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Don't force barcode data to be a JSON object.

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Just send the scanned data to the server, and let it handle it...
This commit is contained in:
Oliver Walters 2020-04-14 23:23:24 +10:00
parent d0f1a4d4ce
commit cd864abfbe
1 changed files with 3 additions and 14 deletions
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17
lib/barcode.dart
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@ -17,20 +17,9 @@ import 'dart:convert';
Future<void> scanQrCode(BuildContext context) async { Future<void> scanQrCode(BuildContext context) async {
QrUtils.scanQR.then((String result) { QrUtils.scanQR.then((String barcode) {
print("Scanned: $result"); print("Scanned: $barcode");
Map<String, dynamic> data;
// Look for JSON data in the result...
try {
data = json.decode(result);
} on FormatException {
showErrorDialog(context, "Bad barcode data", result);
return;
}
showProgressDialog(context, "Querying Server", "Sending barcode data to server"); showProgressDialog(context, "Querying Server", "Sending barcode data to server");
@ -39,7 +28,7 @@ Future<void> scanQrCode(BuildContext context) async {
* It is the responsibility of the server to validate and sanitize the barcode data, * It is the responsibility of the server to validate and sanitize the barcode data,
* and return a "common" response that we know how to deal with. * and return a "common" response that we know how to deal with.
*/ */
InvenTreeAPI().post("barcode/", body: data).then((var response) { InvenTreeAPI().post("barcode/", body: {"barcode": barcode}).then((var response) {
// Close the progress dialog // Close the progress dialog
Navigator.pop(context); Navigator.pop(context);