PHP - Limit Upload Capacity For User (php/mysql)

OK so i have a textfield and i have everything going to the database and then displaying on the page.
Now is there a way to display code on the page but disable it from executing?
There are some break tags in there so i would still like them to be executed on the page.

I'm trying to add a file size limit to my upload form-code (below)

How can I limit the amount of time my while loop to only show the first 4 rows of my SQL table?   I am using this code to loop my whole table:  

while($row = $results->fetch(PDO::FETCH_ASSOC))
                        {
                        echo '
                        <li>Mileage: '.number_format($row["Mileage"]).'</li>
                        <li>Engine size: '.$row["EngineSize"].'cc</li>
                        ';} ?>

I only want to loop through the first 4 rows of my SQL table, I then want to duplicate the same code but start at the 5th row until the 8th row of the table, how can I do this?   Thanks, Nick

Hi,

I am getting frustrated beyond belief at the moment with trying to get a very simple script to run, I am using PHP 5.3.3 and MySQL 5.1 on a Win2k8 server with IIS7.5. Basically my script is connecting to a local database, running a single select query, returning those rows and building up a string from them.

The problem is that I am receiving complete BS responses from PHP that the access is denied for the user being specified. This is complete rubbish since the user can connect via mysql, sqlyog, ASP.NET MVC without issue but for some bizarre reason it is not working via PHP.

The code for the script is here :

Code: [Select]
<?php

$mysql = mysql_connect('127.0.0.1:3306', 'myuser', 'mypass', 'mydatabase');


if (!$mysql) {
  die(mysql_error());
  $content = "<nobr></nobr>";
} else {
  
  $result = mysql_query('SELECT * FROM tblEventGroup'); 
  $content = "<nobr>";
  
  if ($result) {     
      while($row = mysql_fetch_assoc($result)) {
       $content .= "<span>";
       $content .= $row['GroupName'];
       $content .= "</span>";
       $content .= "<a href=\"../Event/EventSearch?groupid=";
       $content .= $row['GroupId'];
       $content .= "\" target=\"_blank\">Book here</a> ";
      }
  }
  
  mysql_close($mysql);
  $content .= "</nobr>";

}

?>
I cannot for the life of me understand what the problem is, the return error is

Access denied for user 'myuser'@'localhost' (using password: YES)

I feel like a tard for having to ask this, but I've spent way too much time trying to figure this out, so i've brought it to you guys.

I have run into a an error running a while loop off a query:

$rowcount= mysql_query("SELECT COUNT(*) FROM `shoutbox`");
$count = mysql_fetch_array($rowcount);
$count = $count['COUNT(*)'];
$said = mysql_query("SELECT * FROM shoutbox ORDER BY date ASC LIMIT ('$count' - 5),'$count' ");

while($row = mysql_fetch_array( $said )){  //line 14 in the code

echo $row['shouter'].':  '.$row['contents'].'<br />'; 

}


Code: [Select]
mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/mahngiel/public_html/shout.php on line 14This script works when I remove the limit counts, but then I have to descend the order and things are not in the position I want them in.  I'm sure the problem is the limit vars, but dunno which way to go to fix it. thanks

Hi. I'm trying to make a sort of a forum, and this is the part of the code that needs to be changed, only I don't know how.
Code: [Select]
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name ORDER BY id DESC";
// OREDER BY id DESC is order result by descending
$result=mysql_query($sql);
?>
<table width='50%' border="0" align="center" cellpadding="3" cellspacing="1">
<?php
$limit=10;
$page=0;
$numrows = mysql_num_rows($result);
$pages = intval($numrows/$limit);
if ($numrows % $limit) {
$pages++;}
$current = ($page/$limit) + 1;
if (($pages < 1) || ($pages == 0)) {
$total = 1;}
else {
$total = $pages;}
$result = mysql_query("SELECT * FROM $tbl_name ORDER BY id DESC LIMIT $page,$limit");
while($rows=mysql_fetch_array($result)){?>
<tr><td><font size="+2"><a href="view_topic.php?id=<? echo $rows['id']; ?>"><? echo $rows['topic']; ?></a><BR></font></td>
<td width='0*'><? echo $rows['datetime']; ?></td></tr>
<tr><td><? echo substr($rows['detail'], 0, 250)."..."; ?>&nbsp;&nbsp;<a href="view_topic.php?id=<? echo $rows['id']; ?>"><? echo "Citeste mai mult";} ?></a>
</td></tr>
<tr><tr><td>
<?
if ($page != 0) { // Don't show back link if current page is first page.
$back_page = $page - $limit;
echo("<a href=\"$PHP_SELF?query=$query&page=$back_page&limit=$limit\">back</a>    \n");}
for ($i=1; $i <= $pages; $i++) // loop through each page and give link to it.
{
 $ppage = $limit*($i - 1);
 if ($ppage == $page){
 echo("<b>$i</b>\n");} // If current page don't give link, just text.
 else{
 echo("<a href=\"$PHP_SELF?query=$query&page=$ppage&limit=$limit\">$i</a> \n");}
}
if (!((($page+$limit) / $limit) >= $pages) && $pages != 1) { // If last page don't give next link.
$next_page = $page + $limit;
echo("    <a href=\"$PHP_SELF?query=$query&page=$next_page&limit=$limit\">next</a>");}
?>
</td></tr></tr><?php
mysql_close();
?>
</table>

I made a database with 4 fields and 11 entries. The script shows the last 10 entries just as it should, but when I click on the 'next' button, nothing happens (it should show the first entry, but it just refreshes the page, and still the last 10 entries/1st page are shown). I think it has something to do with the link, but I can't figure out what exactly.
Can someone help, or at least give me a hint? Thanks in advance. Bye.

Let's say I have 1 table and 100 rows in this table. The table is simple and only contains an Auto Increment ID field and a load_count field.

I am building a commenting system, it's sort of like facebooks comments. I have it right now so it loads the first last 3 comments in the database, and then there's a button to view all comments. I have it working fine, except when you add another comment to it, and then click the view all comments,


This is the script that loads onload showing the last 3 comments


$limit  = 3;

$query5 = mysql_query("SELECT * FROM sub_comments WHERE status_id='$id' ORDER BY id DESC LIMIT $limit");



This is the php script that runs when you click view more




//($start_id would be 3 in this case)

$query5 = mysql_query("SELECT * FROM sub_comments WHERE status_id='$comment_id' ORDER BY id DESC LIMIT $start_id, 1000");

so to try to show you what's happening:

OnLoad--------------------------------|---When you add more comments, (let's just say u add 1 comment before u click view comments)----
comment 9   --------------------------|   comment 10
comment 8   --------------------------|   comment 9
comment 7   --------------------------|   comment 8
------------------------------------------|   comment 7
----to be loaded onclick-------------|------What's loaded onclick---
comment 6   --------------------------|   comment 7
comment 5   --------------------------|   comment 6
comment 4   --------------------------|   comment 5
comment 3   --------------------------|   comment 4
comment 2   --------------------------|   comment 3
comment 1   --------------------------|   comment 2
------------------------------------------|   comment 1


(see how "comment 7" is being shown twice in a row)

I have these two tables... schedule (gameid, homeid, awayid, weekno, seasonno) teams (teamid, location, nickname) This mysql query below gets me schedule info for ALL 32 teams in an array...

$sql = "SELECT
h.nickname AS home,
a.nickname AS away,
h.teamid AS homeid,
a.teamid AS awayid,
s.weekno
FROM schedule s
INNER JOIN teams h ON s.homeid = h.teamid
LEFT JOIN teams a ON s.awayid = a.teamid
WHERE s.seasonno =2014";
$schedule= mysqli_query($connection, $sql);
if (!$schedule) {
die("Database query failed: " . mysqli_error($connection));
} else { 
// Placeholder for data
$data = array();
while($row = mysqli_fetch_assoc($schedule)) {
if ($row['away'] == "") {$row['away']="BYE";}
$data[$row['homeid']][$row['weekno']] = $row['away'];
$data[$row['awayid']][$row['weekno']] = '@ '.$row['home'];
}
}

However, I only want to get info for one specific team, which is stored in the $teamid variable. This should be very easy, right? I have tried multiple things, including this one below (where I added an AND statement of "AND (h.teamid=$teamid OR a.teamid=$teamid)"), but this one still outputs too much...

$sql = "SELECT
h.nickname AS home,
a.nickname AS away,
h.teamid AS homeid,
a.teamid AS awayid,
s.weekno
FROM schedule s
INNER JOIN teams h ON s.homeid = h.teamid
LEFT JOIN teams a ON s.awayid = a.teamid
WHERE s.seasonno =2014
AND (h.teamid=$teamid OR a.teamid=$teamid)";
$schedule= mysqli_query($connection, $sql);
if (!$schedule) {
die("Database query failed: " . mysqli_error($connection));
} else { 
// Placeholder for data
$data = array();
while($row = mysqli_fetch_assoc($schedule)) {
if ($row['away'] == "") {$row['away']="BYE";}
$data[$row['homeid']][$row['weekno']] = $row['away'];
$data[$row['awayid']][$row['weekno']] = '@ '.$row['home'];
}
}

Below is the array that the above outputs. In a nutshell, all I want is that 1st array ([1]) which has, in this example, the Eagles full schedule. It's not giving me too much else and I guess I could live with it and just ignore the other stuff, but I'd rather be as efficient as possible and only get what I need...

Array
(
[1] => Array
(
[1] => Jaguars
[2] => @ Colts
[3] => Redskins
[4] => @ 49ers
[5] => Rams
[6] => Giants
[7] => BYE
[8] => @ Cardinals
[9] => @ Texans
[10] => Panthers
[11] => @ Packers
[12] => Titans
[13] => @ Cowboys
[14] => Seahawks
[15] => Cowboys
[16] => @ Redskins
[17] => @ Giants
)

[27] => Array
(
[1] => @ Eagles
)

[28] => Array
(
[2] => Eagles
)

[4] => Array
(
[3] => @ Eagles
[16] => Eagles
)

[14] => Array
(
[4] => Eagles
)

[15] => Array
(
[5] => @ Eagles
)

[3] => Array
(
[6] => @ Eagles
[17] => Eagles
)

[] => Array
(
[7] => @ Eagles
)

[16] => Array
(
[8] => Eagles
)

[25] => Array
(
[9] => Eagles
)

[11] => Array
(
[10] => @ Eagles
)

[7] => Array
(
[11] => Eagles
)

[26] => Array
(
[12] => @ Eagles
)

[2] => Array
(
[13] => Eagles
[15] => @ Eagles
)

[13] => Array
(
[14] => @ Eagles
)

)

I have a commenting system and i have a limit of a certain number of comments to be shown. What i want to do is have a button on the bottom of the page at the end of the comments that are showing and when you click it ajax loads the next certain number of of rows (but not all of them),and then you click it again and it shows more of them, etc.

So for example.

comment 1
comment 2
comment 3
comment 4
--click button--(loads 4 more)---
comment 5
comment 6
comment 7
comment 8
--click button--(loads 4 more)--
comment 9
comment 10
comment 11
comment 12

etc. until there are no more rows.

what's the best way to do this? (I know how to do the ajax and all, i just need help with the script to select the rows)

Thanks.

Hello all -

I'm fairly new to PHP and have been following some online tutorials to learn more, but hit a wall concerning a form that would allow a user to upload an image to his/her specified directory.

Basically, I'd like the user to have to put in a password to upload. This "password" would actually just be the name of their directory on the server, so if a user put in "michael83" in as their password, the image would upload to "http://www.mysite.com/images/uploaded/michael83/".

Here's my code so far:



<form name="newad" method="post" enctype="multipart/form-data" action="upload.php" onSubmit="return validate_form ( );">
<table>
<tr><td><input type="file" name="image"></td></tr>
<tr><td>&nbsp;</td></tr>
<tr><td>Password:</td></tr>
<tr><td><input type="text" name="password"></td></tr>
<tr><td>&nbsp;</td></tr>
<tr><td><input name="Submit" type="submit" value="Upload"></td></tr>
</table>
</form>

<?php

define ("MAX_SIZE","1536");

function getExtension($str) {
$i = strrpos($str,".");
if (!$i) { return ""; }
$l = strlen($str) - $i;
$ext = substr($str,$i+1,$l);
return $ext;
}

$errors=0;

if(isset($_POST['Submit']))
{

$image=$_FILES['image']['name'];
$dir=$_POST['username'];

if ($image)
{

$filename = stripslashes($_FILES['image']['name']); 

$extension = getExtension($filename);
$extension = strtolower($extension);

if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "pdf") && ($extension != "gif"))
{

echo '<h4>Sorry, your file is an unknown extension.</h4>';
$errors=1;
}
else
{

$size=filesize($_FILES['image']['tmp_name']);

if ($size > MAX_SIZE*1024)
{
echo '<h4>Sorry, you have exceeded the size limit.</h4>';
echo '<p>If you need more help with this, please <a href="#">contact us</a> directly.</p>';
$errors=1;
}

$newname="images/uploaded/".$image_name;

$copied = copy($_FILES['image']['tmp_name'], $newname);
if (!$copied)
{
echo '<h4>Oops, looks like the upload was unsuccessfull.</h4>';
echo '<p>If you continue to have problems, please <a href="#">contact us</a>.</p>';
$errors=1;
}}}}

if(isset($_POST['Submit']) && !$errors)
{
echo "<h4>Your file was uploaded successfully!</h4><br><br>";
echo '<a href="http://www.mysite.com/' . $newname . '">http://www.mysite.com/' . $newname . '</a><br><br>';
}

?>



Any help would be greatly appreciated. Many thanks in advance!

EDIT Note: I would be the one setting the directories up, so if the user enters a "password" (directory) that doesn't exist, the form would return an error.

i have my upload process working that uploads documents to the server and then displays in onto the user page like this:

<?php 
						function find_all_files($dir) 
						{ 
    							$root = scandir($dir); 
    							foreach($root as $value) 
    							{ 
    				    				if($value === '.' || $value === '..') {continue;} 
        							if(is_file("$dir/$value")) {$result[]="$dir/$value";
        							continue;
        							} 
        							foreach(find_all_files("$dir/$value") as $value) 
        							{ 
           							 $result[]=$value; 
       								 } 
	   					 	} 
  					 		 return $result; 
						} 
							$fileupload = 'fileupload';
							$getem = find_all_files($fileupload);
							foreach($getem as $key => $value)
							{
								echo '<a href="'.$value.'">'.$value.'</a><br />';
							}
						?>
						
						
						
						<?php
						if($handle = opendir('members/')) {
						while (false !== ($entry = readdir($handle))) {
							if($entry != "." && $enrty != "..") {
							echo "<a href='download.php?file=".$entry."'>".$entry."</a>\n";
							}
							}
							closedir($handle);
							}
							?>

here they can download the files to their computer however is there a way to only display the file of the user that is logged in through their session?      

...and the file is to be copied into a folder created by me. I was wondering what the file path would be for Windows, since it does not seem to recognise the file path that I have chosen:

Upload Form code:
Code: [Select]
<html>
<head><title>Car Accident Report Uploader</title></head>
<body>
<form action="fileuploadform.php" method="post" enctype="multipart/form-data">
<input type="file" size="45" name ="file"><br>
<input type="submit" value = "Upload Car Accident Report">
</form>
</body>
</html>

Copying Script Code:
Code: [Select]
<?php
if ($_FILES['file']['name'] !="") {
copy ($_FILES['file']['tmp_name'], 'C:\xampp\htdocs\rcm' . $_FILES['file']['name'])
or die ("Could not copy Car Accident Report Form");
}
else {die ("No Car Accident Report Form has been chosen");}
?>
<html>
<head><title>Car Accident Report Form Upload Complete</title></head>
<body><h3>Car Accident Report Form has been uploaded</h3>

<ul>
<li>Sent: <?php echo $_FILES['file']['name']; ?>
<li>Size: <?php echo $_FILES['file']['size']; ?> bytes
<li>Type: <?php echo $_FILES['file']['type']; ?>

<a href = "<?php echo $_FILES['file']['name']; ?>">
Click here to view the Car Accident Report Form</a>
</body>
</html>

If anyone can help me find the relevant file path, it would be appreciated!

Hi all,
I am quite new to PHP and MySql. I know how to upload a file a and save the path in the DB, but now I need to send an automatic email to the user when a new invoice is added to his directory. Please help. Thanks

I need a way a user can upload a file onto my website(it's a shared site, I believe, but I own the domain), I've tried using the cURL library but it seems a bit over my head at the moment. And everything I tried with it would not result in a file being uploaded into any folder.

What can I do what I need? I've searched around and never found anything that worked for me.