好吧,所以我在尝试找出如何将我在localStorage中保存的一些数据传递给我编写的PHP脚本时遇到了一些问题,因此我可以将其发送到服务器上的数据库.我之前找到了一些代码,(https://developer.mozilla.org/en-US/docs/DOM/XMLHttpRequest/Using_XMLHttpRequest),看起来它会起作用,但我没有运气.
这是我保存数据的代码,而不是尝试将其传递给我的PHPscript
function getLocation() {
if (navigator.geolocation) {
navigator.geolocation.getCurrentPosition(initialize, showError, takeSnap);
}
else {
alert("Geolocation is not supported by this browser.");
}
}
function initialize(position) {
var lat = position.coords.latitude,
lon = position.coords.longitude;
var mapOptions = {
center: new google.maps.LatLng(lat, lon),
zoom: 14,
mapTypeId: google.maps.MapTypeId.ROADMAP,
mapTypeControl: true
}
var map = new google.maps.Map(document.getElementById("map-canvas"), mapOptions);
var marker = new google.maps.Marker({
position: new google.maps.LatLng(lat, lon),
map: map,
title: "Current Location"
});
}
function showError(error) {
switch (error.code) {
case error.PERMISSION_DENIED:
alert("User denied the request for Geolocation.");
break;
case error.POSITION_UNAVAILABLE:
alert("Location information is unavailable.");
break;
case error.TIMEOUT:
alert("The request to get user location timed out.");
break;
case error.UNKNowN_ERROR:
alert("An unkown error occurred.");
break;
}
}
function storeLocal(position) {
if (typeof (Storage) !== "undefined") {
var lat = position.coords.latitude,
lon = position.coords.longitude;
localStorage.latitude = lat;
localStorage.longitude = lon;
}
else {
alert("Your browser doesn't support web storage");
}
return
}
function snapShot() {
if (navigator.geolocation) {
navigator.geolocation.getCurrentPosition(storeLocal, showError);
}
else {
alert("Geolocation is not supported by this browser.");
}
var oReq = new XMLHttpRequest();
oReq.onload = reqListener;
oReq.open("post", "snap.PHP?lat=" + localStorage.latitude + "&lon=" + localStorage.longitude, true);
oReq.send();
}
function reqListener() {
console.log(this.reponseText);
}
这是我编写的用于将值保存到数据库中的脚本
<?PHP
// Connecting to the database
MysqL_connect("localhost", "username", "password");
MysqL_select_db("db_name");
$latitude = MysqL_real_escape_string($_GET["lat"]);
$longitude = MysqL_real_escape_string($_GET["lon"]);
// Submit query to insert new data
$sql = "INSERT INTO locationsTbl(LociD, lat, lon ) VALUES( 'NULL', '". $latitude ."', '". $longitude . "')";
$result = MysqL_query( $sql );
// Inform user
echo "<script>alert('Location saved.');</script>";
// Close connection
MysqL_close();
?>
解决方法:
怎么样:
oReq.open("get", "snap.PHP?lat=" + localStorage.latitude + "&lon=?" + localStorage.longitude, true);
(你也有localStorage.lon而不是.longitude)
由于值(字符串)在变量中,您需要连接它们,而不是将它们放在字符串中.
此外,由于您似乎将这些内容传递给PHP以保存到数据库,从语义上讲,您应该使用POST请求…这与AJAX请求的处理方式不同.
在PHP中,您需要使用:
$latitude = $_GET["lat"];
$longitude = $_GET["lon"];
实际获取使用GET请求发送的值.虽然这些值应该被转义以避免sql注入.
另外,我不确定你为什么要设置AJAX请求的onload属性.相反,使用onreadystatechange属性……类似于:
oReq.onreadystatechange = function () {
if (oReq.readyState === 4) {
if (oReq.status > 199 && oReq.status < 400) {
console.log("successful response");
} else {
console.log("Failed response: " + oReq.status);
}
}
};
.readyState属性引用其状态,其中4表示已完成(响应已返回). .status属性是指HTTP状态代码.通常介于200和200之间. 400是好的.我知道我见过人们只检查200(不是范围).
更新:
为了在请求中传递POST参数,您不会将它们附加到URL – 您可以在.send()方法中传递它们.以下是您的代码示例:
oReq.open("POST", "snap.PHP", true);
oReq.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
oReq.send("lat=" + encodeURIComponent(localStorage.latitude) + "&lon=" + encodeURIComponent(localStorage.longitude));
要在PHP中检索它们,您可以使用:
$latitude = $_POST["lat"];
$longitude = $_POST["lon"];
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 dio@foxmail.com 举报,一经查实,本站将立刻删除。