PHP设置值时的查询问题？

Question

快速提问。我有一个适用于我的数据库的 SQL 查询。但是，我确定如何在 PHP 文件中设置参数，以便它适用于我的查询？

有什么想法吗？我在 php 方面设置参数并使我的查询工作时遇到问题。

<?php

/*
 * Following code will list all the products
 */

// array for JSON response
$response = array();

// include db connect class
require_once __DIR__ . '/db_connect.php';

// connecting to db
$db = new DB_CONNECT();

set $orig_lat=32.42; 
set $orig_long=-23.550; 
set $bounding_distance=1;

// get all products from products table
$result = mysql_query(" SELECT * ,((ACOS(SIN($orig_lat * PI() / 180) * SIN(`latitude` * PI() /
180) + COS($orig_lat * PI() / 180) * COS(`latitude` * PI() / 180) *
COS(($orig_long - `longitude`) * PI() / 180)) * 180 / PI()) * 60 *
1.1515) AS `distance` FROM `Location` WHERE ( `latitude` BETWEEN
($orig_lat - $bounding_distance) AND ($orig_lat + $bounding_distance)
AND `longitude` BETWEEN ($orig_long - $bounding_distance) AND
($orig_long + $bounding_distance) ) ORDER BY `distance` ASC limit 1") or die(mysql_error()); 

// check for empty result
if (mysql_num_rows($result) > 0) {
    // looping through all results
    // products node
    $response["Location"] = array();

    while ($row = mysql_fetch_array($result)) {
        // temp user array
       $Location = array();
       $Location["id"] = $row["id"];
       $Location["latitude"] = $row["latitude"];
       $Location["longitude"] = $row["longitude"];



        // push single product into final response array
        array_push($response["Location"], $Location);
    }
    // success
    $response["success"] = 1;

    // echoing JSON response
    echo json_encode($response);
} else {
    // no products found
    $response["success"] = 0;
    $response["message"] = "No Profiles found";

    // echo no users JSON
    echo json_encode($response);
}
?>

Answer 1

这是一个简单的语法问题...与查询相对应的值没有很好地注明。在 PHP 文件中，您不能使用 $ 编写设置和声明变量。

原文:

set $orig_lat=32.42; 
set $orig_long=-23.550; 
set $bounding_distance=1;

解决方案:

$orig_lat=32.42; 
$orig_long=-23.550; 
$bounding_distance=1;