Question

我使用以下指南使用谷歌地图创建商店定位器：https://developers.google.com/maps/articles/phpsqlsearch_v3

到目前为止，我没有收到任何错误，但我无法将数据库中的数据显示为XML。相反，它显示以下内容：

This XML file does not appear to have any style information associated with it. The document tree is shown below.
<locations/>

它只显示文档应该是什么样的结尾，但它没有返回任何数据。

这是我到目前为止的代码：

<?php
require("dbinfo.php");

// Get parameters from URL
$center_lat = $_GET["lat"];
$center_lng = $_GET["lng"];
$radius = $_GET["radius"];

// Start XML file, create parent node
$dom = new DOMDocument("1.0");
$node = $dom->createElement("locations");
$parnode = $dom->appendChild($node);

// Opens a connection to a mySQL server
$connection=mysql_connect ($hostname, $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());
}

// Search the rows in the locations table
$query = sprintf("SELECT address, name, lat, lng, ( 3959 * acos( cos( radians('%s') ) * cos( radians( lat ) ) * cos( radians( lng ) - radians('%s') ) + sin( radians('%s') ) * sin( radians( lat ) ) ) ) AS distance FROM locations HAVING distance < '%s' ORDER BY distance LIMIT 0 , 20",
  mysql_real_escape_string($center_lat),
  mysql_real_escape_string($center_lng),
  mysql_real_escape_string($center_lat),
  mysql_real_escape_string($radius));
$result = mysql_query($query);

$result = mysql_query($query);
if (!$result) {
  die("Invalid query: " . mysql_error());
}

header("Content-type: text/xml");

// Iterate through the rows, adding XML nodes for each
while ($row = @mysql_fetch_assoc($result)){
  $node = $dom->createElement("locations");
  $newnode = $parnode->appendChild($node);
  $newnode->setAttribute("name", $row['name']);
  $newnode->setAttribute("address", $row['address']);
  $newnode->setAttribute("lat", $row['lat']);
  $newnode->setAttribute("lng", $row['lng']);
  $newnode->setAttribute("distance", $row['distance']);
}

echo $dom->saveXML();
?>

我的表名是'locations'。

我之前收到了sql错误，但设法通过纠正我之前做的数据库连接错误来传递它们，所以我确信它已连接到数据库。

感谢任何帮助或建议。

由于

麦克

Answer 1

您的代码没问题。主要问题是谷歌在此链接https://developers.google.com/maps/articles/phpsqlsearch_v3中提供的样本位置不包含25个半径范围内的地址。因此，您需要在URL

中将半径从25更改为28或30

phpsqlsearch_genxml.php?lat=37&lng=-122&radius=28

虽然您的代码可能有一些错误。所以我在这里提供我的代码。

<?php
require("phpsqlsearch_dbinfo.php");

// Get parameters from URL
$center_lat = $_GET["lat"];
$center_lng = $_GET["lng"];
$radius = $_GET["radius"];

// Start XML file, create parent node
$dom = new DOMDocument("1.0");
$node = $dom->createElement("markers");
$parnode = $dom->appendChild($node);


// Search the rows in the markers table
$query = sprintf("SELECT address, name, lat, lng, ( 3959 * acos( cos( radians('%s') ) * cos( radians( lat ) ) * cos( radians( lng ) - radians('%s') ) + sin( radians('%s') ) * sin( radians( lat ) ) ) ) AS distance FROM markers HAVING distance < '%s' ORDER BY distance LIMIT 0 , 20",
  mysqli_real_escape_string($connection,$center_lat),
  mysqli_real_escape_string($connection,$center_lng),
  mysqli_real_escape_string($connection,$center_lat),
  mysqli_real_escape_string($connection,$radius));
$result = mysqli_query($connection,$query);
if (!$result) {
  die("Invalid query: " . mysqli_error($connection));
}

header("Content-type: text/xml");

// Iterate through the rows, adding XML nodes for each
while ($row = @mysqli_fetch_assoc($result)){
  $node = $dom->createElement("marker");
  $newnode = $parnode->appendChild($node);
  $newnode->setAttribute("name", $row['name']);
  $newnode->setAttribute("address", $row['address']);
  $newnode->setAttribute("lat", $row['lat']);
  $newnode->setAttribute("lng", $row['lng']);
  $newnode->setAttribute("distance", $row['distance']);
}

echo $dom->saveXML();
?>

这肯定会解决你的问题。

如果您需要完整的商店定位器代码，请写入注释。

使用php和sql创建一个谷歌商店定位器

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