Problem need solution

[bn1995]

Hi all ,

down you will see my code

what I want to covert the values to selected name .

means : I have these fields in form sex,agemin,agemax,nationality,contryorgin,countrystaying and status

I inserted sex,nationality,countryorgin,countrystaying,status to mysql using

drop-down box ,so all data will be in numbers for example the inserted for male in mysql 2

1)so what i want in this code when I do display data to be in name 2==converted to==> male instead of printing 2 .

2) how can I get another field is not in the form like username to be printed with each row and to be as a hayber link refre to profile.php

3) as you see there are agemin and agemax two fields for age in my sql there is only age ,,,, in this sorting form I have 2 input text boxs agemin & agemax

so there when I enter agemin as 18 & agemax as 30 ,,, will sort all people between these 2 ,,,, how can I manage it in this code.

I know lot of requests but you are the experts.

thank you in advance .

=========

<?php

include('conit.php');
include("MySQLPagedResults.class.php");



$paging_results = new MySQLPagedResults("SELECT count(*) FROM 

users","page","",12,10,"<<","Previous","Next",">>"," | ");

$first_nav = $paging_results->getFirstNav();
$prev_nav = $paging_results->getPrevNav();
$next_nav = $paging_results->getNextNav();
$last_nav = $paging_results->getLastNav();
$pages_nav = $paging_results->getPagesNav();
$offset = $paging_results->currentOffset();
$results_per_page = $paging_results->results_per_page;
$current_page = $paging_results->current_page;
$total_pages = $paging_results->totalPages();
$start_number = $paging_results->getResultNumbersStart();
$end_number = $paging_results->getResultNumbersEnd();




function ee( $var )
{
return mysql_real_escape_string( $var );
} 

function add_sql( $field , $var )
{
global $res;
if ( !empty( $_POST["$var"] ) ) {
if ( $_POST["$var"] != "-1" )
$res .= $field . "='" . ee( $_POST["$var"] ) . "' AND ";
} 
} 

$res = null;


add_sql( "sex","sex" ); // field name in the table, and $_POST variable 

name
add_sql( "age", "agemax" );
add_sql( "age", "agemin" );
add_sql( "nationality", "nationality" );
add_sql( "contryorgin", "contryorgin" );
add_sql( "countrystaying", "countrystaying" );
add_sql( "status", "status" );

$fields = array( "sex" =>"sex"  , "age" => "agemax" , "age" => 

"agemin","nationality" => "nationality","contryorgin" => 

"contryorgin","countrystaying" => "countrystaying","status" => "status" );






if ( !empty( $res ) ){
$Q = " WHERE " . substr( $res , 0, -4 );

}
else
$Q = "";



$res="SELECT * FROM USERS $Q LIMIT $offset,$results_per_page";


$query = mysql_query($res);  

$counter = 1;
while ( $data = mysql_fetch_assoc($query) ) {
print $counter . ":
";
foreach( $fields AS $key => $value ) {
echo "$value: " . $data["$key"] . "
";
} 

print "
<hr>";
$counter++;
} 

echo "<p><b>Page:</b> $current_page of $total_pages &nbsp;|&nbsp; 

<b>Results:</b> $start_number - $end_number<br>";
echo "$first_nav $prev_nav $pages_nav $next_nav $last_nav</p>";




?> 

[djjjozsi]

To solve the first bullet,

lets add the query the another tables in a JOIN:

for example:

$res="SELECT * FROM USERS LEFT JOIN sex ON sex.sex_id=USERS.sex_id $Q LIMIT $offset,$results_per_page";

And in the field list add the mentioned field names.

$fields = array( "username" => "User Name" , "sex" =>"sex"  , "age" => "agemax" , "age" =>

"agemin","nationality" => "nationality","contryorgin" =>

"contryorgin","countrystaying" => "countrystaying","status" => "status" ); 

[bn1995]

thank you djjjozsi,

I tired ur way and its working with ALL when I select ALL =male + female

its working fine

but when I select either female or male error coming out

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in

thats what I did

$res="SELECT * FROM USERS LEFT JOIN sex ON sex.sex=USERS.sex $Q LIMIT 

$offset,$results_per_page";

$query = mysql_query($res);  

$counter = 1;
while ( $data = mysql_fetch_assoc($query) ) {
print $counter . ":
";

foreach( $fields AS $key => $value ) {
echo "$value: " . $data["$key"] . "
";
} 

[nemonoman]

The failed query is failing because of something in

"$Q $offset,$results_per_page"; 

echo "$res" and review the query as it is being sent to the database.