Question

我目前使用LIKE进行SQL搜索查询。然后，这将用于生成HTML格式的返回结果。

但是，如果它包含多个关键字，我想只返回一次，可能是第一行。目前，它会为关键字x返回一次项目，对关键字y再次返回。

目前我正在使用for和while循环来执行每个关键字的搜索。这是最好的解决方法吗？还是有其他方法可以做到这一点？

目前我想保持简单，所以没有solr或lucene等等。

为了清晰起见，我添加了代码，但是大量代码并非无关紧要。代码可以工作，所以不在这方面寻找建议。

<?php
session_start();
include_once('connect_to_sql.php');
include_once('distance.php');
$itemList = '';
if(isset($_POST['words'])){
    $words = $_POST['words'];
} else { 
    $words = "";
}
$queries = preg_split("/[\s,]+/", $words);
for($i = 0; $i < count($queries); $i++){
    $keyword = $queries[$i];
    if($keyword!=""){
if(isset($_SESSION['myLong']) && isset($_SESSION['myLat'])){
    $request_long = $_SESSION['myLong'];
    $request_lat = $_SESSION['myLat'];
    $itemQuery = mysql_query("SELECT *, ( 3959 * acos( cos( radians('$request_lat') ) * 
cos( radians( item_lat ) ) * 
cos( radians( item_long ) - 
radians('$request_long') ) + 
sin( radians('$request_lat') ) * 
sin( radians( item_lat ) ) ) ) 
AS distance FROM offerings WHERE sold = 'no' AND item_desc LIKE '%$keyword%'
ORDER BY distance ASC");
} else {
    $itemQuery = mysql_query("SELECT * FROM offerings WHERE sold = 'no' AND item_desc LIKE '%$keyword%'
    ORDER BY id DESC");
}


//get all items

    if(mysql_num_rows($itemQuery) < 1){
        $itemList = "No items currently on sale.";
    } else {
        while($itemResult = mysql_fetch_array($itemQuery)){
            $from = $itemResult['from_user'];
            $locationQuery = mysql_query("SELECT postcode FROM users WHERE id = '$from' ");
            $locationResult = mysql_fetch_array($locationQuery);
            $se = urlencode($locationResult['postcode']);
            $num = urlencode($itemResult['id']);
            if($itemResult['item_picture'] == ''){
                $img = '<img src="../offerings/placehold.png" />';
            } else {
                $img = '<img src="../offerings/'.$itemResult['from_user'].'/'.$itemResult['item_picture'].'" />';
            }
                    $url = "http://maps.google.com/maps/api/geocode/json?address=".$se."&sensor=false";
                    $jsonData = file_get_contents($url);
                    $data = json_decode($jsonData);
                    $xlat = $data->{'results'}[0]->{'geometry'}->{'location'}->{'lat'};
                    $xlong = $data->{'results'}[0]->{'geometry'}->{'location'}->{'lng'};
                    if(isset($request_lat) && isset($request_long)){
                        $dist = number_format(distance($request_lat, $request_long, $itemResult['item_lat'], $itemResult['item_long']), 2)." miles / ";
                        $distkm = number_format(distance($request_lat, $request_long, $itemResult['item_lat'], $itemResult['item_long'])*1.609344, 2)." km";
                    } else {
                        $dist = $locationResult['postcode']; 
                        $distkm = '';
                    }

        $checkifsold = mysql_query("SELECT * FROM orders WHERE item_id='$num'");
        if(mysql_num_rows($checkifsold) > 0){
            $itemPrice = "SOLD";
        } else {
            $itemPrice = '£'.$itemResult['item_price'];
        }
        $itemList .= '<div class="anItemSearched"><a href="item.php?num='.$num.'">'.$img.'</a>
                        <div class="innerDetail">'.$dist.$distkm.'</div>
                        <div class="innerDetail">'.$itemPrice.'</div>
                        <div class="innerDetail">'.$itemResult['item_name'].'</div>
                        </div>';
        }
    }
    }
    }
    echo $itemList;
?>

谢谢， Cillian

SQL搜索关键字搜索，仅返回一行

0 个答案: