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richiejones24

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Posts posted by richiejones24

    Getting individual data from an array

    in PHP Coding Help

    Posted

    I am having problems displaying individual data I am using  $filename[0] to display the first data, any ideas where i am going wrong. 

     

    


<?php

$active_keys = array();
foreach($_POST['img'] as $key => $filename)
{
    if(!empty($filename))
    {
        $active_keys[] = $key;
    }

echo "". $filename[1];
}





?>

    Problems with an Array

    in PHP Coding Help

    Posted

    I've tried changing my form to this, but its still not working  :(

     

    <!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=utf-8" />

    <title>Untitled Document</title>

    </head>

     

    <body>

    <form id="form1" name="form1" method="post" action="testd.php">

      <p>

        <label>

    <input type="text" name="img[]"><br>

    <input type="text" name="img[]"><br>

    <input type="text" name="img[]"><br>

    <input type="text" name="img[]"><br>

    <input type="text" name="img[]"><br>

        <label>

          <input type="submit" name="Submit" id="Submit" value="Submit" />

        </label>

      </p>

    </form>

    </body>

    </html>

     

     

    but i still keep getting the following error:

     

    Warning: Invalid argument supplied for foreach() in /homepages/22/d378569747/htdocs/testd.php on line 4

     

    Problems with an Array

    in PHP Coding Help

    Posted

    ye this is the form i am going to use

     

     

    <!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=utf-8" />

    <title>Untitled Document</title>

    </head>

     

    <body>

    <form id="form1" name="form1" method="post" action="">

      <p>

        <label>

          <input type="text" name="img[]" id="img[]" />

        </label>

      </p>

      <p>

        <input type="text" name="img2[]" id="img2[]" />

      </p>

      <p>

        <input type="text" name="img3[]" id="img3[]  " />

        <label>

          <input type="submit" name="Submit" id="Submit" value="Submit" />

        </label>

      </p>

    </form>

    </body>

    </html>

     

     

     

     

    Problems with an Array

    in PHP Coding Help

    Posted

    I am trying my hand at using an array i have never done it before, and I am having some problems getting it work work. if someone could point me in the right direction?

     

    Thanks 

    

<?php

$active_keys = array();
foreach($_GET['img'] as $key => $filename)
{
    if(!empty($filename))
    {
        $active_keys[] = $key;
    }
}



?>

    data not being added to MySql DB

    in PHP Coding Help

    Posted

    The script below used to work fine but the HTTP_REFERER was unreliable so i added a re-direct if the HTTP_REFERER was unavailable and now nothing is being added to the MySQL BD, the script runs ok ie it sends the user back to where they come from, any ideas what could be wrong? 

     

    
<?php
require("../include/mysqldb.php");
$uinwish = $_GET['uinwish'];
$wishrefer = $_SERVER['HTTP_REFERER'];
$wishdate  = mktime(0, 0, 0, date("m"), date("d")+3, date("Y"));
$rand_wish_cookie = $_COOKIE["wishtracking"];

$con = mysql_connect("$dbhost","$dbuser","$dbpass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("$dbame", $con);

//DB Stuff start code below

//check if the uin and cookie have already been stored together.
$sql = "select * from Wish_list_guest where usercookie='$rand_wish_cookie' and uinwish='$uinwish'";
$result = mysql_query($sql);


$row = mysql_fetch_row($result);
if (mysql_num_rows($result)!= 1) {

$issetvar = "1";

if (!isset ($wishrefer)){

$host  = $_SERVER['HTTP_HOST'];
$extra = 'profile.php?uin=';
header("Location: http://$host/$extra$uinwish");
exit;
}

if (isset ($wishrefer)){

header("Location: $wishrefer");
exit;
}
}

if ($issetvar = 1) {

mysql_query("INSERT INTO Wish_list_guest (usercookie, date_delete, uinwish, referwish)
VALUES ('$rand_wish_cookie', '$wishdate', '$uinwish', '$wishrefer')");

mysql_close($con);


if (!isset ($wishrefer)){

$host  = $_SERVER['HTTP_HOST'];
$extra = 'profile.php?uin=';
header("Location: http://$host/$extra$uinwish");
exit;
}

if (isset ($wishrefer)){

header("Location: $wishrefer");
exit;
}
}


?>

    Can you put an if and an else inside of an if?

    in PHP Coding Help

    Posted

    hello, might be a stupid question but can you put an if and else inside of if brackets like i have done in this script?

    
<?php
require("../include/mysqldb.php");
$uinwish = $_GET['uinwish'];
$wishrefer = $_SERVER["HTTP_referer"];
$wishdate  = mktime(0, 0, 0, date("m"), date("d")+3, date("Y"));
$rand_wish_cookie = $_COOKIE["wishtracking"];

$con = mysql_connect("$dbhost","$dbuser","$dbpass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("$dbame", $con);


$sql = "select * from Wish_list_guest where usercookie='$rand_wish_cookie' and uinwish='$uinwish'";     //search db has the profile already been added to favorites
$result = mysql_query($sql);


$row = mysql_fetch_row($result);
if (mysql_num_rows($result)!= 1) {    //if 1 of more results found profile already added redirect back!


//Re-direct back script!!
if (!isset ($wishrefer)){

$host  = $_SERVER['HTTP_HOST'];
$uri   = rtrim(dirname($_SERVER['PHP_SELF']), '/\\');
$extra = 'profile.php?uin=';
header("Location: http://$host/$extra$uinwish");
exit;

}


else { header("Location: $wishrefer");
exit;
}

//end redirect script

else {

mysql_query("INSERT INTO Wish_list_guest (usercookie, date_delete, uinwish, referwish)
VALUES ('$rand_wish_cookie', '$wishdate', '$uinwish', '$wishrefer')");


mysql_close($con);

}

if (!isset ($wishrefer)){

$host  = $_SERVER['HTTP_HOST'];
$uri   = rtrim(dirname($_SERVER['PHP_SELF']), '/\\');
$extra = 'profile.php?uin=';
header("Location: http://$host/$extra$uinwish");
exit;

}


else {
header('Location: $wishrefer');
exit;

}


?>

    Displaying MySql Results table side by side

    in PHP Coding Help

    Posted

    need a little help guys! I use the script below to display profile images, trouble is it shows 1 on top of the other, and i need it to double up 2 profile images on top of 2 profile images ect any ideas how i can do this. 

     

    
require("./include/mysqldb.php");

$con = mysql_connect("$dbhost","$dbuser","$dbpass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("$dbame", $con);
$result = mysql_query("SELECT * FROM Search_profiles_up WHERE upgrade_one ='1' ORDER BY RAND() LIMIT 40");


print "<table width=\"293\" height=\"111\" border=\"0\">
  <tr>\n";

while($row = mysql_fetch_array($result))
  {
print "<td width=\"142\"><img src=" . $row['search_small_image'] . " width=\"144\" height=\"169\" /></td>\n";
print "  </tr>\n";
print "  <tr>  \n";
print "<td>" . $row['star'] . "</td>\n";
print "  </tr>\n";
print "  <tr>\n";
print "<td>" . $row['username_search'] . "</td>\n";
print "  </tr>\n";
print "  <tr>  \n";
print "<td>" . $row['phone_search'] . "</td>\n";
print "  </tr>   \n";

  }
print "</table>";

mysql_close($con);
?>

    Use results from MySql Search to Search another MySql Database can it be done?

    in PHP Coding Help

    Posted

    Why would you make a 2nd table, when you could just add a status column to the users table to indicate whether they are "upgrade" users?

    Its complicated but i cant do that it has to run on its own table.

     

    What exactly is complicated?

     

    The User profiles are spread over 5 tables and i use  MySql Join and the username to pull all the information from the tables to make the profile. If i added the upgrade option to one of the tables all i would be able to to is get the information from that table which would be useless i need the username to get the rest of the profile from the different tables.

    Use results from MySql Search to Search another MySql Database can it be done?

    in PHP Coding Help

    Posted

    Something like this?

     

    
$query1 = mysql_query("SELECT key FROM db1 WHERE something = 'something'") or die(mysql_error());

if( mysql_num_rows($query1) >= 1 ) {

   while( $row = mysql_fetch_array($query1) ) {
      $searchKey = $row['key'];
   }

   $query2 = mysql_query("SELECT key FROM db2 WHERE something = '$searchKey'") or die(mysql_error());

} else { 

   echo "nothing found";

}

     

    Thanks For the Reply!

    This is not exactly what i need as the original search will bring multiple results (user names) so the second search I need it to do something like this ("SELECT key FROM db2 WHERE something = '$searchKey' OR something = '$searchKey' OR something = '$searchKey' OR something = '$searchKey'  ")

     

    in some kind of array ive tried using "WHILE" and "FOREACH" but i cant figure out how to do it or whether it can be done.   

    Use results from MySql Search to Search another MySql Database can it be done?

    in PHP Coding Help

    Posted

    Thanks for your reply's, let me clarify what i am trying to do, I have a dating site and i have added the option to upgrade your profile so it is displayed on the homepage so I have created a new MySql table with the users username and the upgrade options they have selected. so what i won't to do is search the upgrade table (SELECT * FROM upgrade WHERE upgrade_one = '1') which will give me all the user names of all the user who have upgraded, then i won't to search the user table with the username (i got from the first search) to pull there profiles of the upgraded users.       

    Print Once While in Loop

    in PHP Coding Help

    Posted

    I am trying to get the code at the bottom of the script to print just once during the loop but it either doesn't print at all or repeats with the loop im am using if (!$i++) to print once and i works the first time i use it.

     

     
foreach($uploadFilename as $key => $myvar)
{
if (!$i++) print "<!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=utf-8\" />
<title>Upload Complete....</title>
</head>

<body>

<body onload=\"document.forms.formname.submit);\">

<form id=\"formname\" name=\"form1\" method=\"post\" action=\"reg5.php\">\n";


echo "<input type=\"hidden\" name=\"image$key\" value=\"";
echo end(explode('/',$myvar));
echo "\">\n";

if (!$i++) print "</form>
</body>
</html>\n";


}

    Multiple Image Upload Help

    in PHP Coding Help

    Posted

    I am trying to write a script to handle multiple image uploads however i cant get it to work i am a total novice in php so please be kind in your replys :-)

     

    
<?php
     
    
$allowed_filetypes = array('.jpg','.gif','.bmp','.png','.JPG','.PNG','.BMP','.GIF'); //These will be the types of file that will pass the validation.
    
$upload_path = '../pic_upload/'; // The place the files will be uploaded to.
     
    
foreach ($_FILES["pictures"]["error"] as $key => $error)
    
{
    
$filename = $_FILES['pictures']['name'][$key];
    
$ext = substr($filename, strpos($filename,'.'), strlen($filename)-1);
     
    
if(!empty($filename))        //is it empty
    
{
    
if ($_FILES["pictures"]["size"][$key] >= 9000000)
    
 die ('Sorry the picture is too Big!');
    
}
     
    
if(!empty($filename))        //Is it empty
    
{
    
if(!in_array($ext,$allowed_filetypes))                                           //Is the file Allowed
    
      die('The file you attempted to upload is not allowed! Pictures only!.');
    
}
     
    
if ($error == UPLOAD_ERR_OK) {
    
        $tmp_name = $_FILES["pictures"]["tmp_name"][$key];
    
        $name = $_FILES["pictures"]["name"][$key];
    
        move_uploaded_file($tmp_name, "$upload_path/$ran$name");
     
    
if (!$i++) print "<!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=utf-8\" />
    
<title>Untitled Document</title>
    
</head>
     
    
<body>
    
<div align=\"center\"><img src=\".../image/8-1.gif\" width=\"100\" height=\"100\" /></div>
    
<form id=\"form1\" name=\"form1\" method=\"get\" action=\"reg.php\">
    
<div align=\"center\">
    
        <input type=\"submit\" name=\"button\" id=\"button\" value=\"Submit\" />
    
        <input type=\"hidden\" name=\"reg\" value=\"6\">
    
  <p>\n";
     
    
print "<input type=\"hidden\" name=\"pic$key\" value=\"$ran$name\">\n";
     
     
    
if (!$i++) print "</p>
    
  <p>
    
    <label>
    
      </div>
    
    </label>
    
  </p>
    
</form>
    
<p> </p>
    
</body>
    
</html>\n";
     
    
} }
    
?>  
    
[code=php:0]
    

						
    
					
    
					
						
    
							

	
    	
		
    
			
				
				
    
					
				
    
			
			
			
		
    
	
    

						
    
					
				
    		
			
    
		
    
	
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			File Upload Problems Noob
		
	
    
	
    in PHP Coding Help
    

    
				
    
				
				
    
					
    
						
						
    
							Posted 
							
														
						
    
						

						
    
							
							

    

    
The following is the basis for the code you are using (gotten from the php.net documentation.) You would put your application level validation logic inside the if(){...} statement, because you can only access and check the uploaded file information when the file was actually uploaded without any errors -
    

     
    

    

    

    <?php
foreach ($_FILES["pictures"]["error"] as $key => $error) {
    if ($error == UPLOAD_ERR_OK) {
        $tmp_name = $_FILES["pictures"]["tmp_name"][$key];
        $name = $_FILES["pictures"]["name"][$key];
        move_uploaded_file($tmp_name, "data/$name");
    }
}
?>
    

    

    

     
    

    
The code you have now makes no sense.
    

    

    

    

     
    

    
Totally Confused now :-s 
    

     
    

    
The code works and uploads the files ok however, the script is for multiple file upload and if the user uploads a file to form field 3 for example and form field 1 is left empty the script dies and no files are uploaded i belive it is this part that is killing it 
    

       if (empty($filename)) {                                                       //Is the field empty
    die ('one of the fields is empty but the rest have been upload');
}

    

    

    

     
    

    
any ideas how to get round this?
    

						
    
					
    
					
						
    
							

	
    	
		
    
			
				
				
    
					
				
    
			
			
			
		
    
	
    

						
    
					
				
    		
			
    
		
    
	
    24. 

	
	
  24. 
		
    

			

			
    	
				
    
					
    
						
						
      
							
							
						
    
					
    
					



    
	
    
		
			
		
			File Upload Problems Noob
		
	
    
	
    in PHP Coding Help
    

    
				
    
				
				
    
					
    
						
						
    
							Posted 
							
														
						
    
						

						
    
							
							

    Total Nood to PHP trying to write a multiple image upload script, however having problem as the image validation kills the script and it wont upload any ideas?
    

     
    

    

    

    $allowed_filetypes = array('.jpg','.gif','.bmp','.png'); //These will be the types of file that will pass the validation.
$upload_path = '../pic_upload/'; // The place the files will be uploaded to.

foreach ($_FILES["pictures"]["error"] as $key => $error)

{

$filename = $_FILES['pictures']['name'][$key];
$ext = substr($filename, strpos($filename,'.'), strlen($filename)-1);

if ($_FILES["pictures"]["size"][$key] >= 700000) {      //Check the Size
   die ('size to big');
}

if (empty($filename)) {                                                       //Is the field empty
    die ('one of the fields is empty but the rest have been upload');
}

if(!in_array($ext,$allowed_filetypes))                                           //Is the file Allowed
      die('The file you attempted to upload is not allowed.');

if ($error == UPLOAD_ERR_OK) {                                                  //Upload File check for errors
       echo"$error_codes[$error]";
       move_uploaded_file($_FILES["pictures"]["tmp_name"][$key],$upload_path . $_FILES["pictures"]["name"][$key]) or die("Problems with upload");
   }


    

    

    

     
    

    
MOD EDIT: code tags added.
    

						
    
					
    
					
						
    
							

	
    	
		
    
			
				
				
    
					
				
    
			
			
			
		
    
	
    

						
    
					
				
    		
			
    
		
    
	
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