PHP MySQL Search

[savagenoob]

I have 1 table that I want to do a search on, with 3 variables. Whats the easiest way to do this, I want to search by either/or.

 

<?php
session_start();
$agency = $_SESSION['SESS_AGENCY'];
$agent = $_SESSION['SESS_MEMBER_ID'];
$firstname = $_GET['First_Name'];
$lastname = $_GET['Last_Name'];
$polnum = $_GET['PolNum'];
$date1 = $_GET['Date1'];
$date2 = $_GET['Date2'];
$date1 = strtotime($date1);
$date2 = strtotime($date2);
$newdate1= date('Y-m-d', $date1);
$newdate2= date('Y-m-d', $date2);
$time2 = " 00:00:01";
$time = " 23:59:59";
$date3 = $newdate2 . $time;
$date4 = $newdate1 . $time2;
$daterange = $_REQUEST['daterange'];
require_once('../config.php');
//Array to store validation errors
$errmsg_arr = array();

//Validation error flag
$errflag = false;

//Connect to mysql server
$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
if(!$link) {
	die('Failed to connect to server: ' . mysql_error());
}

//Select database
$db = mysql_select_db(DB_DATABASE);
if(!$db) {
	die("Unable to select database");
}
    if($daterange = "today"){
        $todaytime1 = " 00:00:01";
        $todaytime2 = " 23:59:59";
        $datetoday = date('Y-m-d');
        $todaydate1 = $datetoday . $todaytime1;
        $todaydate2 = $datetoday . $todaytime2;
        $todayquery = mysql_query("SELECT * FROM clients WHERE First_Name = '%". $firstname ."%' AND Last_Name = '%". $lastname ."%' AND Agency = '$agency' BETWEEN '$todaydate1' AND '$todaydate2' ") or die(mysql_error());
        $numrows = mysql_num_rows($todayquery);
        if($numrows == 0){
            echo "No prospects matching that criteria.";
        }?>

 

This will only return a restult if all 3 return somthing.....

[fenway]

You need LIKE, not =