我正在使用google maps locator代码来获取某个半径范围内的用户(例如:25)。当我在运行后尝试将输出检查为xml文件时,它只显示预定义的echo语句而不是sql数据。我不知道sql是否检索数据或没有? 任何人都知道这个问题请回答。
这是我的PHP代码:
<?php
require("phpsqlajax_dbinfo.php");
header("Content-type: text/xml");
function parseToXML($htmlStr)
{
$xmlStr=str_replace('<','<',$htmlStr);
$xmlStr=str_replace('>','>',$xmlStr);
$xmlStr=str_replace('"','"',$xmlStr);
$xmlStr=str_replace("'",''',$xmlStr);
$xmlStr=str_replace("&",'&',$xmlStr);
return $xmlStr;
}
// Get parameters from URL
$center_lat = $_GET["lat"];
$center_lng = $_GET["lng"];
$radius = $_GET["radius"];
// Opens a connection to a MySQL server
$connection=mysql_connect ("localhost","$username","$password");
if (!$connection) {
die('Not connected : ' . mysql_error());
}
// Set the active MySQL database
$db_selected = mysql_select_db("$database",$connection);
if (!$db_selected) {
die ('Can\'t use db : ' . mysql_error());
}
// Select all the rows in the markers table
$query = sprintf("SELECT gender, username, lat, lng, ( 3959 * acos( cos( radians('%s') ) * cos( radians( lat ) ) * cos( radians( lng ) - radians('%s') ) + sin( radians('%s') ) * sin( radians( lat ) ) ) ) AS distance FROM register HAVING distance < '%s' ORDER BY distance LIMIT 0 , 20",
mysql_real_escape_string($center_lat),
mysql_real_escape_string($center_lng),
mysql_real_escape_string($center_lat),
mysql_real_escape_string($radius));
$result=mysql_query($query);
if (!$result) {
die('Invalid query: ' . mysql_error());
}
// Start XML file, echo parent node
echo "<markers>\n";
// Iterate through the rows, printing XML nodes for each
while ($row ==@mysql_fetch_assoc($result)){
// ADD TO XML DOCUMENT NODE
echo '<marker ';
echo 'name="' . parseToXML($row['username']) . '" ';
echo 'address="' . parseToXML($row['gender']) . '" ';
echo 'lat="' . $row['lat'] . '" ';
echo 'lng="' . $row['lng'] . '" ';
echo 'distance="' . $row['distance'] . '" ';
echo "/>\n";
}
// End XML file
echo "</markers>\n";
?>