I have a database insert. But it inserts a new record twice. No errors or bug report.
Form code is:
<form id="frmfotosend" method="post" is="dmx-serverconnect-form" action="dmxConnect/api/pages/fotowedstrijd/album/insertalb.php" dmx-on:success="notifies1.success('Uw inzending is succesvol geupload en verzonden');frmfotosend.reset()">
Action step:
PhpMyAdmin print:
Anyone who can tell me how this happens and, even more important, how I can fix this?
did you check form and button type ?
Make sure you do not have a submit button running the servre action as well as an onclick or onsuccess or something like that running the same server action twice.
Open the Chrome developer tools, look under network, XHR, and check if the server action is fired once or twice.
This is the button:
<button type="submit" class="btn btn-primary" dmx-on:click="frmfotosend.submit()">Foto's inzenden<span class="spinner-border spinner-border-sm" role="status" dmx-show="state.executing"></span></button>
And this is the form:
<form is="dmx-serverconnect-form" id="frmfotosend" method="post" action="dmxConnect/api/pages/fotowedstrijd/album/insertalb.php" dmx-generator="bootstrap4" dmx-form-type="horizontal" dmx-on:success="notifies1.success('Uw inzeding is succesvol geupload en verzonden');frmfotosend.reset()">
I can not see any anomalies.
<button class="btn btn-primary" dmx-on:click="frmfotosend.submit()">Foto's inzenden<span class="spinner-border spinner-border-sm" role="status" dmx-show="state.executing"></span></button>
try again
anomalies ;
type="submit" and dmx-on:click="frmfotosend.submit()"
use
type="submit or dmx-on:click="frmfotosend.submit()"
Noted. Working now. Thanks a million
Yes the submit button already submits the form and having an extra click action submits it again… so there you have your double records
@George: duly noted. Sunday afternoon rush must have caused this. BTW is there any reply to my pm to you?
