PHP - Getting Date Result, But Not Expecting Any. Why Does It Give Result?

i have a game script and it's always bringing back the same result and i cant see why..

all i get when the script it ran is

[/code]echo '<div class=noticeBox><h2>Sorry! No available member found![/code]

the code is below and can provide the functions and db if needed, but cant see what's wrong.

Code: [Select]
<?php

include '../header.php'; 

echo '<div id=AjaxSubDiv>';
echo '<div id="actionDiv">';

include '../statusbar.php';

echo '<div id=contentBox>';

// Show Fight Result
if( !empty($_GET[action]) AND $_GET[action]=='attack' ) {
echo $str;


// Fight Lists
echo '<P class=pageheader>Fight</P>';
echo '<P>Its time for action! Fight with rival thieves and grab their cash. You will require 1 strength for each fight and you will gain some cash and exp if you win the fight. However you will just earn some exp if you loose!! Each fight will decrease your health so keep an eye on health level because if your health goes below zero you will die and loose everything...</P>';

$res = query("SELECT COUNT(*) FROM `usertable` WHERE `userid`!=$user");
list($total) = mysql_fetch_array($res);
if( $total>0 ) {

$res = query("SELECT * FROM `usertable` WHERE `userid`!=$user LIMIT 0,20");




$viewplayer = Player::getById($row[userid]);

echo '<tr>';
echo '<td width=120px>';
echo '<P class=titleP><a href="#" onclick="load_menu_val(\'AjaxSubDiv\', \'profile\', '.$row[userid].');return false;"><fb:profile-pic uid="'.$row[userid].'" size="square" linked="false" width="75px" class="userImg" /></a></P>';
echo '<P class=titleP><a href="#" onclick="load_menu_val(\'AjaxSubDiv\', \'profile\', '.$row[userid].');return false;"><fb:name uid="'.$row[userid].'" linked="false" /></a></P>';
echo '</td>';

echo '<td width=120px>';
echo '<P><B>Level:</B> '.$viewplayer->level.', '.get_level_title($player->level).'</P>';
echo '</td>';

echo '<td width=120px>';
echo '<P><B>Crew:</B> '.$viewplayer->crew.'</P>';
echo '</td>';

echo '<td width=300px>';
if( $viewplayer->health > 20 ) {
echo '<form id="FightForm'.$row[userid].'" method="POST" class="itemForm">';
echo '<input type=hidden name=user value="'.$user.'">';
echo '<input type=hidden name=opid value="'.$row[userid].'">';
echo '<input type=hidden name=action value="dofight">';
echo '<P><input type=button onclick="do_submit(\'FightForm'.$row[userid].'\',\'http://www.bofs.us/thief/thiefmaster/fight/post_fight.php\', \'actionDiv\'); return false;" value="Fight Now" class="submitButton" /></P>';
echo '</form>';
}
else {
echo '<P class=titleP>Too Weak To Fight</P>';
}
echo '</td>';
echo '</tr>';
}
echo '</table>';
echo '</center>';
}
else {
echo '<div class=noticeBox><h2>Sorry! No available member found!</h2></div>';
print"$res";

}

echo '</div>'; // end of contentBox

echo '</div>'; // end of actiondiv
echo '</div>'; // end of AjaxSubDiv


?>

Hi all,

I am trying to figure out how to calculate 5 working days prior to a given date.

I have done some googling but can only see examples of how to add 5 working days onto a date, such as this:

Code: [Select]
$holidayList = array();
$j = $i = 1;

while($i <= 5)

{
    $day = strftime("%A",strtotime("+$j day"));
    $tmp = strftime("%d-%m-%Y",strtotime("+$j day"));
    if($day != "Sunday" and $day != "Saturday" and !in_array($tmp, $holidayList))
    {
        $i = $i + 1;
        $j = $j + 1;
    }
    else
        $j = $j + 1;
}
    $j = $j -1;
echo strftime("%A, %d-%m-%Y",strtotime("+$j day"));
Does anyone know how to calculate 5 working days prior to a date?

Many thanks,

Greens85

How can I check if a returned mysql value is equal to '' i.e. nothing?

I keep getting an error where the page won't load because the returned value is '' so i need to check for it

Hey Everyone,

I am posting today because I have a site I am working on which uses the xzero classifieds and having a small problem accomplishing something with it. I am trying to relocate the ad count number (of the city the user is currently browsing) to a box in my sidebar which has various site statistics like the following

Online Users: XX
Ads in this city: 0

But my code keeps giving me a 0 (even after I posted some test ads)

Anywho, heres what I have so far but cant seem to make it work

<?php
$country_adcounts = array();
$city_adcounts = array();
$sql = "SELECT ct.cityid, c.countryid, COUNT(*) as adcnt
FROM $t_ads a
INNER JOIN $t_cities ct ON ct.cityid = a.cityid AND ($visibility_condn)
INNER JOIN $t_countries c ON ct.countryid = c.countryid
WHERE ct.enabled = '1' AND c.enabled = '1'
GROUP BY ct.cityid";

$res = mysql_query($sql) or die(mysql_error().$sql);

while($row=mysql_fetch_array($res))
{
$country_adcounts[$row['countryid']] += $row['adcnt'];
$city_adcounts[$row['cityid']] += $row['adcnt'];
}

$adcount = 0+$city_adcounts[$city['cityid']];
?>

and this is the code which should recall the adcount

<?php echo $adcount; ?>

What am I doing wrong? I'm placing the first code on my sidebar and attempting to call the ad count through an echo. So ya, if anyone could tell me why this isn't working or lend a hand in helping me fix it please know I'll greatly appreciate it. 

Thanks.

This code is suppose to echo out ALL the custom pages that are in the database, but it's only show ONE. Why is this?


$query_get = mysql_query("SELECT COUNT(id),id,position,content,title FROM custom_pages ORDER BY id");

while($row = mysql_fetch_assoc($query_get))
{
echo '<table border="0">
<tr><th><p>Title</p><td>'. $row['title'] .'</td></tr>
<tr><th><p>Position</p><td>'. $row['position'] .'</td></tr>
<tr><th><p>Action</p><td><a href="index.php?pages=1&delete='. $row['id'] .'">Delete</a> or <a href="index.php?pages=1&edit='. $row['id'] .'">Edit</a></td></tr>
</table>
';
}

I have a list of ID's returned from a table (even number). I always used a loop to get the next result, but I want to put these straight into a new table

INSERT INTO table VALUES ($ID1, $ID2);

This is recursive till there are no more ID's. How do I get the result then the next result without loops. ID1 and ID2 cannot be null. What is the best way?

i have two tables, and i want to display data from the most recent entry of the two(i.e. display just one entry). i'm joining together 2 tables so that i can get this entry.Both tables contain the date, so i'm ordering the results of the query so that the most recent comes first.
My tables a
TOPIC
topic_id
topic _name
section_sub_id
date

POST
post_id
post_name
section_sub_id
date

I'm using the following code to get the last result, but nothing is being echoed
$join =mysql_query("SELECT topic.*, post.* FROM topic LEFT JOIN post ON topic.section_sub_id=post.section_sub_id WHERE post.section_sub_id = " .$row2['section_sub_id']. " ORDER BY topic.date") or die("Select Error :" . mysql_error());
$latest= mysql_fetch_row($join);
echo $latest['post.date'];
i'm not sure if there is something wrong with the sql, or with the php. ANy help would be great

Hello all.   I am using a premade login script which has some very powerful function built in, the one I am looking at using is as follows:

  /** 
    * query - Performs the given query on the database and 
    * returns the result, which may be false, true or a 
    * resource identifier. 
    */ 
   function query($query){ 
      return mysqli_query( $this->connection, $query); 
   } 

The issue is, displaying the results.  I call is via the following code:

$database->query('SELECT LastName, Firstname FROM Patrons WHERE LastName = "Delude"');

But I have no idea how to display it.  I have attempted to echo it out, but I get the following error in the logs:   PHP Catchable fatal error:  Object of class mysqli_result could not be converted to string    

I have been looking at this code all day trying to find where I had results and didn't free them.  Hopefully your eyes will have better luck.


// if form is filled continue with script
if($full == "TRUE"){

// function to concatinate two individuals string into one string
function fullname($first, $last){
$fullname = "$first"." "."$last";
return $fullname;
}
$fullname = fullname($_POST['firstname'], $_POST['lastname']);


// Check for existing user in rak5
  $sql = "SELECT COUNT(*) FROM rak5 WHERE empid = '$_POST[empid]' AND fullname = '$fullname'";
  $rakfcheck = mysql_query($sql);
  if (@mysql_result($rakfcheck,0,0) > 0) {
  mysql_free_result($rakfcheck);
// check uaid in user table
  $sql = "SELECT COUNT(*) FROM user WHERE UAID = '$_POST[UAID]'";
  $UAIDcheck = mysql_query($sql);

//if UAID is in the user table then check employee id for redundancy
if(@mysql_result($UAIDcheck,0,0) > 0){
mysql_free_result($UAIDcheck);
$sql = "SELECT COUNT(*) FROM user WHERE UAID = '$_POST[UAID]' AND employeeID = '$_POST[empid]'";
  $UAID_Empcheck = mysql_query($sql);

//if there is a UAID emplid match on the user table print redundancy
//else UAID is occupied by another user

mysql_free_result($UAID_Empcheck);
error(' Employee already has a UAID!');

}
else
{
error('UAID is currently active by a different employee');
}
}
//UAID is not in the user table
else{
$newpass = substr(md5($_POST['empid']),0,6);
//insert new user
$sql = "INSERT INTO user SET
             userid = '',
             firstname = '$_POST[firstname]',
             lastname = '$_POST[lastname]',
             email = '$_POST[newemail]',
             notes = '',
             password = ('$newpass'),
             UAID = '$_POST[UAID]',
             BOG = '',
            employeeID = '$_POST[empid]'";

$result = mysql_query($sql);

// Email the new password to the person.
   $message = "G'Day!

Your personal account for the Project Web Site
has been created! To log in, proceed to the
following address:

 http://ciwdvweb1.itd/web2/test2/index.php

Your personal login ID and password are as
follows:

   userid: $_POST[UAID]
   password: $newpass

You aren't stuck with this password! Your can
change it at any time after you have logged in.

If you have any problems, feel free to contact me at

Security Front End GUI creator
";

   $umail = $_POST['newemail'];
   mail($umail,"UAID and password",
        $message, "From: <you@example.com>");

include "MoveOn.html";
}
}
else{
error('either the name or employee id entered is incorrect please resubmit');
}
}
else
{
error('Please fill out the whole form');
}
mysql_close();
endif;
?>



I was wondering if it was my INSERT query but I tried to save the query into a variable and free it but I got another warning saying the variable wasn't the right parameter.

Thank you for your help

i have it set up so the user picks the year and session and then the database shows all the games in that specific year and session and then displays them with a link to their photo galleria and also shows the record(wins - losses - ties) but the way i have it know its not displaying the first result here is the code
Code: [Select]
<?php
 //declares $y as year played and $s as session
 $y = ' ';
 $s = ' '; 
 
 //handles submition of form for picking year and session
 if (isset($_POST['submit'])) {
 $data = mysql_real_escape_string($_POST['session']);
 $exploeded = explode(",", $data);
 $y = $exploeded[0];
 $s = $exploeded[1];
 }

  //Query for the selection menu
    $squery = "SELECT DISTINCT(Year_Played), Sessions FROM pinkpanther_games";
    $sresults = mysql_query($squery) or die("squery failed ($squery) - " . mysql_error());
   
    //checks $y and $s to see if their is a vaule
    if ($y == ' '){$y = '20112012';}
    if ($s == ' '){$s = '2';}
   
    //Query for getting record and for getting all the appropriate info for displaying the games
    $query = " SELECT (SELECT COUNT(id) FROM pinkpanther_games WHERE Year_Played = $y AND Sessions = $s AND Win_Loss = 'Tie' ) AS 'Tie', 
         (SELECT COUNT(id) FROM pinkpanther_games WHERE Year_Played = $y AND Sessions = $s AND Win_Loss = 'Win') AS 'Win',
         (SELECT COUNT(id) FROM pinkpanther_games WHERE Year_Played = $y AND Sessions = $s AND Win_Loss = 'Loss') AS 'Lose',
         Opponent AS Opponent, Score AS Score, Gallery_no AS Gal, Win_Loss AS Record FROM pinkpanther_games WHERE Year_Played = $y AND Sessions = $s";
    $results = mysql_query($query)  or die("Query failed ($query) - " . mysql_error());
    $row_a = mysql_fetch_assoc($results);   
    //Set record variables
    $wins = $row_a['Win'];
    $loss = $row_a['Lose'];
    $tie = $row_a['Tie'];

    //creates from and selection menu for year and session
echo "<tr><td colspan='2'><form method='post' action=''><select name='session' class='txtbox2'>";
while ($srow = mysql_fetch_assoc($sresults)){
echo "<option value='" . $srow['Year_Played'] . "," . $srow['Sessions'] . "'>" . $srow['Year_Played'] . " Session " . $srow['Sessions'] . "</option>";
}
echo "</select><input type='submit' name='submit' value='Go'  class='txtbox2' /></form></td></tr>";
//displays  record
echo "<tr><td colspan='2'>" .$wins . " - " . $loss . " - " . $tie . "</td></tr>";
echo "<tr></tr>";
//displays games and scores in appropriate year and session
while ($row = mysql_fetch_assoc($results)){
$opp = $row['Opponent'];
$score = $row['Score'];
$gal = $row['Gal'];
$wlt = $row['Record'];
//styles games acording to if win lose or tie
if ($wlt == 'Win'){
echo "<tr><td class='win'>";
echo "<a  href='galleries.php?g=$gal'" . $gal . " >" . $opp . "</a>";
echo "</td><td class='win'>";
echo   $score . "";
echo "</td></tr>";
}
if ($wlt == 'Loss'){
echo "<tr><td class='loss'>";
echo "<a  href='galleries.php?g=$gal'" . $gal . " >" . $opp . "</a>";
echo "</td><td class='loss'>";
echo   $score . "";
echo "</td></tr>";
}
if ($wlt == 'Tie'){
echo "<tr><td class='tie'>";
echo "<a  href='galleries.php?g=$gal'" . $gal . " >" . $opp . "</a>";
echo "</td><td class='tie'>";
echo   $score . "";
echo "</td></tr>";
}
}
?>

Hi, I was wondering if I wanted lets say 3 to the power of 10 how can I do this in php?
I searched a few places for some power to operator in PHP but no luck could some one tell me thanks?

Basically the following code works fine, exepct when it comes to the last result it inserts it twice, example:

(3,17),(4,17),(5,17),(5,17)

Any clues as to why? Thanks

Code: [Select]
if(count($addIDs_ary) > 0)
{
$str = "";
foreach($addIDs_ary as $val)
{
$str .= "({$val},{$playerID}),";
if(end($addIDs_ary) == $val)
{
$str .= "({$val},{$playerID})";
}
}
echo $str; // (val,val), (val,val), (val,val) etc..
$query = "INSERT INTO hitlist
(hit_id,player_id) values $str";

Hello, everyone. I am having some trouble figuring this out. I am trying to create a paging bar(so to speak) to display results by page number.
Trying to get the paging bar to go as follows...
Requested Page     The resulting paging bar
Page 1                     1 2 3 4 5
Page 2                     1 2 3 4 5
Page 3                     1 2 3 4 5
Page 4                     1 2 3 4 5
Page 5                     4 5 6 7 8
Page 6                     4 5 6 7 8
Page 7                     4 5 6 7 8
Page 8                     7 8 9 10 11
......
This is what I have right now that I am trying to get to work. I have changed this so many times and been working on it for 2 days. I think I'm just confusing my self. This can't be that hard.. 
Code: [Select]
<?php
echo '<html>';
//---these values changed based on query----
$Page=$_GET["page"]; //the page number requested
$AvailablePages=5; //number of pages available after the page requested

//---calculations---
$StartPage=round($Page/5)*5;
$EndPage=ceil($StartPage/5)*5 ; //the last page number to be displayed in paging
if($StartPage<5) {$StartPage=1;}

//just looking at variables
echo 'page is: ' . $Page . '<br>';
echo '<br>Starting @: ' . $StartPage;
echo '<br>Ending @: ' . $EndPage . '<br>';

//--create the paging bar--
$PagingBar='';
while($StartPage<=$EndPage)
{
if($Page==$StartPage) {
$PagingBar=$PagingBar . $StartPage . '';
}
else {
$PagingBar=$PagingBar . '
<a href="test.php?page=' . $StartPage . '">' . $StartPage . '</a>
';
}
$StartPage++;
}

//---show the paging bar--
echo $PagingBar;
echo '</html>';
?>

$name = $name . "<option value=". $row['name'] . ">" .$row['name'] . "</option>";

but the name is "Mr. Robert",it does have dot symbol...how to overcome? please help

I'm self-studying PHP and put to myself to a task. But I can't figure this out. It's about putting a database result into array.

There is this database called books with columns: isbn, author,title, price. The DB has records. I want to output certain records of certain columns:for example echo $books['title"];. I also to use functions.

I used the following code to titles listing for example:

function db_connect() {
   $result = new mysqli('localhost', 'root', 'password', 'books');
   if (!$result) {
      return false;
   }
   $result->autocommit(TRUE);
   return $result;
}
function get_book_details($isbn) {
  // query database for all details for a particular book
  if ((!$isbn) || ($isbn=='')) {
     return false;
  }
  $conn = db_connect();
  $query = "select * from books where isbn='".$isbn."'";
  $result = @$conn->query($query);
  if (!$result) {
     return false;
  }
  $result = @$result->fetch_assoc();
  return $result;
}
$book = get_book_details($isbn);
foreach($book as $isbn => $qty){
echo $book['title'];
}


I have this error: Warning: Invalid argument supplied for foreach() in ..\foreach.php on line 41
I'm guessing that $book is not array that's why I get this error message. But the big problem is as you can see in the function
function get_book_details($isbn)
inside the function it already gets the result into array
$result = @$result->fetch_assoc();