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    If no result display 0 in table results

    in PHP Coding Help

    Posted

    Sorry Zane

     

    There is a table of item a user has in their shop.

    They can click a button next to the pricing field in their shop that will set the price for them by searching the database of all users shops for that item, retrieving the cheapest price and setting their price $1 less.

     

    The trouble is, when it comes across an item that noone else has in stock it doesnt know what to do and simply just uses the previous items price.

    If no result display 0 in table results

    in PHP Coding Help

    Posted

    Heres the setup

     

    Virtual pet site, users shop.

     

    They can click on an icon that will price their item 1 point less than the lowest existing.

     

    Trouble is, if no-one else has that item in stock, the icon prices the item the same price as the previous item listed.

     

    Example

     

    Chocolate Bar $1

    Red Hat $1 <--- theres none of these in any other shop therefor it needs to return $0

     

    Here is my code.

     

    $findResults = mysql_query("SELECT * FROM usershops_items2 WHERE price > '0' AND item_id = '$find_item2[id]' ORDER BY price ASC LIMIT 1"); 

while ($getResults = mysql_fetch_array($findResults)) 
{ 
 $cheap = $getResults[price];
 $cheap2 = $cheap -1;
}
$cheap = "<span onclick=\"MM_setTextOfTextfield('item_price[$id]','','$cheap2')\"><img src=images/items/item_747.gif width=24 height=24></span>";

     

    I have tried adding in if statements in different places but nothing works.

    Compare 2 tables

    in PHP Coding Help

    Posted

    Thank you for your advice although I managed to get it working with the below code.

     

    Theres probably a much neater way to do it though.

     

    $sql_query = "SELECT id,item_name FROM items2 WHERE item_type = 'intel_inc'";
$result = mysql_query($sql_query);
if(mysql_num_rows($result))
{
 while($row = mysql_fetch_row($result))
 {


$books_id = $row[0];
$find_pet = fetch("SELECT item_id FROM read2 WHERE pet_id = '$id' AND item_id = '$books_id'");
$pets_books_id = $find_pet[item_id];
if ($pets_books_id == $books_id) 
   {
 ?>
 <style>
 .book2 img{
  opacity:1.0;
  filter:alpha(opacity=100); /* For IE8 and earlier */
 </style>
 <?
  echo ("<div class='book book2'><a href=/books/$books_id.php><img src='$base_url/images/items/item_$books_id.gif' width='64px' height='64px'><br>$row[1]</a> </div> ");
   }
  else 
   {
   ?>
   <style>
   .book3 img{
 opacity:0.2;
 filter:alpha(opacity=20); /* For IE8 and earlier */
   </style>
   <?
  echo ("<div class='book book3'><img src='$base_url/images/items/item_$books_id.gif' width='64px' height='64px'><br>$row[1] </div> ");
   }



 }
}
//if no fields exist
else
{
 echo "no values in the database";
} 
?>

    Compare 2 tables

    in PHP Coding Help

    Posted

    Using mysql and php 5

     

    Table 1 is a list of books.

     

     

    Table 2 is a list of userids and the books read.

     

    USER_ID | ITEM_ID

    ________________

     

    1 | 345

    1 | 456

    2 | 345

    2 | 457

    2 | 347

     

    My page displays images (opaque) of all books available from Table 1

     

    In the while statement that displays the books, I would like to check Table 2 for a match and if the user has read it, the image is not opaque and also contains a link.

     

    I have no clue on the best way to do this.

     

    Below is my code so far.

     

    
$sql_query = "SELECT id,item_name FROM items2 WHERE item_type = 'intel_inc'";
$result = mysql_query($sql_query);
if(mysql_num_rows($result))
{
 while($row = mysql_fetch_row($result))
 {
  $get_books_read = mysql_query("SELECT item_id FROM read2 WHERE user_id = '$id'");

  if( IF THERE IS A MATCH )
   {
  echo ("<div class='book_yes'><img src='$base_url/images/items/item_$row[0].gif' width='64px' height='64px'><br>$row[1]</div> ");
   }
  else 
   {
  echo ("<div class='book_no'><a href='#'><img src='$base_url/images/items/item_$row[0].gif' width='64px' height='64px'><br>$row[1]</a></div> ");
   }
 }
}
//if no fields exist
else
{
 echo "no values in the database";
}

     

    Thanks in advance for any help offered.

    Shout Box - Report a User with Auto Ban on 3 reports

    in PHP Coding Help

    Posted

    Hi Guys

     

    My site has a shout box, now and again people post inappropriate content and if a moderator is not online to mute them it doesnt look good for other users.

     

    I have written a snippet of code to report an offender. My goal is that if 3 different people report an offender the site auto mutes them. (manual mute function already exists for moderators)

     

    I setup a table that records the reporters name the offenders name the post id the message content and the time stamp.

     

    I have never been good at counting num rows etc and need some help in what direction to go in.

     

    My current code that inserts the reported content.

     

    mysql_query("INSERT INTO `chicka_pets`.`records_sbreports` (`id`, `offender`, `reporter`, `post_id`, `post_content`, `timestamp`) 
VALUES ('', '$offender', '$username', '$id', '$sbmessage', '$timestamp')");

     

    I am guessing that somewhere after that I need some sort of count numrows on the post id where reporter is NOT the same and then an if numrows >2 set offender to mute

     

    my problem is how to do the num rows bit where the reporters are unique?

     

    Any help is much appreciated

     

     

     

    if else display

    in PHP Coding Help

    Posted

    Thank you so much

     

    num rows worked a treat

     

    Just in case anyone else has a similar problem here tis the code

     

    <FORM ACTION="<?php echo $_SERVER['PHP_SELF']; ?>" METHOD=get>
<input type="text" name="band" value="<?php echo $_GET['band']; ?>" /> 
<input type="text" name="bust" value="<?php echo $_GET['bust']; ?>" /> 
<INPUT TYPE=submit VALUE="Calculate">
</FORM>


<?php
if(isset( $_GET['band']) && isset($_GET['bust']) ){
$band = $_GET['band'];
$bust = $_GET['bust'];

$query = sprintf("SELECT size FROM bra_calculator  WHERE band='%s' AND bust='%s'", mysql_real_escape_string($band), mysql_real_escape_string($bust));
$result = mysql_query($query);
$num_rows = mysql_num_rows($result);


if (!$num_rows) {
echo "N/A";
}
else {
while ($row = mysql_fetch_assoc($result)) {
   echo "$row[size]";
}
}
mysql_free_result($result);
}

    if else display

    in PHP Coding Help

    Posted

    I have written a bra calculator, when the user enters their measurements their bra size is displayed however as the code I have used to do this came from a help tutorial I do not understand how it was constructed as I am quite a newb.

     

    The form works perfectly when returning data however if there is no size for a particular measurement it just displays nothing. I would like it to display N/A

     

    I have tried adding various if elses but without success.

     

    Any help is much appreciated

     

    <FORM ACTION="<?php echo $_SERVER['PHP_SELF']; ?>" METHOD=get>
<input type="text" name="band" value="<?php echo $_GET['band']; ?>" /> 
<input type="text" name="bust" value="<?php echo $_GET['bust']; ?>" /> 

<INPUT TYPE=submit VALUE="Calculate">
</FORM>


<?php
$band = $_GET['band'];
$bust = $_GET['bust'];

$query = sprintf("SELECT size FROM bra_calculator  WHERE band='%s' AND bust='%s'", mysql_real_escape_string($band), mysql_real_escape_string($bust));
$result = mysql_query($query);

if (!$result) {
   echo "no";
}
else {
while ($row = mysql_fetch_assoc($result)) {
   echo $row['size'];
}
}
mysql_free_result($result);
?>

    displaying a message to a user without it being in the URL

    in PHP Coding Help

    Posted

    Hi all,

     

    I am in need of a bit of advice on the best way to display a message on screen to a user.

     

    This is a little difficult to explain so please bare with me. I am using php5 and mysql

     

    My worry is to do with XSS etc

     

     

    Current Scenario

     

    My website is a pet gaming site where often a user is able to claim a free gift or prize.

     

    User clicks a button or link on Page A to claim a free gift.

     

    This redirects the user to Page B where all the magic happens and the gift is chosen and then redirects the user back to Page A without them even seeing what happened with a message of "Congrats you won a blah blah"

     

    Now my problem is, the way the message is handled.

     

    Page A Code

     

    <a href=$base_url/PageB.php>link</a>

     

    Page B Code

     

    	die(header(error("PageA.php","message here")));

     

    Result URL is

     

    $base_url/PageA.php&error=message here

     

    I am concerned that a malicious user could inject the error= variable

     

     

    what would be the best way to do this without having to have the information in the URL?

     

     

    Warning: mysql_num_rows(): - TEARING HAIR OUT

    in PHP Coding Help

    Posted

    Heres a doosey - well for me anyway.

     

    I have a cron file to clear out high scores and award a trophy to the top 3 players.

     

    Now if I run the Bubble Pop cron job (top code) it works fine however if I run the Pyramid one(bottom code) it throws back an error at me

     

    Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/kabooc/public_html/ghetto_cron.inc.php on line 384

     

    Line 384 is  if(mysql_num_rows($result))

     

    Both are the exact same code bar the second one has a WHERE clause.

     

     

    if ($run['Bubblepop'])
{
$sql_query = "SELECT username,MAX(score) as total FROM highscores_bubblepop GROUP BY username ORDER BY 'total' DESC LIMIT 0 , 3";
//store the SQL query in the result variable
$result = mysql_query($sql_query);

if(mysql_num_rows($result))
{
$x =110;
while($row = mysql_fetch_row($result))
{
mysql_query("INSERT INTO trophy (user_name, trophy_id) VALUES ('$row[0]','$x')");
mysql_query("INSERT INTO records (record_type, record_entry, userid) VALUES ('Bubble Pop Trophy','$row[0] was awarded a $x trophy','1')");
$x ++;
}
}
mysql_query("DELETE FROM highscores_bubblepop");
}

     

    if ($run['Trophy Pyramid Solitaire'])
{
$sql_query = "SELECT username,MAX(score) as total FROM highscores WHERE gameID = '106' GROUP BY username ORDER BY 'total' DESC LIMIT 0 , 3";
//store the SQL query in the result variable
$result = mysql_query($sql_query);

if(mysql_num_rows($result))
{
$x =101;
while($row = mysql_fetch_row($result))
{
mysql_query("INSERT INTO trophy (user_name, trophy_id) VALUES ('$row[0]','$x')");
mysql_query("INSERT INTO records (record_type, record_entry, userid) VALUES ('Trophy Pyramid Solitaire','$row[0] was awarded a $x trophy','1')");
$x ++;
}
}

}

     

    I have looked through google and most errors I find the solution is to...

    double check database tables - done

    check spelling - done

    check connection - done

     

    is there an alternative way to do what I am trying and what is it? or is there some sort of blatantly obvious error and im just not seeing it?

    GROUP BY and SORT BY causing issues

    in PHP Coding Help

    Posted

    I worked it out

     

    $sql_query = "SELECT username,MAX(score) as total FROM highscores_bubblepop group by username ORDER BY `total` DESC LIMIT 0 , 10";

     

    Had to order by total not score and it now works perfectly.

     

    Hopefully this helps others as I know its taken me a long long time to find the answer :)

     

    Thank you all who helped

    GROUP BY and SORT BY causing issues

    in PHP Coding Help

    Posted

    Hi me again!

     

    I now have beta testers on the site and it seems its not working as it should after all :(

     

    $sql_query = "SELECT username,MAX(score) as total FROM highscores_bubblepop group by username ORDER BY `score` DESC LIMIT 0 , 50";


//store the SQL query in the result variable
$result = mysql_query($sql_query);

if(mysql_num_rows($result))
{
//output as long as there are still available fields
$x =1;

while($row = mysql_fetch_row($result))
{
$current_avatar = fetch("SELECT * FROM members_profiles2 WHERE username = '$row[1]'");
echo ("<div class=\"game_high_scores_content\" style=\"background-image: URL($base_url/images/avatars/$current_avatar[avatar].gif);\">$row[0] Scored $row[1]</div>");
$x ++;
}
}

//if no fields exist
else
{
echo "no values in the database";
} 
mysql_close($con);

     

    Result

     

    2random4me Scored 2097

    lollypopchicka Scored 1541

    loulou2370 Scored 603

    plutoheart Scored 1662

    maryann1950 Scored 2126

    twinkle Scored 386

    faeryofavalon Scored 632

    GROUP BY and SORT BY causing issues

    in PHP Coding Help

    Posted

    woot nvm i added ORDER BY `score` DESC

     

    $sql_query = "SELECT username,MAX(score) as total FROM highscores_bubblepop group by username ORDER BY `score` DESC LIMIT 0 , 50";

     

    joe Scored 777

    lollypopchicka Scored 269

    fred Scored 115

     

    I could kiss you!!! thank you soo much I have been trying to fix this problem for years on and off.

     

     

    GROUP BY and SORT BY causing issues

    in PHP Coding Help

    Posted

    That kinda works

     

    How would I then only include a players top score or 1 distinct entry?

     

    Query

    $sql_query = "SELECT distinct `username`,`score` FROM `highscores_bubblepop` ORDER BY `username`,`score` DESC LIMIT 0 , 50";

     

     

    Results

     

    fred Scored 115

    fred Scored 111

    joe Scored 777

    joe Scored 234

    lollypopchicka Scored 269

    lollypopchicka Scored 262

     

    On a personal note I just read your blog entry on selling your Mac, priceless!!

    GROUP BY and SORT BY causing issues

    in PHP Coding Help

    Posted

    order by score and username give the below results

     

    fred Scored 111

    fred Scored 115

    joe Scored 234

    lollypopchicka Scored 262

    lollypopchicka Scored 269

    joe Scored 777

     

    $sql_query = "SELECT distinct `username`,`score` FROM `highscores_bubblepop` ORDER BY `score`,`username` DESC LIMIT 0 , 50";

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