Jump to content

ryanfilard

Members
 View Profile  See their activity

  • Posts

    252

  • Joined

  • Last visited

 Content Type 

Posts posted by ryanfilard

    MySQL-Link not valid

    in PHP Coding Help

    Posted

    this is all of my php code

     

    <?php require_once('Connections/news.php'); ?>
<?php
$id = $_REQUEST['id'];
if (!function_exists("GetSQLValueString")) {
function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "") 
{
  if (PHP_VERSION < 6) {
    $theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue;
  }

  $theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($theValue) : mysql_escape_string($theValue);

  switch ($theType) {
    case "text":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;    
    case "long":
    case "int":
      $theValue = ($theValue != "") ? intval($theValue) : "NULL";
      break;
    case "double":
      $theValue = ($theValue != "") ? doubleval($theValue) : "NULL";
      break;
    case "date":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;
    case "defined":
      $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue;
      break;
  }
  return $theValue;
}
}

mysql_select_db($database_news, $news);
$query_news = "SELECT * FROM news WHERE id = '$id'";
$news = mysql_query($query_news, $news) or die(mysql_error());
$row_news = mysql_fetch_assoc($news);
$totalRows_news = mysql_num_rows($news);

mysql_select_db($database_news, $news);
$query_arguments = "SELECT * FROM argue WHERE pid = '$id'";
$arguments = mysql_query($query_arguments, $news) or die(mysql_error());
$row_arguments = mysql_fetch_assoc($arguments);
$totalRows_arguments = mysql_num_rows($arguments);
?>

    MySQL-Link not valid

    in PHP Coding Help

    Posted

    What is wrong?

     

    Warning: mysql_select_db(): supplied resource is not a valid MySQL-Link resource in /home/rweekly/public_html/2012/view_arguments.php on line 41

     

    Warning: mysql_query(): supplied resource is not a valid MySQL-Link resource in /home/rweekly/public_html/2012/view_arguments.php on line 43

     

    mysql_select_db($database_news, $news);
$query_news = "SELECT * FROM news WHERE id = '$id'";
$news = mysql_query($query_news, $news) or die(mysql_error());
$row_news = mysql_fetch_assoc($news);
$totalRows_news = mysql_num_rows($news);

mysql_select_db($database_news, $news);
$query_arguments = "SELECT * FROM argue WHERE pid = '$id'";
$arguments = mysql_query($query_arguments, $news) or die(mysql_error());
$row_arguments = mysql_fetch_assoc($arguments);
$totalRows_arguments = mysql_num_rows($arguments);

    Need Help With Else If Script

    in PHP Coding Help

    Posted

    This does not work correctly and I need help with it. If $rating is equal to zero it displays 4 stars on this page http://rwdev.whekle.com/view.php?id=4571

    what I am doing wrong, it is probably a simple mistake.

    if($rating <= '1.5'){
   $star1 = "yes";
}elseif($rating <= '2.5'){
    $star2 = "yes";
}elseif($rating <= '3.5'){
     $star3 = "yes";
}elseif($rating <= '4.5'){
       $star4 = "yes";
}elseif($rating <= '5.0'){
   $star5 = "yes";
}

    Why Won't My Database Update?

    in PHP Coding Help

    Posted

    No here is my form that submits.

     

    <form method="post" action="view.php?id=<?PHP echo $_REQUEST['id']; ?>"> 
                              <input name="star2" type="radio" class="star" value="1"/>
<input name="star2" type="radio" class="star" value="2"/>
<input name="star2" type="radio" class="star" value="3"/>
<input name="star2" type="radio" class="star" value="4"/>
<input name="star2" type="radio" class="star" value="5"/> 
<input type="hidden" name="formcheck" value="rating">
<input name="rate" type="submit" id="rate" value="Rate"> 
</form>

    Why Won't My Database Update?

    in PHP Coding Help

    Posted

    I am trying to make a rating system and this script is not working.

     

    $ratingget = $_REQUEST['star2'];
mysql_select_db($database_news, $news);
$query_ratingsys = "UPDATE news SET votes = votes +1 AND rating = rating +'$ratingget' WHERE id = '$id'";
$ratingsys = mysql_query($query_ratingsys, $news) or die(mysql_error());

    Celbre - My Website

    in Website Critique

    Posted

    I spent over 3 months coding and planing this website with other and I expect it to be done in about 1 more week. I was contacted by a school teacher with 200 students in total, she was asking me if it is okay for students to register. I said yes because I am adding a lot of new privacy features. She is going to use it in about a month. Do you think it is ready for action?

     

    http://www.celbre.com

     

    Demo User: ptest

    Demo Pass: ptest1

    PHP SQL Question

    in PHP Coding Help

    Posted

    I am trying to make my database search only search where private = '$priiv'(Usually 0,1, or 2) but it does not work

     
$query_search = "SELECT * FROM users WHERE username LIKE '%$idea%' OR fname LIKE '%$idea%' OR lname LIKE '%$idea%' OR tags LIKE '%$idea%' AND private = '$priiv'";

     

    What am i doing wrong?

    Android App - PHP Phactory

    in Miscellaneous

    Posted

    I am developing an android application called PHP Phactory I am still working on the name. The program is for android phones and tablet but would be best suited for a tablet. With this program you can write and execute php scripts. Is this something that you would use?

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.