Question

有人可以帮助您使用此代码。代码基本上要求做两件事之一。如果输入了邮政编码，则代码执行半径搜索，并在数据库中搜索包含数组中邮政编码的任何行。但是，如果输入城市/街道，我希望代码绕过这个并只搜索包含它们的详细信息。守则如下。

    $postcode = str_replace( '+', '%20', $postcode );

$conn = mysql_connect('127.0.0.1','user','pass') or die('db connect error:'.mysql_error());
mysql_select_db('database', $conn) or die('could not select database');

$sqlstring = "SELECT * FROM postcodelatlng WHERE postcode ='".$postcode."'";
$result = mysql_query($sqlstring);   

$row = mysql_fetch_assoc($result);

$lng = $row["longitude"] / 180 * M_PI;
$lat = $row["latitude"] / 180 * M_PI;

mysql_free_result($result);    

$sqlstring2 = "SELECT DISTINCT 
  postcodelatlng.postcode,
  (
    6367.41 * SQRT(
      2 * (
        1- COS(RADIANS(postcodelatlng.latitude)) * COS(".$lat.") * (
          SIN(RADIANS(postcodelatlng.longitude)) * SIN(".$lng.") + COS(RADIANS(postcodelatlng.longitude)) * COS(".$lng.")
        ) - SIN(RADIANS(postcodelatlng.latitude)) * SIN(".$lat.")
      )
    )
  ) AS Distance 
FROM
  postcodelatlng AS postcodelatlng 
HAVING Distance <= '".$radius."'
ORDER BY Distance ";

$result1 = mysql_query($sqlstring2) or die('query failed: ' . mysql_error());
    while($row = mysql_fetch_array($result1)){
        if ($row['postcode'] == 0){

            $sqlstring3="SELECT * FROM property_details WHERE  street = '".$postcode."' OR address3 = '".$postcode."' OR county = '".$postcode."' ";
    $result2 = mysql_query($sqlstring3) or die('Query Failed: ' . mysql_error());
        while($row2 = mysql_fetch_array($result2)) {
            echo ' 
    <div class="container">
        <div class="row">
            <a href="viewprop.php?id='.$row2[0].'">
                <div class="col-md-12"> 
                    <div class="col-md-12 col-md-offset-0 inner col-xs-12" align="justify">
                        <div class="col-md-2 col-xs-12">
                            <img src="images/rentals/1.jpg" width="80%" height="80%" />
                        </div>
                        <div class="col-md-8 col-xs-12">
                            <h2>'. $row2[13]. '</h2>
                        </div>
                        <div class="col-md-2 col-xs-12">
                            <h4 class="rbgclr2">£'. $row2[2]. ' Per Month</h3>
                            <h5 class="rbgclr1">£'. $row2[3]. ' Deposit</h4>
                            <br />
                            <p>Bedrooms = '. $row2[4]. '</p>
                        </div>
                    </div>
                </div>
            </a>
        </div>
    </div>
            <br />';
        }
        }
        else {
            $sqlstring4="SELECT * FROM property_details WHERE postcode = '".$row['postcode']."' ";
    $result3 = mysql_query($sqlstring4) or die('Query Failed: ' . mysql_error());
        while($row2 = mysql_fetch_array($result3)) {
            echo ' 
    <div class="container">
        <div class="row">
            <a href="viewprop.php?id='.$row2[0].'">
                <div class="col-md-12"> 
                    <div class="col-md-12 col-md-offset-0 inner col-xs-12" align="justify">
                        <div class="col-md-2 col-xs-12">
                            <img src="images/rentals/1.jpg" width="80%" height="80%" />
                        </div>
                        <div class="col-md-8 col-xs-12">
                            <h2>'. $row2[13]. '</h2>
                        </div>
                        <div class="col-md-2 col-xs-12">
                            <h4 class="rbgclr2">£'. $row2[2]. ' Per Month</h3>
                            <h5 class="rbgclr1">£'. $row2[3]. ' Deposit</h4>
                            <br />
                            <p>Bedrooms = '. $row2[4]. '</p>
                        </div>
                    </div>
                </div>
            </a>
        </div>
    </div>
            <br />';
        }
    }
    }




mysqli_close($con); 

?>

为什么我的代码没有显示结果

0 个答案: