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3 Posted Topics
Re: If javascript grabs the response as a comma separated string, sResponse, you do a arArray = sResponse.split(",") arArray is an array. So now you create your Select dropdown with each arArray[] element. If you have a variable number of dropdown elements, you will have to for-loop through the array while … ![]() | |
Re: Try hard coding a value into Line 33. It looks like the variable $voteModel might not be interpreted. In line 31, $voteModel = $_POST["vote_".$ID]; shouldn't that be $voteModel ="vote_".$_POST[$ID]; ? | |
Re: It looks like Chrome does not automatically create the object "uploadUid", and you have to use the function GetElementbyID, as in var myObj=document.getElementbyID("upLoadUid"); : : : formData.append('uploadUid', myObj.value); |