I want my script to get the latitude and longittude coordinates from a GPS(mobile phone) and then insert them into a google map and display the location of the Mobile phone. Well , i am able to retrieve the Lat,Long coordinates from the GPS but it won't insert them as the çentre'of the map displayed. My code : <script> $(document).ready(function(){ navigator.geolocati
我想让我的脚本从GPS（手机）获取纬度和长度坐标，然后将它们插入谷歌地图并显示手机的位置。 那么，我可以从GPS中检索Lat，Long坐标，但不会将它们插入到显示的地图的çentre。 我的代码： <script> $(document).ready(function(){ navigator.geolocation.getCurrentPosition(useposition); }); function useposition(position){ lat = position.coords.latitude;
I found this link on Stack Overflow Google Maps - How to get the distance between two point in metre? which shows the following solution in regards to finding a km distance given two different latitude and longitude values: <script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false&libraries=geometry"></script> var p1 = new google.maps.LatLng(45.463688,
我发现这个链接堆栈溢出谷歌地图 - 如何获得米之间的两点之间的距离？ 其中给出了关于在给定两个不同纬度和经度值的情况下找到km距离的以下解决方案： <script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false&libraries=geometry"></script> var p1 = new google.maps.LatLng(45.463688, 9.18814); var p2 = new google.maps.LatLng(46.0438317, 9.75936230000002); alert(ca
I Have this script the construct a JSON: echo "{"; echo ""name": "flare","."n"; echo ""children": ["."n"; foreach($tree as $tag => $customer){ foreach($customer as $customerId => $order){ echo "{"name":"".$customerId."","; echo ""children": ["."n"; foreach($order as $orderId => $orderTotal){ echo "{"name": "".$orderId."", "size": "".$order
我有这个脚本构造一个JSON： echo "{"; echo ""name": "flare","."n"; echo ""children": ["."n"; foreach($tree as $tag => $customer){ foreach($customer as $customerId => $order){ echo "{"name":"".$customerId."","; echo ""children": ["."n"; foreach($order as $orderId => $orderTotal){ echo "{"name": "".$orderId."", "size": "".$orderTotal.""},";
I am trying to pretty print JSON currently in PHP, I have looked at the threads like Pretty-Printing JSON with PHP this but it does not work. I am on PHP 7.0.8-0ubuntu0.16.04.3 Code: <form method="POST"> <textarea name="json_data" id="json_data"> <?php if(isset($_POST['json_data'])){ echo json_encode($_POST['json_data'], JSON_PRETTY_PRINT
我试图漂亮地打印目前在PHP中的JSON，我已经看过像PHP这样的漂亮打印JSON的线程，但它不工作。 我在PHP 7.0.8-0ubuntu0.16.04.3 码： <form method="POST"> <textarea name="json_data" id="json_data"> <?php if(isset($_POST['json_data'])){ echo json_encode($_POST['json_data'], JSON_PRETTY_PRINT | JSON_UNESCAPED_SLASHES); } ?>
I'm trying to pass JSON constants from a handler down to an object that I want to serialize into JSON. When I try the following code: class AJAXHandler { public function getPrettyPrint() { $jh = new JSONHandler(); $jh->getJSON(JSON_PRETTY_PRINT | JSON_FORCE_OBJECT); } } class JSONHandler { protected $id; protected $name; public function getJSON($json
我试图从一个处理程序传递JSON常量到我想要序列化成JSON的对象。 当我尝试下面的代码： class AJAXHandler { public function getPrettyPrint() { $jh = new JSONHandler(); $jh->getJSON(JSON_PRETTY_PRINT | JSON_FORCE_OBJECT); } } class JSONHandler { protected $id; protected $name; public function getJSON($json_constants) { if (isset($json_constants)) {
Hard to understand (json_encode) i'm using the code: <?php $query = mysql_query("SELECT * FROM messages ORDER BY ID"); while($fetch = mysql_fetch_assoc($query)) { $titel = $fetch[title]; $post = array('items' => array( 0 => array('title' => "$title", 'description' => "$title"))); echo json_encode($post); } ?> Output: {"items":[{"title":"title
很难理解（json_encode）我使用的代码： <?php $query = mysql_query("SELECT * FROM messages ORDER BY ID"); while($fetch = mysql_fetch_assoc($query)) { $titel = $fetch[title]; $post = array('items' => array( 0 => array('title' => "$title", 'description' => "$title"))); echo json_encode($post); } ?> 输出： {"items":[{"title":"title","description":"tit
How can I do JSON pretty print from MySQL? I used JSON_PRETTY_PRINT in my code but it is not printing what I am expecting. My current script is: <?php //open connection to mysql db $connection = mysqli_connect("127.0.0.1","root","Kunal@7890","testdb") or die("Error " . mysqli_error($connection)); //fetch table rows from mysql db $sql = "select id,title,profilepic,created_at,url from new
我怎样才能做MySQL的JSON漂亮打印？ 我在我的代码中使用了JSON_PRETTY_PRINT ，但它没有打印我期待的内容。 我目前的脚本是： <?php //open connection to mysql db $connection = mysqli_connect("127.0.0.1","root","Kunal@7890","testdb") or die("Error " . mysqli_error($connection)); //fetch table rows from mysql db $sql = "select id,title,profilepic,created_at,url from news"; $result = mysqli_query
return Response::json(array( 'status' => 200, 'posts' => $post->toArray() ), 200); Using the code above I returned data in json format. I have seen other api's that return json giving it back in formatted view. Like: http://api.steampowered.com/ISteamNews/GetNewsForApp/v0002/?appid=440&count=3&maxlength=300&format=json But mine is returning it in one line.
return Response::json(array( 'status' => 200, 'posts' => $post->toArray() ), 200); 使用上面的代码我以json格式返回数据。 我已经看到其他的api返回json的格式化视图。 喜欢： http://api.steampowered.com/ISteamNews/GetNewsForApp/v0002/?appid=440&count=3&maxlength=300&format=json 但我的回归是一条线。 如何用laravel格式化生成json？ 更新 我明天才能测试代码。 所以我
This question already has an answer here: Pretty-Printing JSON with PHP 20 answers I am assuming that you are displaying "HTML" to the user and the formatting that you wish to show is in HTML, then n wont cause new lines... use " <br/>" . And then your single and double quotes wont matter, either. 尝试将您的新行字符转换为<br> HTML元素，因为无论您用于解析并显
这个问题在这里已经有了答案： 漂亮的打印JSON与PHP 20答案 我假设你正在向用户显示“HTML”，并且你希望显示的格式是HTML格式，那么 n不会导致新行...使用“ <br/>" 。 然后你的单引号和双引号也不重要。 尝试将您的新行字符转换为<br> HTML元素，因为无论您用于解析并显示JSON，显然都不会遵守n新行： $displaystring = nl2br($displaystring); $displaystring = "Welcome to Polio Buddy. Please select.
This question already has an answer here: Pretty-Printing JSON with PHP 20 answers Use JSON_PRETTY_PRINT flag (PHP 5.4+): $json_str = json_encode($data, JSON_PRETTY_PRINT); Since you're having a JSON-string, you can first decode it into an object, and then re-encode it. Example: $str = '{"name":"John","age":"12","Location":"U.S.A"}'; echo json_encode(json_decode($str), JSON_PRETTY_PR
这个问题在这里已经有了答案： 漂亮的打印JSON与PHP 20答案 使用JSON_PRETTY_PRINT标志（PHP 5.4+）： $json_str = json_encode($data, JSON_PRETTY_PRINT); 既然你有一个JSON字符串，你可以先将它解码成一个对象，然后对它进行重新编码。 例： $str = '{"name":"John","age":"12","Location":"U.S.A"}'; echo json_encode(json_decode($str), JSON_PRETTY_PRINT); 输出： { "name": "John", "age": "12",