Question

这个位置变量在我的代码中工作正常。 http://vince.netau.net

var locations = [
  ["John Doe", "145 Rock Ridge Road, Chester, NY ", "41.314926,-74.270134", "http://maps.google.com/mapfiles/ms/icons/blue.png"],
  ["Jim Smith", "12 Williams Rd, Montvale, NJ ", "41.041599,-74.019554", "http://maps.google.com/mapfiles/ms/icons/green.png"],
  ["John Jones", "689 Fern St Township of Washington, NJ ", "40.997704,-74.050598", "http://maps.google.com/mapfiles/ms/icons/yellow.png"],
 ];

现在我的下一步是代替上面的静态数据，我想从我的mysql数据库中检索数据并将其用于var位置。我有这个php，它以与上面完全相同的格式从mysql中吐出数据。 http://vince.netau.net/db-connect-test.php

<?php
$servername = "";
$username = "";
$password = "";
$dbname = "";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT id, name, address, lat, lng, Icon FROM markers";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
    echo '["'  . $row["name"]. '"'.    ', '. '"' . $row["address"].'"'.', '. '"'. $row["lat"].','. $row["lng"].'"'.', '. '"'. $row["Icon"]. '"]'. ','. "<br>";
    }
} else {
    echo "0 results";
}
$conn->close();
?>

我只是不知道如何让它发挥作用。非常感谢有关如何执行此操作的任何帮助。我是初学者。

由于

更新 - 大卫，我尝试了你写的代码。在获取html中的位置数据方面，我还不确定这对我有什么作用？

<?php
    //open connection to mysql db
    $connection = mysqli_connect("","","","") or die("Error " . mysqli_error($connection));

    //fetch table rows from mysql db
    $sql = "select * from markers";
    $result = mysqli_query($connection, $sql) or die("Error in Selecting " . mysqli_error($connection));

    $location = array();;
if ($result->num_rows > 0) {
// output data of each row
$key = 0;
while($row = $result->fetch_assoc()) {
$locations[$key][] = $row["name"];
$locations[$key][] = $row["address"];
$locations[$key][] = $row["lat"];
$locations[$key][] = $row["lng"];
$locations[$key][] = $row["Icon"];
$key++;
}
}
else {
echo "0 results";
}

echo json_encode($locations); 


    //close the db connection
    mysqli_close($connection);
?>

以下是结果，有点搞砸了。

[[＆＃34; John Doe＆＃34;，＆＃34; 147 Rock Ridge Road，Chester，NY＆＃34;，＆＃34; 41.314926＆＃34;，＆＃34; -74.270134＆＃ 34;，＆＃34; http：//maps.google.com/mapfiles/ms/icons/blue.png"]，[＆＃34; Jim Smith＆＃34;，＆＃34; 14 Williams Rd， Montvale，NJ＆＃34;，＃34; 41.041599＆＃34;，＆＃34; -74.019554＆＃34;，＆＃34; http://maps.google.com/mapfiles/ms/icons/green .png＆＃34;]，[＆＃34; John Jones＆＃34;，＆＃34; 691 Fern St Township of Washington，NJ＆＃34;，＆＃34; 40.997704＆＃34;，＆＃34; - 74.050598＆＃34;＆＃34; HTTP：//maps.google.com/mapfiles/ms/icons/yellow.png"]]

Answer 1

在服务器端，您应该使用json_encode将数据插入页面。

e.g

<?php
echo "<script> var data = '" + json_encode($locations) + "'; </script>";
?>

在客户端，

var locations = JSON.parse(data);

Answer 2

我看到你正在尝试获取位置数组，因此存储为multidimentional数组会很有帮助

$location = array();;
if ($result->num_rows > 0) {
$key = 0;
// output data of each row
while($row = $result->fetch_assoc()) {
$locations[$key][] = $row["name"];
$locations[$key][] = $row["address"];
$locations[$key][] = $row["lat"];
$locations[$key][] = $row["lng"];
$locations[$key][] = $row["Icon"];
$key++;
}
}
else {
echo "0 results";
}
//echo json_encode($locations); 
<?php
echo "<script> var locations  = '" . json_encode($locations) . "'; </script>";
?>

Answer 3

因此，我们假设php文件现在为我提供了以前在html中var = locations下硬编码的确切位置数据。

我需要在html中使用什么新代码，以便调用php文件，硬编码数据将替换为从mysql表中提取的php位置结果。

谢谢

从mysql数据库中检索javascript变量

3 个答案: