JS:
document.addEventListener("deviceready",onDeviceReady,false);
function onDeviceReady() {
document.getElementById("xbutton").addEventListener("click",savecheck,false); }
function savecheck(){
var checkb = document.getElementById("MedA");
var medAHistory = JSON.parse(localStorage.getItem("MedA") || "[]");
medAHistory.push(checkb.checked);
localStorage.setItem("MedA", JSON.stringify(medAHistory));
}
html:
<input type="checkbox" id="MedA" name="Med" value="A">Medication A
<input id="xbutton" type="button" onClick="savecheck()">
So basically when I submit the form, the status of the checkbox will be saved to localstorage without overwriting. Right now when I click Submit button, duplication of the checkbox status will be saved to localstorage. E.g if I unchecked the box and submit, value in localstorage will be [false, false] even if I only submit once.
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