[[SOLVED] array still not working properly]

Also,,, the array will still not let a .jpg be uploaded..... gif bmp png upload just fine. I tried jpg and jpeg,, but still get the same thing,, no upload.  Oh yeh,, before I forget, how much data can be stored in a mysql db?

[[SOLVED] array still not working properly]

Sorry to bring this post back from the dead, but I was wondering if the line marked below can be eliminated? Couldn't I just have $fileType in the IF statement?

<?php
$allowed = array('image/bmp','image/x-png','image/jpg','image/gif');
$imageType = $fileType;    // this line here
if (!in_array($imageType,$allowed))
?>

[[SOLVED] array still not working properly]

everything works great except for jpg files. the array will not let me upload jpg files

<?php
$allowed = array('image/bmp','image/x-png','image/jpg','image/gif');
$imageType = $fileType;
if (!in_array($imageType,$allowed))
?>

[[SOLVED] array still not working properly]

Thanks a bunch. My headache is now gone.

[[SOLVED] array still not working properly]

Just trying to limit file types on upload. This is not hard to do(maybe it is). No matter what file I choose, i get the error saying that it is the wrong type of file. Here is the code,,,,again...

<?php
session_start();
include ("upload_db_info.php");
if (!empty($_POST['upload']))
{
extract($_POST);
   	if(isset($_POST['upload']) && $_FILES['upload_file']['size'] < 500000)
{
    	$fileName = $_FILES['upload_file']['name'];
       	$tmpName = $_FILES['upload_file']['tmp_name'];
       	$fileSize = $_FILES['upload_file']['size'];
       	$fileType = $_FILES['upload_file']['type'];
    	if ( file_exists($tmpName))
    	{
    		$content = file_get_contents($tmpName);
    	}
    }


$allowed = array('.gif','.bmp');
$fileName = $_FILES[$fileName]['name'];
$imageType = strtolower(substr($fileName,-4));
if (!in_array($imageType,$allowed))
{
   		unset($_SESSION['uploadcomplete']);
	$_SESSION['uploaderror'] = "<font color=red><font size=2>Please select a valid picture format under 500,000 bytes(.5 				 									 megabytes)";
	header("location: http://www.----------.com");
	exit();
}


$user = mysql_real_escape_string($user);      
$trade = mysql_real_escape_string($trade);
$picname = mysql_real_escape_string($picname);
$fileName = mysql_real_escape_string($fileName);
    $fileSize = (int)$fileSize;
    $fileType = mysql_real_escape_string($fileType);
    $content  = mysql_real_escape_string($content);
    $descrip = mysql_real_escape_string($_POST["descrip"]);
$trade = mysql_real_escape_string($_POST["trade"]);
$picname = mysql_real_escape_string($_POST["picname"]);
$query = "INSERT INTO UploadedFiles (name, size, type, content, user, descrip, trade, picname)VALUES('$fileName', 						 			 '$fileSize', '$fileType', '$content', '$user', '$descrip', '$trade', '$picname')";        
   	$result = mysql_query($query)or die (mysql_error());
    unset($_SESSION['uploaderror']);
$_SESSION['uploadcomplete'] = "Your picture was uploaded to our system.";
    header("location: http://www.------------.com");
    exit();
}
?>

[Up all night, still can't limit file types on upload]

Not really sure what you mean by that. I'll try that next.

[Up all night, still can't limit file types on upload]

Thanks guys, I'll give it a shot.

[Up all night, still can't limit file types on upload]

I would only like to be able to upload gif jpeg png jpg  Any other file type would get an error. Right now if the file is larger than .5mb, i will will recieve an error, which is good, that is working correctly. Just the files types I can't seem to get. Here is the script again.

<?php
session_start();
include ("upload_db_info.php");

if (!empty($_POST['upload']))
{
extract($_POST);
   		
if(isset($_POST['upload']) && $_FILES['upload_file']['size'] < 500000)
{
    	$fileName = $_FILES['upload_file']['name'];
        $tmpName = $_FILES['upload_file']['tmp_name'];
        $fileSize = $_FILES['upload_file']['size'];
        $fileType = $_FILES['upload_file']['type'];
    	if ( file_exists($tmpName) )
    	{
    		$content = file_get_contents($tmpName);
    	}
    }
    else
{
	unset($_SESSION['uploadcomplete']);
	$_SESSION['uploaderror'] = "<font color=red><font size=2>Please select a valid picture format under 500,000 bytes(.5 				 									 megabytes)";
	header("location: http://www.---------.com");
	exit();
}

$user = mysql_real_escape_string($user);      
$trade = mysql_real_escape_string($trade);
$picname = mysql_real_escape_string($picname);

$fileName = mysql_real_escape_string($fileName);
    $fileSize = (int)$fileSize;
    $fileType = mysql_real_escape_string($fileType);
    $content  = mysql_real_escape_string($content);
    $descrip = mysql_real_escape_string($_POST["descrip"]);
$trade = mysql_real_escape_string($_POST["trade"]);
$picname = mysql_real_escape_string($_POST["picname"]);

$query = "INSERT INTO UploadedFiles (name, size, type, content, user, descrip, trade, picname)VALUES('$fileName', '$fileSize',                                        '$fileType', '$content', '$user', '$descrip', '$trade', '$picname')";        
    $result = mysql_query($query)or die (mysql_error());
    
unset($_SESSION['uploaderror']);
$_SESSION['uploadcomplete'] = "Your picture was uploaded to our system.";
    header("location: http://www._________-.com");
    exit();
}
?>

[View images from mysql]

ok, here it is. i have a db with table that consists of pics, description and the username of who uploaded. i would like each user to view a page that shows which pic they uploaded. so i think i need to find all the images in the database with the field of their username. right? any ideas

[problem with echo]

bump ttt

[problem with echo]

with escaping the single quotes or even using double quotes, i don't get any errors, but when i click on the text field, my initial value is not removed.

[problem with echo]

sorry premiso,, i skipped your post,,, my bad

[problem with echo]

don't understand why that would make a difference. here is the code with the onfocus for the text boxes removed. everything loads just fine.

<table width="950" border="0" align="center" cellpadding="0" cellspacing="0">
  <tr>
    <th height="297" scope="col"><div id="header"><img src="images08/header.png" alt="header" width="421" height="47" /></div>
      <div id="apDiv1">
        <?php
        session_start();
        if(isset($_SESSION['auth']))
        {
      	  echo $_SESSION['message'];
        }
        else
        {
      	   echo '<form name="form1" method="post" action="login.php">
			 <input name="username" type="text" id="username" value="User ID"  />
			 <input name="password" type="password" id="password" value="Password"  />
			 <input type="submit" name="submit" id="submit" value="Login">
			 <br />
			 Not registered? <a href="register.php">Register Now!</a>
			 </form>';
  	    }
	    ?>
    </div></th>
  </tr>
</table>

i get the error when the 2 lines are changed to this

 

<input name="username" type="text" id="username" value="User ID" onfocus="this.value='';" />

<input name="password" type="password" id="password" value="Password" onfocus="this.value='';" />

 

[problem with echo]

i removed the php tags.here i what i got.. still an error.

<table width="950" border="0" align="center" cellpadding="0" cellspacing="0">
  <tr>
    <th height="297" scope="col"><div id="header"><img src="images08/header.png" alt="header" width="421" height="47" /></div>
      <div id="apDiv1">
        <?php
        session_start();
        if(isset($_SESSION['auth']))
        {
      	  echo $_SESSION['message'];
        }
        else
        {
      	   echo '<form name="form1" method="post" action="login.php">
			 <input name="username" type="text" id="username" value="User ID" onfocus="this.value='';"  />
			 <input name="password" type="password" id="password" value="Password" onfocus="this.value='';" />
			 <input type="submit" name="submit" id="submit" value="Login">
			 <br />
			 Not registered? <a href="register.php">Register Now!</a>
			 </form>';
  	    }
	    ?>
    </div></th>
  </tr>
</table>

still getting an error with this line

<input name="password" type="password" id="password" value="Password" onfocus="this.value='';" />

 

[problem with echo]

here is the full code for the table i have everything in

<?php // this start tag is not in my code,, just put it there for you
<table width="950" border="0" align="center" cellpadding="0" cellspacing="0">
  <tr>
    <th height="297" scope="col"><div id="header"><img src="images08/header.png" alt="header" width="421" height="47" /></div>
      <div id="apDiv1">
        <?php
        session_start();
        if(isset($_SESSION['auth']))
        {
      	  echo $_SESSION['message'];
        }
        else
        {
      	   echo '<?php 
	   		 <form name="form1" method="post" action="login.php">
			 <input name="username" type="text" id="username" value="User ID" onfocus="this.value='';"  />
			 <input name="password" type="password" id="password" value="Password" onfocus="this.value='';" />
			 <input type="submit" name="submit" id="submit" value="Login">
			 <br />
			 Not registered? <a href="register.php">Register Now!</a>
			 </form>';
  	    		 ?>
	}
	    ?>
    </div></th>
  </tr>
</table>
?> // same with this tag

now you can get a better look at why it is not working.

[problem with echo]

this doesn't work

echo '<?php 
	   		 <form name="form1" method="post" action="login.php">
			 <input name="username" type="text" id="username" value="User ID" onfocus="this.value='';"  />
			 <input name="password" type="password" id="password" value="Password" onfocus="this.value='';" />
			 <input type="submit" name="submit" id="submit" value="Login">
			 <br />
			 Not registered? <a href="register.php">Register Now!</a>
			 </form>';
  	    		 ?>

but this does work,, what is going on?

<?php
<form name="form1" method="post" action="login.php">
<input name="username" type="text" id="username" value="User ID" onfocus="this.value='';"  />
<input name="password" type="password" id="password" value="Password" onfocus="this.value='';" />
<input type="submit" name="submit" id="submit" value="Login">
<br />
Not registered? <a href="register.php">Register Now!</a>
</form>
?>

when i echo out the form i get an error

Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING, expecting ',' or ';' in D:\-----\-----\html\------\index.php on line 35

here is line 35

<input name="username" type="text" id="username" value="User ID" onfocus="this.value='';"  />

should something be different when echoing out the form while using the onfocus

 

 

[php for this?]

checking for the username and password would be inside of my login.php

i'm just looking for the login form if no one is logged in and a welcome statement if someone is logged in. this is what i have so far. i don't think that i am echoing out the form correctly though.

<?php
session_start();
if(isset($_SESSION['auth']))
{
      echo "Welcome $username  Log Out";    //Need to make Log Out a link
}
else
{
      echo "<form name="form1" method="post" action="login.php">
  <input type="text" name="username" id="username">
  <input type="text" name="password" id="password">
  <input type="submit" name="submit" id="submit" value="Login">
  </form>"
}
?>

[php for this?]

i know about sessions and <?php

                                    if(isset($_SESSION['loggedIn'])) {

                                    // show relevant information

                                    }else{

                                    // show login form

                                    }

                                    ?>

i see some code with EOD, is that something i should look into also

[php for this?]

i want to display the username password text fields if a user is not logged in.

and once the user is logged in, something like welcome $username would be there instead of the login text fields.

[looking for some help, i don't want to just use someones script]

so far i have a page where users can register and login. i would like the user to be able to upload a picture, name of the picture, and a small description of that picture into a mysql database. i see a lot of scipts online that i could just copy and use but i dont feel that is the correct way of learning. could someone point me in the right direction? thanks again.

[anyone see what is wrong with this simple image upload script]

Unable to open file at location...... that is the error now.

<?php 
include "msdata.php";

$PSize = filesize($location); 
$fh = fopen($location, "r") or die("Unable to open file at location.");
$location = addslashes(fread($fh, $PSize)); 

mysql_connect($host,$username,$password) or die("Unable to connect to MySQL."); 
mysql_select_db($db_name) or die("No database found."); 
mysql_query("INSERT INTO $tbl_name(pic)VALUES('$location')") or die("Operation unable to perform."); 

?>

[anyone see what is wrong with this simple image upload script]

Warning: fread(): supplied argument is not a valid stream resource in D:\Hosting\3388298\html\mylahstone\upload.php on line 5

 

 

what exactly do these 2 lines accomplish??

$PSize = filesize($location);

$location = addslashes(fread(fopen($location, "r"), $PSize));

[anyone see what is wrong with this simple image upload script]

<?php 
include "msdata.php";

$PSize = filesize($location); 
$location = addslashes(fread(fopen($location, "r"), $PSize)); 

mysql_connect($host,$username,$password) or die("Unable to connect to MySQL."); 
mysql_select_db($db_name) or die("No database found."); 
mysql_query("INSERT INTO $tbl_name(pic)VALUES('$location')") or die("Operation unable to perform."); 

?>

[upload script question]

sorry,,,,, but any suggestions for a good starting point. i guess it would be something similar to ebay.

[upload script question]

Could anyone point me in the right direction for a decent upload script. And then also linking a member back to what he/she uploaded. After a user uploads some pictures(picture, name, description of pic), they could go to a page to see just what they uploaded. Or they could also just view all the pics that all the users uploaded. Is this just way to difficult to do....When a user uploads a picture they get to pick from a list of categories that the picture would go in.