PHP - Combine Results From Multiple Members Who Have Like Categories Using Or

Hi,

Please help

I need to make categories with su-categories
I made in my database tables like this

table = categories

cid      |     cat_id    |   parent_list    | categ_name                 | type
1               0                   1                  General category            c
2               1                   1,2               sub category                   s
3               1                   1,3               sub category                   s
4               1                   1,2               sub category                   s
5               1                   1,5               sub category                   s
6               5                   1,5,6            sub sub category            s
7               0                   7                  General category2          c
8               7                   7,8               sub category                   s

I need to view this in dropdown menu using parent_list column

I have been going in circles about how to handle the issue that I have come up with and was hoping that someone could provide a different method. Basically I want to be able to create an infinite number of categories and allow them to be assigned to other categories then making them subcategories/sub sub categories if you will. My initial plan was to use a method I have used before by just saving the parent id of the subcategories and recursively going through the data. However, I am not to fond of this because of the inefficiency that it provides. I did some research and found that method is called the "Adjacency List". I also found another way to do it called "Nested Sets" which seems to be much more efficient but its a little harder to understand / maintain. I was hoping that someone here could offer alternatives to both of these methods that will allow me to keep the efficiency of "Nested Sets" but providing the ease of use / maintenance of the "Adjacency List".

Thanks in advance for any help.

I've read through a few examples of how to paginate query results, but with my currently php skills I don't understand them, or they do not fit into my code.  I wondered if I could make my own, and the logic behind this code makes sense to me, but it's still not working.  Here is the code:

<?php

if (!$Start) {
$Start = 0;
}
else {
     $Start=$_POST['Start'];
}

$crittercount = 120;

$numpages = $crittercount / 30;
$pagei = 1;
$pagereit = 1;

echo '<p><center>
<FORM method=POST action=testpagination.php>
<SELECT NAME="Start">
        <OPTION>Page 1';
        while ($pagereit < $numpages) {
        $pagei = $pagei + 30;
        echo '<OPTION value='.$pagei.'>';
        $pagereit = $pagereit + 1;
        echo "Page: " .$pagereit;
        }
        echo '</SELECT>';
echo'<input type=submit value=Go></FORM><p>';

echo $Start;

?>

The test site is he
http://lab-lib.com/felishorns/Felishorns/websitetest/testpagination.php

$crittercount is a variable that changes depending on the main $query.  This query is invisible, and is used to pull a count for all the results in total.  I am modifying it with $Start to create a second query which the viewer sees.

$queryF = $query . " LIMIT " . $Start . ", 30";

The idea is that $Start changes depending on the value selected, thereby changing the query.  In this test code, I've simply echoed the $Start value of that page, to see if the code is working, and it's not changing.  What's wrong with this code?  Is it not possible to have a form lead to the same link of the page it's on?  Does the variable not become updated?

Thanks in advance.

i have done this code but i want to limit to 10 records per page, i cant get the start from and limint number on page correct, it only echos out the links, 1 and 2 i have 18 data inputs in my data base so that bit is corrent by its not echoing the data out please help   i think the problem is im my query but cant put my finger on it

<?php
include 'connect.inc';
if (isset($_GET["page"]))
{
$page  = $_GET["page"];
}
else
{
$page=1;
};
$start_from = ($page-1) * 10;
$result = mysql_query ("SELECT * FROM images ODER BY date ASC $start_from,10");

$num_rows = mysql_num_rows($result);
echo $num_rows;
?>
<table>
<?php
while ($row = mysql_fetch_assoc($result))
{
?>
            <tr>
            <td><? echo $row['title']; ?></td>
            <td><? echo $row['description']; ?></td>
            </tr>
<?php
};
?>
</table>
<?php
$rs = mysql_query("SELECT COUNT(id) FROM images");
$row = mysql_fetch_row($rs);
$totalrecords = $row[0];
$totalpages = ceil($totalrecords / 10);
 
for ($i=1; $i<=$totalpages; $i++) {
            echo "<a href='hello.php?page=".$i."'>".$i."</a> ";
};
?>

thanks matt

I am pulling some info from a mysql table, checking some logic, and displaying in on a page. The way this is written now it shows all of the items in a one column table, I simply want to break it up into multiple columns (ideally by specifying say 20 items per column). I know I need a counter and a loop but I'm just not sure how it all fits in with my existing program. Any help would be greatly appreciated.

Here is my code
Code: [Select]
$result2 = mysql_query("SELECT * FROM propertyids ORDER BY propertyname ASC");
while($row2 = mysql_fetch_array($result2)){
${'resp'.$row2['id']} = curl_multi_getcontent(${'ch'.$row2['id']});
${'status'.$row2['id']} = strpos(${'resp'.$row2['id']},$httpvar);
if(${'status'.$row2['id']} === false) {
echo "<tr><td>";
echo "<img src='images/down.png'> <a href='" . $row2['primaryurl'] . "' target='_blank'>" . $row2['propertyname'] . "</a>";
echo "</td></tr>";
}
else {
echo "<tr><td>";
echo "<img src='images/up.png'> <a href='" . $row2['primaryurl'] . "' target='_blank'>" . $row2['propertyname'] . "</a>";
echo "</td></tr>";
}
}
echo "</td></table>";

Hello Everyone was wondering if I could get some help with the following code? I am querying a database for results of listings that are in a database these listings are displayed on the page in a form. I am wanting each listing to be on a different page. Below is my code.
Code: [Select]


$lim=1;
if (!isset($s) || $s < 1 || !is_numeric($s)) {
            $s = 1;
        }


    $start = ($s - 1) * $lim;
     $sql = "select id,bussimg,imagewidth,imageheight,email,usridm,company,businesscategory,address1,address2,state,city,zip,website,email,repname,description,phonenumber,country,status from $approvecheckbusinesses where usridm='$user_id'";
    $result=db_query($sql);
     



        $countpages = $sql;
        $sql = $sql . " order by id asc limit $start, $lim";
       $result=db_query($sql);
        $pages = ceil(mysql_num_rows(mysql_query($countpages)) / $lim);
     
   


$result=db_query($sql);
for ($i = 0; $i < mysql_num_rows($result); $i++) {
$Listid= mysql_result($result, $i, "id");

$usridm= mysql_result($result, $i, "usridm");
$CompanyName= mysql_result($result, $i, "company");
$realname= mysql_result($result, $i, "repname");
$email= mysql_result($result, $i, "email");
$BusinessCategory= mysql_result($result, $i, "businesscategory");
$status= mysql_result($result, $i, "status");
   



echo ("FORM IS TO BE DISPLAYED HERE");

}

if ($pages > 1) {
            echo("<p align=left style='font-size: 85% color=white'>");
            for ($i = 1; $i <= $pages; $i++) {
                echo("[");
                if ($i == $s) {echo("<b>");}
                else {echo("<a id=home_offerLink href='index.html?EditMemberListing&user_id=$user_id&s=$i'>");}
                echo("Page $i");
                if ($i == $s) {echo("</b>");}
                else {echo("</a>");}
                echo("]  ");
            }
            echo("</p>")


Page Numbers here using the above code.. The problem I seem to be running into is that it only displats the first record. The page numbers show up page 1 page 3
page 2 and three are blank there is no mysql error or anything for some reason I only get that first result out of three

Hello I have a question. I'm trying to perform multiple queries based off the results from a query.  Is this possible ?

$result = mysql_query("SELECT id FROM sometable");

for each id returned

$result1 = mysql_query("SELECT * FROM someothertable WHERE id=result from $result query");

Any help would be GREAT

I have a script that runs a MySQL query and loops through a bunch of results.   For each person's name, I have a short bio that I want to show/hide by clicking on the person's name.  I have a basic start, but I can't figure out how to show/hide them individually. When I click on someone's name, it shows/hides all of the bio's.  I do have an individual ID for each person, and could use that somehow, but I'm scratching my head trying to figure out how to accomplish that.  Any ideas?   http://jsfiddle.net/APA2S/3644/  

Hello everyone,

Very new to coding - enjoying it but struggling! I think I'm trying to do something pretty common but I seem to have come up against a complete wall now and after hours/days searching the internet and reading books I'm completely stuck! I'm  trying to write some code to search a MySQL database of products, then display the results. For some search results there will be lots of products so I want to display 10 products on the first page then allow visitors to go to the next page to see another 10, and so on - a type of pagination, as they should then be able to click back to see the last page etc.
I've got to the point of being able to display the first 10 search results, but I can't figure out at all how to create some kind of page scrolling/pagination system.
Please, does anybody have any ideas??
I've attached my code, I hope this is the correct way of doing things here.
Many thanks for your time!

The PHP search code...
Code: [Select]

<?php
//opens connection to mysql server
$dbc = mysql_connect('localhost');
if (!$dbc) {
die('Not connected :' . mysql_error());
}
echo "Connected to mysql database<br />";
//select database
$db_selected = mysql_select_db("NAME_OF_DATABASE", $dbc);
if (!$db_selected)
{
die ("Cannot connect :" . mysql_error());
}
echo "Connected to database<br /><hr />";
echo "Here are your results";

$term = $_POST['term'];
$category = $_POST['category'];
$brand = $_POST['brand_name'];
 
  
$sql = mysql_query("SELECT * FROM products where product_name like '%$term%' AND category_name like '%$category%' AND brand_name like '%$brand%' LIMIT 0, 10");
 
{ 
while ($row = mysql_fetch_array($sql)){
echo "<table border='1' width='100%'> ";
echo "<tr>"; 
echo "<td style='vertical-align:top' width='25%'>" . '<img src="', $row['image_url'], '" alt="', $row['product_name'], '"  width="100" height="100" />' . "</td>";
echo "<td style='vertical-align:top' width='50%'>" . $row['product_name'];
echo "<br />";
echo  "<span style='font-size: 10px'>" . $row['description'] . "</span>" . "</td>"; 
echo "<td style='vertical-align:top' width='25%'>" . $row['price'];
echo "<br />";
echo "<br />";
echo "<hr />";
echo "$row['merchant_name'] </td>";
echo "</tr>";
    }
echo "</table>"; 
}
?>

I have to make a mail form (which is already coded) that will pull peoples names and email address from a database based on certain form parameters. I have the form done and am so close on finishing it up but am stuck on some mail() issues.

I have the form set up to use:
mail($sendTo, $theSubject, $theMessage, $headers);

Using this format, I know that I can send this email to multiple recipients at once doing the following:
mail('Name1 <email1@na.com>, Name2 <email2@na.com>', $theSubject, $theMessage, $headers);

So I have my mySQL query all set up and it looks like so:
$query="SELECT * FROM operations_forms WHERE department='$initiatedBy'";
$result=mysql_query($query);

while( $row = mysql_fetch_assoc( $result ) ) {

$name = $row["name"]; 
$email = $row["email"];

$recip = $name . " <" . $email .">, ";
$sendTo = $recip;

}

The problem I am having is that my $sendTo variable only is passing on one record, not a combination of all of them. I am at a loss here on how to spit out all the variables and then combine them into one. I can use the above and produce an echo call that looks correct... but it doesn't function correctly on the back end.

Any help? 

Thanks!

I have a simple (1 table, 14 fields) database. I'm using 9 separate search boxes from my search page :
http://www.cinemaposter.com/database.html
There are currently 162 entries. Hitting "submit" throws them all out, same with valid input into any box, with the exception of "country" - example  : entering "F" gives 20results; "PL" gives 34 results.
Here's my code (to simplify, I only went down to the 4th box ('country' - the last and only one giving correct results).  I'm one sorry newbie.... please help !
Here's the code :
------------------

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"
"http://www.w3.org/TR/1998/REC-html40-19980424/loose.dtd">
<html>
<head>
   <title>Search Results</title>
   <meta name="generator" content="BBEdit 6.5">
   <style type="text/css">
td
{
   color: #000000;
   font-family: arial;
   font-size: 10px;
}
</style>
</head>
<body>
<?php
$link = mysql_connect("xxxx", "xxxx", "xxxx") or die("Could not connect");
///print "Connected successfully<BR>";
mysql_select_db("xxxx") or die("Could not select database");


{
$Author = $_POST['Author'];
$sql = "SELECT * FROM posters WHERE Author LIKE '%$Author%'";
}

{
$Title = $_POST['Title'];
$sql = "SELECT * FROM posters WHERE Title LIKE '%$Title%'";
}

{
$Originaltitle =$_POST['Originalitle'];
$sql="SELECT * FROM posters WHERE Originaltitle LIKE '%$Originaltitle%'";
}

{
$Country =$_POST['Country'];
$sql="SELECT * FROM posters WHERE Country LIKE '%$Country%'";
}


$result = mysql_query ($sql) or die("Query failed : " . mysql_error());
$numrs = mysql_num_rows($result);

if ($numrs > 0){

if ($numrs==1){
   print ("<p><font face=\"Arial\"><b>There is $numrs record in our database.</B> <BR><BR>");
   }elseif($numrs>1){
   print ("<p><font face=\"Arial\"><b>There are $numrs records in our database. </B><BR><BR>");//Apparently, he wan't to be able to search again...</b><form method \"POST\" action=\"$PHP_SELF\">Search Again:");
   }
   
   
   
   
  echo("<a href='javascript:history.go(-1)'>Go Back</a></p>");
  print("<P align=center>Click on the column heading links to sort the results.<br><br />");   
   
print ("<TABLE CELLPADDING=\"1\" CELLSPACING=\"2\" BORDER=\"0\">\n");
PRINT ("<TR BGCOLOR=\"#FFFF77\">
<TD COLSPAN=2><P align=center>Author</td>
<TD><P align=center>Title</TD>
<TD><P align=center>Original Title</TD>
<TD><P align=center>Country</TD>
<TD><P align=center>Director</TD>
<TD><P align=center>Cast</TD>
<TD><P align=center>Year</TD>
<TD><P align=center>Original Year</TD>
<TD><P align=center>Photo</TD>
<TD><P align=center>Producer</TD>
<TD><P align=center>Circulation</TD>
<TD><P align=center>Size</TD>
<TD><P align=center>Print type</TD>
<TD><P align=center>Remarks</TD></TR>");

$z=1;
      while ($line = mysql_fetch_array($result)) {
      
         $num=$z++;
         //Note here that I'm assigning the found objects new names (indicated by lowercase) to that php wont get confused.....
         $auth=$line['Author'];
         $title=$line['Title'];
         $originaltitle=$line['Originaltitle'];
         $country=$line['Country'];
         $director=$line['Director'];
         $cast=$line['Cast'];
         $year=$line['Year'];
         $originalyear=$line['Originalyear'];
         $producer=$line['Producer'];
         $circ=$line['Circulation'];
         $size=$line['Size'];
         $print=$line['Print'];
         $remarks=$line['Remarks'];
         $pict=$line['Picture'];


         //Set the formatting
         $bgcolor = ($i++ & 1) ? '#d3d3d3' : '#c0c0c0';
         
                           
      PRINT("<TR BGCOLOR=$bgcolor>
      <TD>$num.</TD>
      <TD>$auth</td>
      <TD>$title</TD>
      <TD>$originaltitle</TD>
      <TD><P align=center>$country</TD>
      <TD>$director</TD>
      <TD>$cast</TD>
      <TD><P align=center>$year</TD>
      <TD><P align=center>$originalyear</TD><TD>");
// This is the test to find do we have a picture
if ($pict) {
// If there is a picture, display it:
print ("<P align=center><a href='zdjecie.php?zdj= $pict&autor=$auth&opis=$title&orig=$originaltitle&rok=$year' target='_blank'>
<img src=$pict height=\"150\"></a>");
} else {

// We don't have a picture, so we print some text instead:
print ("<P align=center>no picture available");
}

print ("</TD><TD>$producer</TD>
<TD><P align=center>$circ</TD>
<TD>$size</TD>
<TD>$print</TD>
<TD>$remarks</TD></TR>");
            
         }
      PRINT("</TABLE><BR><font face=\"Arial\"><a href=\"javascript:history.go(-1);\">Search Again</a></font>");
   }else {
   print ("<font face=\"Arial\">Sorry, there were no results for your search. <a href=\"javascript:history.go(-1);\">Please try again</a></font>");
   }
//mysql_close($link);
?>                                                                     
</body>
</html>

Hi all, Im really struggling to build an update multiple records form, Ive built my table which calls all data i need from the db in a looped region which works fine I just cant understand how to then input these back into the table in the correct rows, if this makes sense? Ive been reading this tutorial but Its not very informative its more of a copy and paste http://www.phpeasystep.com/mysql/10.html

Hi Everyone,

Well I have a bit of a problem... I have created a search page that I would like to return results from my MySQL database. However for some reason I can not get my page to display the results. Instead I am getting an error page. Below is the code I am using:

In addition to this, should I wish for a user to be able to edit the returned results by clicking a link, how would I do that?

<?php
session_start();
?>
<link rel="stylesheet" type="text/css" href="css/layout.css"/>
<html>

<?php
$record = $_POST['record'];
echo "<p>Search results for: $record<br><BR>";

$host = "localhost";
$login_name = "root";
$password = "P@ssword";

//Connecting to MYSQL
MySQL_connect("$host","$login_name","$password");

//Select the database we want to use
mysql_select_db("schedules_2010") or die("Could not find database");

$result = mysql_query("SELECT * 
FROM schedule_september_2010 
WHERE 
champ LIKE '%$record%' ") 
or die(mysql_error());  


// keeps getting the next row until there are no more to get
while($row = mysql_fetch_array( $result )) {
        // Print out the contents of each row
echo "<br><p>Your Schedule<BR></p><br>";
echo "<table border=1>\n";
echo "<tr>
<td bgcolor=#444444 align=center><p><b>Champ</p></td>
<td bgcolor=#444444 align=center><p><b>Date</p></td>
<td bgcolor=#444444 align=center><p><b>Start Time</p></td>
<td bgcolor=#444444 align=center><p><b>End Time</p></td>
<td bgcolor=#444444 align=center><p><b>Department</p></td>
<td bgcolor=#444444 align=center><p><b>First Break</p></td>
<td bgcolor=#444444 align=center><p><b>Second Break</p></td>
<td bgcolor=#444444 align=center><p><b>Login ID</p></td>
</tr>\n";
do {




printf("<tr>
<td><p>%s</p></td>
<td><p>%s</p></td>
<td><p>%s</p></td>
<td><p>%s</p></td>
<td><p>%s</p></td>
<td><p>%s</p></td>
<td><p>%s</p></td>
<td><p>%s</p></td>
</tr>\n"
, $row["champ"]
, $row["date_time"]
, $row["start_time"]
, $row["end_time"]
, $row["department"]
, $row["first_break"]
, $row["second_break"]
, $row["login_id"]
);

} while ($row = mysql_fetch_array($result));
echo "</table>\n";
} else {
echo "$champ No Records Found"; 
} 

mysql_free_result($result);
mysql_close($con);
?>
</html>
 

Hi,

having problems getting checkboxes to display all reuslts when a user selects more than one check box say in the category section and one in the location section - see page http://www.partyco.co.uk/event-and-party-venues/ - submit to see reults page:

I managed to get it to display reults if the user only seletc either a right or left column option OR one of each - BUT not when thet select multiple categories and one location - and ideas how to do this ? putting it into an array perhaps - but how - new to some of this....

here is the code for the reults page:
Code: [Select]
<?php
$location = $_POST[location];
$category = $_POST[category];
?>

<?php

$Link = mysql_connect("xxxxxxxxx", "xxxxxxxxx", "xxxxxxxx") or die(mysql_error());
mysql_select_db("xxxxxxxx") or die(mysql_error());

// selects db listings when location not given
if (empty($location)) {
$query = "SELECT * FROM venues WHERE category = '$category' order by title";
$result = mysql_query($query) or die(mysql_error());

// selects db listings when category not given
} elseif(empty($category)) {
$query = "SELECT * FROM venues WHERE location = '$location' order by title";
$result = mysql_query($query) or die(mysql_error());

// selects db listings when both given
}else {
$query = "SELECT * FROM venues WHERE category = '$category' and location = '$location' order by title";
$result = mysql_query($query) or die(mysql_error());
}

while($row = mysql_fetch_array($result)){
echo "<div class=\"resultsShort\" style=\"margin-bottom:10px;\">";
echo "<h2 id=\"resultsHeading\">";
echo $row['title'];
echo "</h2>";
echo "<p class\"resultspara\">". nl2br($row['description']). "</p>";
echo "<h4 style=\"margin:5px 0 0 0; padding:0;\">Contact details</h4>";
echo "<p class\"resultspara\">". nl2br($row['contact']). "</p>";
echo "<div style=\"float:left; width:124px; height:40px; margin:10px 15px 0 0;\">";
echo "<a href=\"/party-supplier-resources/email-supplier.php?title=". $row['title']. "&email=" . $row['email']. "&location=" . $row['location']. "&category=" . $row['category']. "\" title=\"contact this venue here\">";
echo "<img src=\"/images/email-supplier.jpg\" width=\"124\" align=\"right\" height=\"35\" alt=\"contact this supplier button\" border=\"0\" /></a>";
echo "</div>";
echo "</div>";
}


?>


<?php include("../include/shareLinks.php"); ?>


<div id="popupContact">
<a id="popupContactClose" title="close this window">close x</a>
<h1>Supplier Directory Enquiry Form</h1>
<?php include("../include/enquiryform.php"); ?>
</div>
<div id="backgroundPopup"></div>


<?php
mysql_close ($Link);
?>
Any help appreciated!

Aaron

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