[英]Show JSON Response (lat, lon) in google map
我正在尝试构建一个从数据库(mysql)获取用户数据(纬度,经度)的应用,然后在Google地图上显示其位置。
这是代码示例:
function initMap() {
var mapDiv = document.getElementById('map');
var map = new google.maps.Map(mapDiv, {
center: {lat: 33.7167, lng: 73.0667},
zoom: 11
});
makeRequest('get_locations.php', function(data) {
//alert("abc");
var data = JSON.parse(data.responseText);
//window.alert(data);
for (var i = 0; i < data.length; i++) {
//display(data[i]);
displayLocation(data[i]);
}
});
}
function makeRequest(url, callback) {
var request;
if (window.XMLHttpRequest) {
request = new XMLHttpRequest(); // IE7+, Firefox, Chrome, Opera, Safari
} else {
request = new ActiveXObject("Microsoft.XMLHTTP"); // IE6, IE5
}
request.onreadystatechange = function() {
if (request.readyState == 4 && request.status == 200) {
callback(request);
}
}
request.open("GET", url, true);
request.send();
}
get_locations.php
<?php
include('connection.php');
$l= array();
$result = mysqli_query($con,"SELECT * FROM users");
while ($row = mysqli_fetch_assoc($result)) {
$l[] = $row;
}
$j = json_decode($l, true);
//echo $j;
?>
我尝试在代码上找到问题的一些方法,但是失败了
在回声$ j的get_location.php中; 它告诉我正确的答案
在此行之前, var data = JSON.parse(data.responseText); alert(“ abc”)有效,但此后不起作用,所以我认为问题出在这一行,但不知道为什么会发生,知道吗?
我认为主要问题是在您需要相反的json_encode
的情况下滥用json_decode
function initMap() {
var mapDiv = document.getElementById('map');
var map = new google.maps.Map(mapDiv, {
center: {lat: 33.7167, lng: 73.0667},
zoom: 11
});
makeRequest('get_locations.php', function(data) {
if( data ){
var data = JSON.parse( data );
for( var n in data ){
var obj=data[ n ];
console.log( '%o', obj );
displayLocation.call(this,obj);
}
}
});
}
function makeRequest(url, callback) {
var request;
request = window.XMLHttpRequest ? new XMLHttpRequest() : new ActiveXObject("Microsoft.XMLHTTP");
request.onreadystatechange = function() {
if (request.readyState == 4 && request.status == 200) {
callback( request.responseText );
}
}
request.open("GET", url, true);
request.send();
}
<?php
include('connection.php');
$l= array();
$result = mysqli_query( $con, "SELECT * FROM users" );
while ( $row = mysqli_fetch_assoc( $result ) ) {
$l[] = $row;
}
/* encode array as a json object */
$json = json_encode( $l );
/* send the correct content-type header */
header('Content-Type: application/json');
/* send the data */
exit( $json );
?>
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