Error Fetching Data From a PHP Server Running MySQL with Angular JS
Question
I am getting a [$http:baddata] error when my Angular app calls my php file. I'm not sure where the error is triggering exactly, but I believe it's a problem constructing the json object! I've attached all relevant code. Any help would be amazing. Thanks in advance :)
Angular documentation shows this
This is the PHP file which connects to my database and attempts to return a JSON formatting string back to the calling file
<?php
echo '<script>console.log("Inside php file!")</script>';
header("Access-Control-Allow-Origin: *");
header("Content-Type: application/json; charset=UTF-8");
require_once('mysqli_connect.php');
$query = "SELECT rideid, destination, price, capacity, userid, origin, departDate, vehicle FROM rides";
$result = @mysqli_query($dbc, $query);
$outp = "";
while($rs = $result->fetch_array(MYSQLI_ASSOC)) {
if ($outp != "") {$outp .= ",";}
$outp .= '{"Origin":"' . $rs["origin"] . '",';
$outp .= '"Destination":"' . $rs["destination"] . '",';
$outp .= '"Price":"'. $rs["price"] . '"}';
$outp .= '"Capacity":"'. $rs["capacity"] . '"}';
}
$outp ='{"records":['.$outp.']}';
//echo '<script>console.log(JSON.stringify('.$outp.'))</script>';
$conn->close();
echo($outp);
This is the Angular app making the request to the PHP file
<html>
<head>
<!-- Include Angular -->
<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.6.6/angular.min.js"></script>
</head>
<body>
<!-- Ride Table -->
<div ng-app = "rideAppTest" ng-controller= "rideCtrlTest">
<table>
<tr ng-repeat = "x in names">
<td>{{x.Origin}}</td>
<td>{{x.Destination}}</td>
<td>{{x.Price}}</td>
<td>{{x.Capacity}}</td>
</tr>
</table>
</div>
<!-- End Ride Table -->
<!--Ride Table Script-->
<script>
var app = angular.module('rideAppTest', []);
app.controller('rideCtrlTest', function($scope, $http) {
$http.get("angularFilter.php")
.then(
function (response) {
$scope.names = response.data.records;
},
function(data){
console.log("Error!" + data);
}
);
});
</script>
</body>
</html>
2 answers
solution1
2 ACCPTED 2017-09-30 21:01:45
Don't construct the JSON the way you are doing, first collect everything inside an array or an object then do echo json_encode($result);
Something like this:
$outp = [];
while($rs = $result->fetch_array(MYSQLI_ASSOC)) {
$outp[] = $rs;
}
echo json_encode(["records" => $outp]);
solution2
0 2017-09-30 21:24:06
Thank to @Madhav for the help. This is what ended up working for me :)
while($rs = mysqli_fetch_array($result)) {
$myArray[] = array(
'Origin' => $rs["origin"],
'Destination' => $rs["origin"],
'Price' => $rs["price"],
'Capacity' => $rs["capacity"]
);
}
echo json_encode(["records" => $myArray]);
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.