just want people's opinions as to the best way (if there is one). I have code with a basic form that submits data to a MySQL database. So when someone submits data the first time, I "clean it up" by doing... Code: [Select] $_POST['data'] = trim(mysql_prep($_POST['data'])); .. and then submit that info into a "varchar" mySQL field. Then if the user comes back to edit it, the form comes up and the data they previously entered is pulled into the field this way (I'm leaving out the MySQL to pull the data, obviously)... Code: [Select] <input type="text" name="field" value="<?php echo $data;?>"> The problem is that if someone entered this originally... Code: [Select] Here is "some" data with apostrophes ...Then when I echo that back into the value of the text field, it would only show... Code: [Select] Here is "...and then cuts off because the quotation mark in the data conflicted with the quotation mark after value= Is htmlentities the answer here, or is there some other/better way? FYI... Code: [Select] function mysql_prep($value) { $magic_quotes_active = get_magic_quotes_gpc(); $new_enough_php = function_exists("mysql_real_escape_string") ; //i.e. PHP >= v4.3.0 if($new_enough_php) { //PHP v4.3.0 or higher //undo any magic quote effects so mysql_real_escape_string can do the work if($magic_quotes_active) { $value = stripslashes($value) ;} $value = mysql_real_escape_string($value); } else { //before php v4.3.0 // if magic quotes aren;t already on then add slashes manually if(!magic_quotes_active) { $value = addslashes($value); } // if magic quotes are active, then the slashes already exist } return $value; } How do I display the actual errors in IIS 7.5? If I miss a semicolon, I get: "HTTP Error 500.0 - Internal Server Error" absolutely useless. In prior versions, I could see the line and get to the PHP error. How do I display PHP errors? I've added: set_ini('display_errors', '1'); but it doesn't help. "BACK" or REFRESH: Preventing database interaction / code execution how to prevent database interaction / code execution when user presses back or refresh button? can i detect? can i disable back/refresh? hey gurus, i am a newbie php coder.. i am learning by example. what i am trying to do is write a piece of code which will alter 3 tables (user, bonus_credit, bonus_credit_usage) ---------------------------------------------------------------- the table structure that will be used is as follows: user.bonus_credit user.ID bonus_credit.bonusCode bonus_credit.qty bonus_credit.value bonus_credit_usage.bonusCode bonus_credit_usage.usedBy ---------------------------------------------------------------- so lets say, in bonus_credit i have the following bonusCode = 'facebook' (this is the code they have to type to redeem the bonus qty = '10' ( number of times the bonusCode can be redeemed, but same person can't redeem it more than once) value = '5' (this is the amount of bonus_credit for each qty) Now, I need to write a code that check to see if the code has been redeemed in the bonus_credit_usage table and if the user.ID exists in this table as bonus_code_usage.usedBy, then give an error that its already been used and if it hasn't been used, then subtract 1 from qty, add ID to usedBy and then add the value to the bonus_credit ----------------------- i have started the steps just to create a simple textbox and entering a numeric value to bonus_credit, and that works.. but now i have to use JOIN and IF and ELSE.. which is a little too advanced for me.. so i'd appreciate a guide as i write the code. if(isset($_REQUEST['btnBonus'])) { $bonus_credit = addslashes($_REQUEST['bonusCode']); $query = "update user set bonus_credit=bonus_credit+'".$bonus_credit."' where id='".$_SESSION['SESS_USERID']."'"; echo "<script>window.location='myreferrals.php?msgs=2';</script>"; mysql_query($query) or die(mysql_error()); } I am trying to run a php code on my site but it kept on saying page cannot be found. Any idea why its saying that??? Hi, I am generating certificates in php i can display the name in arabic but it will not print on the certicate. It can show the English names though! is there a special code to be able to show that? code:
<!doctype html>
<html lang="en">
<head>
<!-- Required meta tags -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1, shrink-to-fit=no">
<!-- Bootstrap CSS -->
<link rel="stylesheet" href="https://stackpath.bootstrapcdn.com/bootstrap/4.4.1/css/bootstrap.min.css" integrity="sha384-Vkoo8x4CGsO3+Hhxv8T/Q5PaXtkKtu6ug5TOeNV6gBiFeWPGFN9MuhOf23Q9Ifjh" crossorigin="anonymous">
<title>Certificate Generator</title>
</head>
<body>
<center>
<br><br><br>
<h3>Certificate Generator</h3>
<br><br><br><br>
<form method="post" action="">
<div class="form-group col-sm-6">
<input type="text" name="name" class="form-control" id="name" placeholder="Enter Name Here...">
</div>
<button type="submit" name="generate" class="btn btn-primary">Generate</button>
</form>
<br>
<?php
if (isset($_POST['generate'])) {
$name = strtoupper($_POST['name']);
$name_len = strlen($_POST['name']);
if ($name == "" ) {
echo
"
<div class='alert alert-danger col-sm-6' role='alert'>
Ensure you fill all the fields!
</div>
";
}else{
echo
"
<div class='alert alert-success col-sm-6' role='alert'>
Congratulations! $name on your excellent success.
</div>
";
//designed certificate picture
$image = "certi.png";
$createimage = imagecreatefrompng($image);
//this is going to be created once the generate button is clicked
$output = "certificate.png";
//then we make use of the imagecolorallocate inbuilt php function which i used to set color to the text we are displaying on the image in RGB format
$white = imagecolorallocate($createimage, 205, 245, 255);
$black = imagecolorallocate($createimage, 0, 0, 0);
//Then we make use of the angle since we will also make use of it when calling the imagettftext function below
$rotation = 0;
//we then set the x and y axis to fix the position of our text name
$origin_x = 200;
$origin_y=240;
//we then set the x and y axis to fix the position of our text occupation
$origin1_x = 120;
$origin1_y=90;
//we then set the differnet size range based on the lenght of the text which we have declared when we called values from the form
if($name_len<=7){
$font_size = 25;
$origin_x = 190;
}
elseif($name_len<=12){
$font_size = 30;
}
elseif($name_len<=15){
$font_size = 26;
}
elseif($name_len<=20){
$font_size = 18;
}
elseif($name_len<=22){
$font_size = 15;
}
elseif($name_len<=33){
$font_size=11;
}
else {
$font_size =10;
}
$certificate_text = $name;
//font directory for name
$drFont = dirname(__FILE__)."/developer.ttf";
//function to display name on certificate picture
$text1 = imagettftext($createimage, $font_size, $rotation, $origin_x, $origin_y, $black,$drFont, $certificate_text);
imagepng($createimage,$output,3);
?>
<!-- this displays the image below -->
<img src="<?php echo $output; ?>">
<br>
<br>
<!-- this provides a download button -->
<a href="<?php echo $output; ?>" class="btn btn-success">Download My Internship Certificate</a>
<br><br>
<?php
}
}
?>
</center>
<footer>
<center><p>Built with ❤ by <a href="https://olawanlejoel.github.io/portfolio/">Olawanle Joel</a></p></center>
</footer>
<!-- Optional JavaScript -->
<!-- jQuery first, then Popper.js, then Bootstrap JS -->
<script src="https://code.jquery.com/jquery-3.4.1.slim.min.js" integrity="sha384-J6qa4849blE2+poT4WnyKhv5vZF5SrPo0iEjwBvKU7imGFAV0wwj1yYfoRSJoZ+n" crossorigin="anonymous"></script>
<script src="https://cdn.jsdelivr.net/npm/popper.js@1.16.0/dist/umd/popper.min.js" integrity="sha384-Q6E9RHvbIyZFJoft+2mJbHaEWldlvI9IOYy5n3zV9zzTtmI3UksdQRVvoxMfooAo" crossorigin="anonymous"></script>
<script src="https://stackpath.bootstrapcdn.com/bootstrap/4.4.1/js/bootstrap.min.js" integrity="sha384-wfSDF2E50Y2D1uUdj0O3uMBJnjuUD4Ih7YwaYd1iqfktj0Uod8GCExl3Og8ifwB6" crossorigin="anonymous"></script>
</body>
</html>
Edited May 22, 2020 by ramiwahdan typo error when I test this script on browser. This is all that I see. invalid username or password Code: [Select] <?php //start session session_start(); include 'functions.php'; if (loggedin()) { header("Location: userarea.php"); exit(); } if ($_POST['login']) { //get data $username = $_POST['username']; $password = $_POST['password']; $rememberme = $_POST['rememberme']; } if (username&&$password) { $login = mysql_query("SELECT * FROM users WHERE username='$username'"); while ($row = mysql_fetch_assoc($login)) { $db_password = $row['password']; if (md5($password)==$db_password) $loginok =TRUE; else $loginok = FALSE; if ($loginok ==TRUE); { if($rememberme=="on") setcookie("username",$username, time()+7200); else if ($rememberme=="") $_SESSION['username']=$username; header("Location: userarea.php"); exit(); } } } else { die("invalid username or password"); } ?> <form action="login.php" method="POST"> username:<br /> <input type="text" name="username" /><br /> password:<br /> <input type="password" name="password" /><br /> <input type="checkbox" name="rememberme">remember me<br /> <input type="submit" name="login" value="Log in" /> </form> I'm teaching myself how to use php and mysql... I created a form to insert data into a mysql database, all went well and when I returned to the form it was blank. I then asked a friend to test it out aswell and when he loaded the form page, the data that entered before him was displayed? (it was still filled out with my info). Even after he refreshed it was still there... Hows this possible? How would I clear the form after its been submitted if it doesnt do it automatically? Thanks Drew |
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