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me102

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Posts posted by me102

  1. Hello I have the following code from php websites

    $myfile = fopen("/config.php", "a+") or die("Unable to open file!");
    
    while(!feof($myfile)) {
      echo fgets($myfile) . "<br>";
     
      $txt = "Mickey Mouse\n";
      fwrite($myfile, $txt);
    }
    

    inside config.php is

    <?php  
      define('script_path', 'files');
    ?>

    How can I replace files with another value? any help would be great thanks.

  2. Hello I have a rating script what I need is a faster and better way of getting the title from the first query so i dont have to run 2 queries.

     

     

    $q1 = "select ItemID, sum(Rating) as mr from dd_rating group by ItemID order by mr desc limit 0,4";
    $r1 = mysql_query($q1) or die(mysql_error());
    while($a1 = mysql_fetch_array($r1))
    {
    $q2 = "select ItemTitle from dd_items where ItemID = '$a1[0]' ";
    $r2 = mysql_query($q2) or die(mysql_error());
    while($a2 = mysql_fetch_array($r2)){

  3. Hello I am having a little bit of a problem when I try to upload a directory and its subdirectorys i keep getting errors because the done exist but how would I go about making them,

     

    This is what I have,

     

    <?php
          
          
    
        
        //Source Files, Where to put them
        $src_dir = "/source_codes/";
        $dst_dir = "/haha/";
        
        // Start the ftp connection 
        $conn_id = ftp_connect($ftp_server) or die("<span style='color:#FF0000'><h2>Couldn't connect to $ftp_server</h2></span>");  
        $login_result = ftp_login($conn_id, $user, $passwd); 
      
         // check connection
         if ((!$conn_id) || (!$login_result)) {
              echo "FTP connection has failed!";
              echo "Attempted to connect to $ftp_server for user $ftp_user_name";
              die;
         } else {
              echo "<br>Connected to $ftp_server, for user $user<br>";
         }
         
         
         
        function ftp_copy($src_dir, $dst_dir) {
         global $conn_id;
    
        $d = dir($src_dir);
    
        while($file = $d->read()) {
            if ($file != "." && $file != "..") {
                if (is_dir($src_dir."/".$file)) {
                    
                    if (!@ftp_chdir($conn_id, $dst_dir."/".$file)) {
                        ftp_mkdir($conn_id, $dst_dir."/".$file);
                    }
    
                      ftp_copy($src_dir."/".$file, $dst_dir."/".$file);
                      
                }else {
    
                $upload = ftp_put($conn_id, $dst_dir."/".$file, $src_dir."/".$file, FTP_BINARY);
                
                 // check upload status
                 if (!$upload) {
                     echo "FTP upload has failed!";
                 } else {
                     echo "Uploaded $source_file to $ftp_server as $destination_file";
                 }
    
                }
            }
        }
    
    $d->close();
    }
    
         
    
    ftp_copy($src_dir, $dst_dir);
    
    
    // close the FTP stream
    ftp_close($conn_id);
    ?>
    
    
    
    

  4. Basicly theres input box where people put there database details then on the next page it check the database details and if everythings ok it adds the details to the database. My problem is the 2 database connections are not working together and I was wondering what I could do?

     

    include(config.php)

     

    $testconnect = @mysql_connect($mysql_host,$mysql_username,$mysql_password);
    
    if(!$testconnect ){    echo "error can not connect to database";  }else{ $add = mysql_query(" INSERT INTO DATABASE () VALUES () }

     

    config.php

    
    $database_a = mysql_connect($host,$username,$password);
    
    $database_1 = mysql_select_db($database, $database_a);
    
    

     

    Any ideas?

  5. Hello I have a script thats checks to see if I am able to connect to the database or not. If I am unable to it should display an image called refused and if I am able to connect it should display the accept image. Everything works fine in my script but I cant seem to get ride of my mysql warning saying its unable to connect to the database.

     

    Heres my code

         
    
    $connect = mysql_connect($showcart['mysql_host'],$showcart['mysql_username'],$showcart['mysql_password']);
           if(!$connect){
    $tconnect = "<img src=\"images/refuse.png\">";
           }else{
             $tconnect = "<img src=\"images/accept.png\">";
           }
    

     

    I just need the warnings to go away if the script cant connect to the database. Any ideas?

  6. Can someone tell me whats wrong with this?

     

     

    <?php
    
    $total = mysql_query("SELECT SUM(amount) as amount, FROM `transaction` WHERE `user_id` = '$user_id'") or die(mysql_error());
    while($row = mysql_fetch_array($total)){
    extract($row);
    
    echo $amount;
    
    }
    
    ?>
    
    

     

    Syntax is wrong. Remove "," from before the "FROM"

     

    Thank you.

  7. Can someone tell me whats wrong with this?

     

     

    <?php
    
    $total = mysql_query("SELECT SUM(amount) as amount, FROM `transaction` WHERE `user_id` = '$user_id'") or die(mysql_error());
    while($row = mysql_fetch_array($total)){
    extract($row);
    
    echo $amount;
    
    }
    
    ?>
    
    

  8. There is one sure way of finding out, try it and see if it works.  ::)

     

     

    There is another way ::)::)::)::)::)::)::) but ill keep it to myself..... anyways I did try it and didnt work which is why I asked in the first place.

  9. Hello I have a database which is filled with numbers which have + or - in it.

     

     

    looks like this,

     

    +10.00

    -100.00

    +45.00

     

    Now I have been working with php for a long time and dont have a clue how I would get the total.

  10. When I submit data to mysql from a textarea it does not add the line breaks and what not.

     

    textarea

     

    Blah blah

     

    Some details down here.

     

    Mysql input looks like

     

    Blah blah Some details down here.

     

    Mysql input should look like,

     

    Blah blah

     

    Some details down here.

     

  11. My Problem, $empl is set alot of times not always the same amount or same value,

     

    The Input

    $empl = "Message 123 <br>";

    $empl = "Message 1234 <br>";

    $empl = "Message 12345 <br>";

    $empl = "Message 123456 <br>";

     

    I need to display every $empl.

     

    Output should be,

     

    Message 123

    Message 1234

    Message 12345

    Message 123456

     

    any ideas?

     

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