PHP - Displaying Even And Odd Sql Results

Hi

I have got results being displayed after clicking the search button in a form on my home page but it brings up all the results which is ok but how do I get onlt the results a user searches for for example a location or property type etc as its for a property website

The coding is below for the results page

Also sorry how do I add a background image to the php page, I tried using css but wouldn't work

Code: [Select]
<style type="text/css">
body {background-image:url('images/greybgone.png');}
</style>

<?php
 
mysql_connect ("2up2downhomes.com.mysql", "2up2downhomes_c","mD8GsJKQ")  or die (mysql_error());
mysql_select_db ("2up2downhomes_c");

echo $_POST['term'];

$sql = mysql_query("select * from properties where typeProperty like '%$term%' or location like '%$term%'");

while ($row = mysql_fetch_array($sql)){
    echo 'Type of Property: '.$row['typeProperty'];
    echo '<br/> Number of Bedrooms: '.$row['bedrooms'];
    echo '<br/> Number of Bathrooms: '.$row['bathrooms'];
echo '<br/> Garden: '.$row['garden'];
echo '<br/> Description: '.$row['description'];
echo '<br/> Price: '.$row['price'];
echo '<br/> Location: '.$row['location'];
echo '<br/> Image: '.$row['image'];
    echo '<br/><br/>';
    }
 
?>

Code: [Select]
<?php 
$result = mysql_query("SELECT * 
FROM FamilyTbl
INNER JOIN PeopleTbl ON (FamilyTbl.Name_ID = PeopleTbl.NameID) 
WHERE FamilyTbl.House_ID = '$address'
ORDER BY NameLast, NameFirst
") OR die(mysql_error());

WHILE ($row = mysql_fetch_array($result) )  { 

   echo $row[NameLast]. ", ". $row[NamePrefix]. " ". $row[NameFirst]. " ". $row[NameMiddle]. $row[NameSuffix]. " "; 
}
?>

OK, some of these queries return A LOT of names.  I'd like to be able to display them in columns in a table like so:
Code: [Select]
Charne, Mr. Michael   Glanger, Mrs. Karin      Kling, Mr. Wayne 
Charne, Mrs. Suzette  Glanger, Mr. Trevor      Lazarow, Mrs. Fiona 
Charney, Mrs. Linda   Jochelson, Mrs. Barbara  Lazarow, Mr. Mark 
Charney, Mr. Norman   Jochelson, Mr. Neil      Norton, Mr. Charles 
Cohen, Mr. Brendan    Karlan, Mr. Dennis       Norton, Mrs. Jodi 
Cohen, Mrs. Joanna    Karlan, Mrs. Helen       Roy, Mr. Michael 
Flekser, Mrs. Jean    Kling, Mrs. Danielle     Roy, Mrs. Nicki 
Frysh, Dr. Howard     Kling, Mrs. Melanie      Tsafrir, Mrs. Lauren 
Frysh, Mrs. Sandra    Kling, Mr. Nevil         Tsafrir, Mr. Thomer 

That way it reads top to bottom THEN left to right.

math-wise, it's simple to set up:
Code: [Select]
$num_results = number_of_$results;
$numcols = 3;
$numrows = trunc($numresults/numcols);
<table>
for row_loop=1 to numrows
<tr>
for col_loop = 1 to numcols
<td>
echo result(col_loop-1)*(numrows)+row_loop
</td>
next col_loop
</tr>
next row_loop
</table>

Anyone want to give this a shot?

Revraz already directed me to http://www.phpfreaks.com/forums/index.php/topic,95426.0.html but that displayed the data from left to right then up to down like so:
Code: [Select]
Charne, Mr. Michael   Charne, Mrs. Suzette  Charney, Mrs. Linda 
Charney, Mr. Norman   Cohen, Mr. Brendan    Cohen, Mrs. Joanna 
Flekser, Mrs. Jean    Frysh, Dr. Howard     Frysh, Mrs. Sandra 
Glanger, Mrs. Karin   Glanger, Mr. Trevor   Jochelson, Mrs. Barbara 
Jochelson, Mr. Neil   Karlan, Mr. Dennis    Karlan, Mrs. Helen 
Kling, Mrs. Danielle  Kling, Mrs. Melanie   Kling, Mr. Nevil 
Kling, Mr. Wayne      Lazarow, Mrs. Fiona   Lazarow, Mr. Mark 
Norton, Mr. Charles   Norton, Mrs. Jodi     Roy, Mr. Michael 
Roy, Mrs. Nicki       Tsafrir, Mrs. Lauren  Tsafrir, Mr. Thomer 

It's a good temp solution, but if anyone is good with for loops, I'd appreciate the help, thanks!

-Dave

Hi everyone.

I'm stuck on the following query. I need to display all the fields listed below on a page, but linked via communications.CommID.  I'd appreciate any assistance you can provide. thank you.

Code: [Select]

<?php
 $result = mysql_query("SELECT records.NameFirst_1, records.NameLast_1, records.CompanyName, records.CompanyBranch, records.CompanyReferenceNumber, records.CaseOwnerSelect, communications.ConversionType, communications.Contact, communications.ContactFrom, communications.CommID, communications.ContactPosition, communications.ContactTelephone, communications.ContactEmail, communications.ContactFax, communications.CallDate, communications.CallTime, communications.ActionTextField
FROM records INNER JOIN communications ON records.IDNumber = '$IDNumber'")

 or die(mysql_error()); 
 $row = mysql_fetch_array($result);
?>

Hi, fairly new to PHP over the last couple weeks. Been having a problem with certain queries. I have a database with football results, games, teams etc. I can filter these using drop down and that's all well and good. The problem I'm having is displaying the data via gameweek. I've been asked to display the table like so - Gameweek1 will display week1 teams, results etc. Gameweek2 will display week2... and so on. I can manage to do this in a drop down. But I've been asked to display this using links like "Previous, 1, 2, 3 Next". I've tried pagination but I couldn't figure it out. Can anyone point me in the right direction? If I need a GET() method, how would I go about coding that so it will be used in a link(s)? Been searching and searching to find an answer but to no avail...

//Database connection etc...

$gameweek = "SELECT * FROM games WHERE gameweek= 1";
    
    //if(isset($_GET['gameweek']))
    //{
    //    $gameweek = $_GET['gameweek'];
    //    
    //}
    //....
    $result=mysqli_query($connection, "select * from games WHERE gameweek= 1");
 //Print table and table headings...
    
    mysqli_close($connection);
?>
<a href="http://weeks.php?gameweek=2">Week 2</a>
<a href="http://weeks.phpgameweek=3">Week 3</a>

</body>
</html>

I'm trying to pull results from a database based on where the user is located based upon the variables $usr_lat & $usr_lng, and search for by a radius of x amount of miles/km (need to make it optional). I can't seem to find exactly what I'm looking for on google so I thought I'd asked here. Any help would be appreciated.

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

I would like assistance in displaying a company name in my web search results.   Below is my code:

<html>

  <head></head>

  <body>

  <?php if (!isset($_POST['q'])) { ?>





<img src="/wvb-logo-slogen.png" border="0" />

    <h2>Search</h2>

    <form method="post" action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>">

      <input type="text" name="q" size="30" />

    </form>



  <?php } else { ?>



<img src="/wvb-logo-slogen.png" border="0" />

    <h2>Search Results</h2>



  <?php

    try {

      // create object

     // $swish = new Swish('/usr/local/apache/htdocs/swish/index.swish-e');

        $swish = new Swish('/var/www/html/pdf2/index.swish-e');





      // get and run query from command-line

      $queryStr = htmlentities($_POST['q']);

      $result = $swish->query($queryStr);

  ?>



   Found <?php echo $result->hits; ?> match(es) for '<?php echo $queryStr; ?>'.



  <?php

      // iterate over result set

      // print details for each match

      while($r = $result->nextResult()) {

   ?>



  <p>

    <?php echo $r->swishreccount; ?>

      <strong>

      <a href="<?php echo '/pdf2', ltrim($r->swishdocpath, '.') ;  ?>">

         <?php echo $r->swishdocpath; ?>

      </a>

   </strong> (sco  <?php echo $r->swishrank; ?>) <br/>

<?php echo $r->swishdocpath; ?><br />

<?php

//Split a filename  by .

$filenames = explode(".", $r->swishdocpath);

//get 3 chars from $filenames to $country

$country = substr($filenames[1],1,3);

echo 'Country name: '.$country."<br />";

//$filenames[2] = explode(".", $r->swishdocpath);

$year = substr($filenames[2],0,4);

echo 'Year: '.$year;

//$filenames = explode(".", $r->swishdocpath);

$wvbnumber = substr($filenames[1],1,12);

echo 'WVB Number: '.$wvbnumber;



?>



  </p>



 <?php

      }

    } catch (Exception $e) {

      die('ERROR: ' . $e->getMessage());

    }

  }

  ?>



  </body>

</html>

 

As you can see I am able now display the WVB number, country name and year. But my question to anyone is how would I display the company name that it corresponds to?

The names of the company are located in a .csv file called active_colist.csv and it is under the /var/www/html directory.

This is what my /var/www/html directory looks like:

[root@zeus wvbadmin]# cd /var/www/html
[root@zeus html]# ls -l
total 2140
-rw-r--r--. 1 root root 2110323 May 14 23:39 active_colist.csv
-rw-r--r--. 1 root root    6678 Apr 30 13:25 favicon.ico
-rw-r--r--. 1 root root   17256 May  5 16:02 h1
-rw-r--r--. 1 root root     113 Apr 29 16:45 hello.php
-rw-r--r--. 1 root root      19 Apr 24 23:53 info.php
drwxr-xr-x. 2 root root   12288 May 12 15:30 pdf2
lrwxrwxrwx. 1 root root      20 May  5 15:46 pdf3 -> /home/wvbadmin/pdf3/
-rw-r--r--. 1 root root    1227 May  6 16:33 pdfsearch2.php
-rw-r--r--. 1 root root    1204 May  6 15:13 pdfsearch3.php
-rw-r--r--. 1 root root    1625 May 19 23:27 pdfsearch.php
-rw-r--r--. 1 root root    1838 Apr 30 13:10 search.php
-rw-r--r--. 1 root root   10077 May 12 11:38 wvb-logo-slogen.png


What is in the .csv file is the first 12 characters that correspond to the company name. What PHP code would I use to grab the first 12 characters of search results match them up with what is in the .csv file and display the proper company name?

This is what is inside of the .csv file:

WVB_NUMBER,PRIMARY_SHORT_COMPANY_NAME,CURR_ISO_CNTRY_OF_INCORP
"AIA000030001","THE NATIONAL BANK OF ANGUILLA","AIA"
"AIA000030003","ANGUILLA ELECTRICITY COMPANY","AIA"
"AIA000030005","KMT-HANSA","AIA"
"ALB000040001","BURSE E TIRANES","ALB"
"ANT000020000","RORENTO","ANT"
"ANT000030001","INTRUM JUSTITIA","SWE"
"ANT000030002","ORTHOFIX INTERNATIONAL","ANT"
"ANT000030004","HAL TRUST","ANT"
"ARE000030001","MASHREQBANK","ARE"
"ARE000030002","COMMERCIAL BANK OF DUBAI","ARE"   Any assistance would be greatly appreciated.

Hi,

I've used this code before, but have had to make some modifications and am now getting a mysql fetch array error:

 Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/tesolcla/public_html/test/results10k201010.php on line 134

If anyone can help, it would be much appreciated.

Code: [Select]

<?php
$dbcnx = @mysql_connect('localhost', 'MYUSERNAME', 'MYPASSWORD');
//$dbcnx = @mysql_connect('localhost', 'root', 'mysql');
if (!$dbcnx) {
  exit('<p>Unable to connect to the ' .
      'database server at this time.</p>');
}

if (!@mysql_select_db('MYDATABASE')) {
//if (!@mysql_select_db('rac')) {
  exit('<p>Unable to locate the results ' .
      'database at this time.</p>');
}
$asc_on = '<img src="images/results_sorting/Asc.gif" border="0" />';
$asc_off= '<img src="images/results_sorting/AscOff.gif" border="0" />';
$desc_on = '<img src="images/results_sorting/Desc.gif" border="0" />';
$desc_off= '<img src="images/results_sorting/DescOff.gif" border="0" />';

$sortfield = isset($_GET['sort']) ? $_GET['sort'] : '4';
$sorttype = isset($_GET['type']) ? $_GET['type'] : '1';

for($i=1; $i<5; $i++)
{
if($i==$sortfield)
{
if ($sorttype==1)
$srt[$i] = $asc_on.'<a href="?sort='.$i.'&type=2">'.$desc_off.'</a>';
else
$srt[$i] = '<a href="?sort='.$i.'&type=1">'.$asc_off.'</a>'.$desc_on;
}
else
{
$srt[$i] = '<a href="?sort='.$i.'&type=1">'.$asc_off.'</a><a href="?sort='.$i.'&type=2">'.$desc_off.'</a>';
}
}

$fields = array("firstname", "lastname", "time", "position");
$sorts = array("ASC", "DESC");

$field = $fields[$sortfield-1];
$sort = $sorts[$sorttype-1];

$field = $field=="" ? $fields[4] : $field;
$sort = $sort=="" ? $sorts[0] : $sort;


$sql = mysql_query("SELECT firstname, lastname, time, position FROM 10k_results ORDER BY $field $sort");


echo "<table border='1'  align='center' bordercolor='#000000' CELLPADDING=5 cellspacing='0' STYLE='font-size:13px'>";
echo "<tr bgcolor='#008000' STYLE='color:white'> <td>*</td><td><H3>First name $srt[1]</h3></td> <td><H3>Lastname $srt[2]</H3></td>  <td><H3>Time $srt[3]</H3></td>  <td><H3>Position $srt[4]</H3></td></tr>";
// keeps getting the next row until there are no more to get

 $row_counter = 1; //create row counter with default value 0

// Print out the contents of each row into a table


 while ($row = mysql_fetch_array($sql))

{   
 // Print out the contents of each row into a table   
echo "<tr>\n";    


echo "</td><td>";
echo $row_counter++; 
echo "</td>";
echo "<td>{$row['firstname']}</td>\n";    
echo "<td>{$row['lastname']}</td>\n";   
echo "<td>{$row['time']}</td>\n";    
echo "<td>{$row['position']}</td>\n";   
echo "</tr>\n";
}
echo "</table>";


?>

am total newbie to programming, apart from knowing SQL, the thing is i have been given a MYSQL database containing various information about kids diseases and a web interface written in php to create reports from the database that can be accessed via the interface. there are almost 25 different variables that need to be computed in the report, i have written sql queries to compute all these values, but i don't know anything about PHP, if i have all these queries isn't there a way for me to combine all these sql queries to be display results on a webpage and come up with this report without writing PHP code?

Thanks for your help

very sorry if this is too basic

Hello everyone,

So what I'm trying to do is have a dropdown menu displaying a number of <options> for people to select and to update that selection to the database, easy enough right? But I want that option to be displayed as the "selected" option when the page is revisited or refreshed and I just can't figure it out!!! (Permission to bang head on desk?) It would seem like it sould be a really basic thing to do but it's got me completely and a lot of menus around the site are going to rely on this so I came to you guys for help.

A simple example would be like the facebook edit profile page, the user selects whether they are Male or Female, the database gets updated and when you return the option you selected before is the one that appears as if selected="selected" had been done. I've tried everything I can think of (all be it from a learners perspective) with no joy, ive managed to get the database connection sorted, the tables done, the login with unique id $_SESSION, logout etc... so then when I got to this I thought... easy LOL yeah right.

Some of this probably doesnt even make sense but I'll show you the kind of things I've tried...


            <select name="gender" size="1" id="gender">
              <option value="male" <?php if ($gender == "male") {echo 'selected="selected"';} ;?>>Male</option>
              <option value="female" <?php if ($gender == "female") {echo 'selected="selected"';} ;?>>Female</option>
            </select>



OR


     <select name="gender" id="gender">
              <option value="" selected="<?php if (!isset($gender)) {echo "selected";} ;?>">Select</option>
              <option value="male" selected="<?php if ($gender == "male") {echo "selected";} else {echo "";} ;?>">Male</option>
              <option value="female" selected="<?php if ($gender == "female") {echo "selected";} else {echo "";} ;?>">Female</option>
            </select>


OR



            <select name="gender" size="1" id="gender">
              <option selected="<?php if (!isset($gender)) {echo "selected";} ;?>">Select</option>
              <option value="<?php if ($gender == "Male") {echo "selected";} else {echo "male";} ;?>">Male</option>
              <option value="<?php if ($gender == "Female") {echo "selected";} else {echo "female";} ;?>">Female</option>
            </select>


OR

            <select name="gender" id="gender">
              <option value="male"><?php if ($gender == "male") {echo "Male";} ;?></option>
              <option value="female"><?php if ($gender == "female") {echo "Female";} ;?></option>
            </select>


Honestly man, I've got no idea.


The other thing is, I have more than 1 dropdown menu in the same form (5 in total) and if I use 2 or more selecting different options as I go I get a blank screen.  And one more, if I have selected Male and it updates the users row and I resubmit Male again it's blank screen time again, lol.

Any help would be tremendous and greatly appreciated.

Thanks very much, Learner

P.S  Man!

Hey guys,
Having a slight problem with part of the code in my index.php file

Code: [Select]
mysql_select_db('db_name', $con);

$result = mysql_query("SELECT * FROM spy ORDER BY id desc limit 25");
$resulto = mysql_query("SELECT * FROM spy ORDER BY id desc");
$count = mysql_num_rows($resulto);
while($row = mysql_fetch_array($result))
  {
?>
<div class="contentDiv">Someone is looking at <?=$row[title];?> Stats for  "<a href="/<?=$row[type];?>/<?=$row[code];?>/<?=$row[city];?>"><?=$row[code];?> <?=$row[city];?></a>"</div>
<?}?>
</div>
  <div id="login"></div>
          <? include("footer.php"); ?>
</div>
</body>
</html>
I'm getting the following error when viewing the file

Code: [Select]
Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in ~path to file/index.php on line 70

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in ~path to file/index.php on line 71

where lines 70 and 71 are

$count = mysql_num_rows($resulto);
while($row = mysql_fetch_array($result))

Any ideas on how to fix this?

Hi,
This has been baffling me for a couple hours now and i cant seem to figure it out.
I have some code which creates an array and gets info from a mysql database and then displays in a list. This works great but after adding more and more rows to my database the list is now becoming quite large and doesnt look great on my site.
Is it possible to split the list into multiple columns of about 25 and if possible once 3 or 4 columns have been created start another column underneath.
To help explain i would be looking at a layout as follows:
Code: [Select]
line 1      line 1      line 1
line 2      line 2      line 2
...      ...      ...
line 25      line 25      line 25

line 1      line 1      line 1
line 2      line 2      line 2
...      ...      ...
line 25      line 25      line 25Im guessing there should be some sort of if statement to check how many items are being displayed and to create a new column if necessary. Is this correct?

Thanks,
Alex

Hey all,

First off, if this is in the wrong section I apologize. I wasn't sure if it should be here or the mySQL section.

What's going on is, I'm in the process of learning the Prepared Statement way of doing things and am changing / updating my code to reflect the changes. Everything was going fine until I attempted to do what I could do using old MySQL methods and that is display the queried results on the same page.


I can place a query and display the results as they should be displayed if I only use one block of code. However, if I try to do any additional queries on the same page, they get killed and do not display anything even though I know the query is fine because I can test the exact same syntax below one a different page and it works.

Here's a code snippet for an example:
Code: [Select]

Code:
<table>
<tr>
<td>
// The below code will display a selection box containing various strings such as "hello world", "great to be here", "Wowserz", "this is mind blowing" etc. that are stored in the database.

<?php

   echo "<select = \"SpecialConditions\">";
   if($stmt->num_rows == NULL){
      echo "No results found.";
   }else{
      while($stmt->fetch()){
   
       echo "<option value=\"$specialId\">$specialcondition</option>";
   }
}
   echo "</select>";
?>
</td>
<td>
// If I place another fetch query below the above fetch() query, this one will not show up. This one is supposed to display values 1 - 20 that have been stored in the DB.

<?php
echo "<select = \"NumberSets\">";
   if($stmt->num_rows == NULL){
      echo "No results found.";
   }else{
      while($stmt->fetch()){
      
          echo "<option value=\"".$numbers."\">".$numbers."</option>";
      }
   }
echo "</select>";

?>

</td>
</tr>
</table>


What am I doing wrong with this? When I use regular SQL queries I can display multiple results on the same page.

The results are being pulled from two separate joined tables but I don't think that's the issue.

Hello,
 
I have a quick question about methods for retrieving records from a mysql table and displaying
them as a links

For example,  imagine I have three tables called countries, cities and city_info.
I'd like to be able to select a country and have a list of that country's city names returned as links. 
I'd then like to be able to click on the link for London, say, and that would trigger a mysql query to retrieve
the entry in city_info about London.

Are there any functions that allow this?

If anyone could point me in the right direction for further research I'd be grateful.

Thanks. 

Alright,
I need to find a way to display the top ten duplicated rows for a table in my database. I have tried multiple methods but I have failed. Can you please assist me with this problem. I currently run a query in phpmyadmin to get the result, but i cant figure out how to properly code the php script.