2016-12-07 19 views
-1

我需要输入我的latlngs价值为locations类似:与latlngs值多标记从数据库

var locations = [ 
['Bondi Beach', -33.890542, 151.274856], 
['Coogee Beach', -33.923036, 151.259052], 
['Cronulla Beach', -34.028249, 151.157507]  
['Manly Beach', -33.80010128657071, 151.28747820854187], 
['Maroubra Beach', -33.950198, 151.259302] 
]; 

这是我的PHP代码:

$result = $conn->query("select address, lat, lang from user where phoneno = '" . $phoneno. "'"); 

$outp = "["; 
while($rs = $result->fetch_array(MYSQLI_ASSOC)) { 
if ($outp != "[") {$outp .= ",";} 
$outp .= '{"address":' . $rs["address"] . ','; 
$outp .= '"lat":"' . $rs["lat"] . '",'; 
$outp .= '"lang":"' . $rs["lang"]  . '"}'; 
} 
$outp .="]"; 

而且这是在URL输出值:

[{"address":Tampines Street 86, 
    Singapore,"lat":"1.3498584","lang":"103.9273744"}] 
+2

你可以请详细说明您的问题 – Nitesh

+0

看一看@json_encode http://php.net/manual/en/function.json-encode.php –

+0

@Nitesh基本上我婉从我的PHP输出调入我的HTML作为var位置的值 – Xinee

回答

0

您需要从您的php中获取这些值并填充wi瘦HTML。 Here is an sample.

+0

我似乎无法使它工作。它也需要在一个循环中 – Xinee

+0

该链接的示例在JavaScript中完成 – Xinee

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