如何使用php将javascript对象保存到mysql?现在我正在使用googlemapapi,它会返回一个响应对象
directionsService.route({
origin: document.getElementById('start').value,
destination: document.getElementById('end').value,
travelMode: 'BICYCLING'
}, function(response, status) {
if (status === 'OK') {
directionsDisplay.setDirections(response); //The Return Response
//This function call ajax below to send response Obj with a key to store to MySQL
storeRouteDB(storeLocations, response);
showSteps(response, markerArray, display, map);
} else {
window.alert('Directions request failed due to ' + status);
}
});
我使用ajax将objet发送到我的后端php
$.ajax({
type: "GET",
url: 'http://localhost:8080/storedRouteDB.php',
async: false,
data: {key: key, value: value},
error: function(){
console.log("Error in Ajax");
},
success: function (data) {
result = data;
}
});
在我的php中,我做到了:
$key = $_REQUEST['key'];
$value = $_REQUEST['value'];
echo checkDB($key, $value);
function checkDB($key, $value) {
$dbServerName = "localhost";
$dbUserName="root";
$dbPassword="";
$dbName="myDB";
....
// Trying to zip the object to String and save it as char to MYSQL
$value_zip = serialize($value);
$sql = "INSERT INTO storedRoute (locations, objects) VALUES ('$key', '$value_zip')";
但我得到的结果是:
Uncaught TypeError: Cannot read property 'b' of undefined
at _.n.intersects (js?key=AIzaSyAbBYjTy8g4-dGYAl4_mHmWDVoWGEziq6c&callback=initMap:147)
at i (jquery.min.js:2)
at jt (jquery.min.js:2)
at jt (jquery.min.js:2)
at jt (jquery.min.js:2)
at jt (jquery.min.js:2)
at Object.<anonymous> (jquery.min.js:2)
at Function.each (jquery.min.js:2)
at jt (jquery.min.js:2)
at jt (jquery.min.js:2)
如何修复此错误?或者有没有其他更好的方法通过使用php将javascript对象存储到mysql?谢谢您!
1条答案
按热度按时间siotufzp1#
我通常会这样做:
然后可以访问中的对象
$_POST['data']
. 也许你也想用json_decode
. 让我知道它是否有用。