inspect data (url variable) before inserting to mysql - php

My php script is posting bad urls that ajax send to it. When I try to filter the url and cancel the insert into mysql, it doesn't work. What I want to do is verify the $link variable if it a link. If not, don't post data to mysql. Can i know what I did wrong and how to fix it? Thank you :)
Here is my code
$con = mysql_connect("localhost","root","");
if (!$con) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("database", $con);
$link = $_POST['new'];
$name = $_POST['name'];
$size = $_POST['size'];
$cat = $_POST['cat'];
// PHP 5.3.5-1ubuntu7.2
$link = mysql_real_escape_string($link);
$name = mysql_real_escape_string($name);
$size = mysql_real_escape_string($size);
$cat = mysql_real_escape_string($cat);
if (filter_var($link, FILTER_VALIDATE_URL)) {} else {
echo "URL is NOT valid";
mysql_close($con);
exit();
}
$check = mysql_query("SELECT link FROM links WHERE link = '{$link}';");
if (mysql_num_rows($check) == 0) {
// insert
$sql="INSERT INTO links (link, name, size, category) VALUES ('$link','$name','$size','$cat')";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
echo "1 record added. Redirecting!";
mysql_close($con);
}

try this one:
if(filter_var($link, FILTER_VALIDATE_URL, FILTER_FLAG_SCHEME_REQUIRED) === false){
echo "URL is NOT valid";
mysql_close($con);
exit();
}
Your are right xxxhxtxtp://example.com will pass as valid url.
Try it with preg_match instead.
if (!preg_match("#^http(s)?://[a-z0-9-_.]+\.[a-z]{2,4}#i", $link)) {
echo "URL is NOT valid";
mysql_close($con);
exit();
}
With this regex you need to have http(s) scheme. So you can search for another regex if that don't work for you.

Is this what you're trying to do? Perhaps you need a regex solution.
if (filter_var($link, FILTER_VALIDATE_URL)) {
$check = mysql_query("SELECT link FROM links WHERE link = '{$link}';");
if (mysql_num_rows($check) == 0) {
// insert
$sql="INSERT INTO links (link, name, size, category) VALUES ('$link','$name','$size','$cat')";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
echo "1 record added. Redirecting!";
mysql_close($con);
}
} else {
echo "URL is NOT valid";
mysql_close($con);
exit();
}

Related

Find method in SQL on my PHP page finds everything, even if the record is null

Ok so I wrote some code to find records on a test database, it works if there is a record and does display the data, if there is no record it still says that it found stuff. It should say it did not. It even finds stuff that is not in the database but obviously has no data to display, its annoying.
I need a new pair of eyes.
I think the error is here:
$sql = "SELECT * FROM Kittenzz
WHERE KittenID='".$_POST['KittenID']."';";
$result = mysql_query($sql, $connection);
But just in case here is the full code minus the login credentials to the db.
<?php
if(isset($_POST['Find']))
{
$connection = mysql_connect("Login Info Deleted");
// Check connection
if (!$connection)
{
echo "Connection failed: " . mysql_connect_error();
}
else
{ //else 1
//select a database
$dbName="Katz";
$db_selected = mysql_select_db($dbName, $connection);
//confirm connection to database
if (!$db_selected)
{
die ('Can\'t use $dbName : ' . mysql_error());
}
else
{ //else 2
if ($_POST[KittenID]=='')
{
$OutputMessage = 'Must add a Kitten-ID';
}
else
{//exception else
$sql = "SELECT * FROM Kittenzz
WHERE KittenID='".$_POST['KittenID']."';";
$result = mysql_query($sql, $connection);
while($row = mysql_fetch_array($result))
{
$Name = $row['Name'];
$KittenID = $row['KittenID'];
$KittenAge = $row['KittenAge'];
$Email = $row['Email'];
$Comments = $row['Comments'];
$Gender = $row['Gender'];
$Passive = $row['Passive'];
$Playful = $row['Playful'];
$Activity = $row['Activity'];
}
if ($result)
{
$OutputMessage = 'Record Found';
//echo "<p>Record found<p>";
}
else
{
$OutputMessage = 'RECORD NOT FOUND';
}
}//exception else
}//else 2 end
}//else 1 end
mysql_close($connection);
}
?>
if ($result)
{
$OutputMessage = 'Record Found';
}
There is your mistake, that means if the query executed successfully (even with 0 records) you are saying records found. You should only say that if the number of records returned are more than 0.
if (mysql_num_rows($result)>0)
{
$OutputMessage = 'Record Found';
}
But the bigger problem with your code can be solved by this reading
How can I prevent SQL injection in PHP?
This may happen, because if $_POST['KittenID'] is empty, the sql query would look like : SELECT * FROM Kittenzz WHERE KittenID=""; you have to change the above if statement to:
if (!isset($_POST[KittenID]) || empty($_POST[KittenID]) || $_POST[KittenID]=='')
{
$OutputMessage = 'Must add a Kitten-ID';
}

PHP - reload new record upon page refresh

I have a little problem with my PHP code and database. Every time i refresh the page, a new empty row is being added to the database also when i open the page, what is the problem?
<?php
if (isset($_POST)) {
$con = mysql_connect("localhost","dwarfmaster","");
if (!$con) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("my_dwarfmaster", $con);
$name = $_POST['name'];
$release_year = $_POST['release_year'];
$publisher = $_POST['publisher'];
$genre = $_POST['genre'];
$sql = "INSERT INTO gamelist (name, release_year, publisher, genre)
VALUES
('$_POST[name]','$_POST[release_year]',
'$_POST[publisher]','$_POST[genre]')";
if (!mysql_query($sql, $con)) {
die('Error: ' . mysql_error());
}
echo "1 record added";
mysql_close($con);
}
?>
to check if the request method is post, use:
if ($_SERVER['REQUEST_METHOD'] == 'POST')
if(isset($_POST)) will always return true
or simply change
if (isset($_POST))
to
if (isset($_POST['name']))
you can add the following code-:
if( !empty(array_filter($_POST)) ) { . . Code to insert data into data base
from form . . }
if (isset($_POST))
will always return true !
try
if (isset($_POST['name'])&&isset($_POST['release_year'])
&&isset($_POST['publisher'])&&isset($_POST['genre']))
instead to check each time for each and every variable !

headers not working PHP

Page 1 abc.html.. on submit it will jump to this PHP page .
This is page PHP1.php here i am trying to validate user input if name and id in in data he will be forwarded to fill out second part of registration if not it will just give error.
<?php
session_start();
$_SESSION["acb"] = "good";
$_SESSION['team'] = $_POST['team_name'];
$con = mysql_connect("localhost", "user", "password");
if (!$con)
{die('Could not connect: ' . mysql_error());}
mysql_select_db("mydbName");
if(isset($_POST['team_name'],$_POST['id'])){
$team_name = mysql_real_escape_string($_POST['team_name']);
$id = mysql_real_escape_string($_POST['id']);
if (!empty($team_name)) {
$result= mysql_query("SELECT COUNT(`teamname`) FROM `table` WHERE `teamname`='$team_name' AND `id`='$id'");
$team_result = mysql_fetch_row($result);
if ($team_result[0] == '0') { //if does not exist print failed.
echo 'Varification failed';
} else {
header('Location: http://www.abc.com/REGISTERpart2.php');
}} } ?>
RegisterPART2.php is where i am checking my session exist or not (the one i started in last file). if not i want to redirect back to form one and fill that first then come to registration part 2
`<?php
session_start();
$name = $_SESSION['team']; //a value stored in session which i used on this page
if (($_SESSION["abc"] !== 'good')) {
header('Location: http://www.abc.com/page1.html'); //take back to stage 1 coz user did not fill first part.
}
else{
echo $name. 'you have completed register process part one you may continue!';
}
?>
If you're using the new MySQL version (MySQLi), so the first page will become:
<?php
session_start();
$_SESSION["acb"] = "good";
$_SESSION['team'] = $_POST['team_name'];
$con = new mysqli("localhost", "user", "password", "mydbName");
if (!$con) {
die('Could not connect: ' . $con->error());
};
if (isset($_POST['team_name'],$_POST['id'])) {
$team_name = $con->real_escape_string($_POST['team_name']);
$id = $con->real_escape_string($_POST['id']);
if (!empty($team_name)) {
$result = $con->prepare("SELECT COUNT(`teamname`) FROM `table` WHERE `teamname`='$team_name' AND `id`='$id'");
$result->execute();
$result->bind_result($one,$two,$three,$etc);
$result->fetch();
if (empty($one) and empty($two) and empty($three) and empty(etc)) { // may be and/or (pick one)
echo 'Varification failed';
} else {
header('Location: http://www.abc.com/REGISTERpart2.php');
}
}
}
?>
You may use the following alternative to header.
prinf('<script>window.location = "URL HERE"</script>');
It should do the same thing as header does.

how to remove the header error

im having a problem with my code in uploading and displaying images.. well I am planning to redirect the page after the upload process is done so I used a header function but gave warning and errors and unfortunately failed the upload.. how can I remove it? here's the code..
<?php
//connect to the database//
$con = mysql_connect("localhost","root", "");
if(!$con)
{
die('Could not connect to the database:' . mysql_error());
echo "ERROR IN CONNECTION";
}
$sel = mysql_select_db("imagedatabase");
if(!$sel)
{
die('Could not connect to the database:' . mysql_error());
echo "ERROR IN CONNECTION";
}
//file properties//
$file = $_FILES['image']['tmp_name'];
echo '<br />';
/*if(!isset($file))
echo "Please select your images";
else
{
*/for($count = 0; $count < count($_FILES['image']); $count++)
{
//$image = file_get_contents($_FILES['image']['tmp_name']);
$image_desc[$count] = addslashes($_POST['imageDescription'][$count]);
$image_name[$count] = addslashes($_FILES['image]']['name'][$count]); echo '<br \>';
$image_size[$count] = #getimagesize($_FILES['image']['tmp_name'][$count]);
$error[$count] = $_FILES['image']['error'][$count];
if($image_size[$count] === FALSE || ($image_size[$count]) == 0)
echo "That's not an image";
else
{
// Temporary file name stored on the server
$tmpName[$count] = $_FILES['image']['tmp_name'][$count];
// Read the file
$fp[$count] = fopen($tmpName[$count], 'r');
$data[$count] = fread($fp[$count], filesize($tmpName[$count]));
$data[$count] = addslashes($data[$count]);
fclose($fp[$count]);
// Create the query and insert
// into our database.
$results = mysql_query("INSERT INTO images( description, image) VALUES ('$image_desc[$count]','$data[$count]')", $con);
if(!$results)
echo "Problem uploding the image. Please check your database";
//else
//{
echo "";
//$last_id = mysql_insert_id();
//echo "Image Uploaded. <p /> <p /><img src=display.php? id=$last_id>";
//header('Lcation: display2.php?id=$last_id');
}
//}
}
mysql_close($con);
header('Location: fGallery.php');
?>
the header function supposedly directs me to another page that would make a gallery.. here is the code..
<?php
//connect to the database//
mysql_connect("localhost","root", "") or die(mysql_error());
mysql_select_db("imagedatabase") or die(mysql_error());
//requesting image id
$image = mysql_query("SELECT * FROM images ORDER BY id DESC");
while($row = mysql_fetch_assoc($image))
{
foreach ($row as $img) echo '<img src="img.php?id='.$img["id"].'">';
}
mysql_close();
?>
I have also a problem with my gallery .. some help will be GREAT! THANKS! :D
The header() function must be called before any other echo or die calls which produce output.
You may could buffer your outputs if you need the output, but in your case it makes no difference because the output will never be shown to the user. The browser will read the redirect and navigate to the second page.
<?php
//connect to the database//
$con = mysql_connect("localhost","root", "");
if(!$con) {
// this output is okay the redirect will never be reached.
die('Could not connect to the database:' . mysql_error());
// remember after a die this message will never be shown!
echo "ERROR IN CONNECTION";
}
$sel = mysql_select_db("imagedatabase");
if(!$sel) {
die('Could not connect to the database:' . mysql_error());
echo "ERROR IN CONNECTION"; // same here with the die!
}
//file properties//
$file = $_FILES['image']['tmp_name'];
// OUTPUT
// echo '<br />';
// removed out commented code
for($count = 0; $count < count($_FILES['image']); $count++)
{
$image_desc[$count] = addslashes($_POST['imageDescription'][$count]);
$image_name[$count] = addslashes($_FILES['image]']['name'][$count]);
// OUTPUT
// echo '<br \>';
$image_size[$count] = #getimagesize($_FILES['image']['tmp_name'][$count]);
$error[$count] = $_FILES['image']['error'][$count];
if($image_size[$count] === FALSE || ($image_size[$count]) == 0)
// you may better use a die if you want to prevent the redirection
echo "That's not an image";
else
{
// Temporary file name stored on the server
$tmpName[$count] = $_FILES['image']['tmp_name'][$count];
// Read the file
$fp[$count] = fopen($tmpName[$count], 'r');
$data[$count] = fread($fp[$count], filesize($tmpName[$count]));
$data[$count] = addslashes($data[$count]);
fclose($fp[$count]);
// Create the query and insert
// into our database.
$results = mysql_query("INSERT INTO images( description, image) VALUES ('$image_desc[$count]','$data[$count]')", $con);
if(!$results) // use die
echo "Problem uploding the image. Please check your database";
// OUTPUT
// echo "";
}
}
mysql_close($con);
header('Location: fGallery.php');
?>
Above I marked every output for you and also removed all outcomments lines.
You've got a header error because you printed out <br /> before the header function. In order to use the header function you can't print out any information before it. That's why you're getting the error.
Regarding your gallery the foreach loop is unnecessary. You can change the code to this:
while($row = mysql_fetch_assoc($image)) {
echo '<img src="img.php?id='.$row["id"].'">';
}
You can use ob_start() to get data in buffer.

Refresh PHP page once only when called

I have a php which would check for certain value if it exists in a mysql database. If the value does not exists, it would simply add the value and refresh the page once to load the page again and now it has a value in the database, would go ahead to add other values. How do I refresh page just once when it is called ?
<?php
$sname = "W3 schools C# tutorials";//$_POST["sitename"];
$stype = "C#";//$_POST["sitetype"];
$saddy = "www.w3schools.com";//$_POST["siteaddress"];
$scomm = "W3 schools C# tutorials";//$_POST["sitecomment"];
$conn = mysql_connect("localhost","root","password");
if(!$conn){
die("Could not connect: ".mysql_error());
} else {
mysql_select_db("bookmarks",$conn);
$rs = mysql_query("select TypeId from bookmarktypes where TypeName = '$stype'");
$row = mysql_fetch_array($rs);
if($row > 0 ){
//Data found, continue to add...
} else {
//No data... insert a valid one
$rs = mysql_query("insert into bookmarktypes (TypeName) values ('$stype')");
if (!$rs){
die('Error: ' . mysql_error());
} else {
//echo "inserted new type data...";
}
//echo "</html>";
}
}
mysql_close($conn);
//Refresh page once
?>
There's the comment to refresh page below after mysql close command.
Refresh it right after insert with
header('Location: url here');
exit;
Btw, read a little about sql injections
Also - mysql_close() is pointless there.
if(check=1)
{
echo "\"<meta http-equiv=\"refresh\" content=\"2;url=http://yourwebsite.com/\">\"\n";
}
if you need to print the data that you just have entered try this
header('Location: YourShowDataPage.php?id='.$_POST['id_dataEntered'])
mi apologizes if is wrong , im a begginer

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