How to connect to existing sqlite database in phonegap - phonegap-build

Question

I use phonegap for create an android application.
I did these steps :
1. Wrote html, css and jquery mobile codes.
2. Include phonegap.js into .
3. Create the database via sqlite manager firefox extension and copy to root the project directory (beside index.html).
4. use this code:
<script>
$(function() {
onDeviceReady();
queryDB();
$("#links").niceScroll();
});
</script>
<script type="text/javascript" charset="utf-8">
document.addEventListener("deviceready", onDeviceReady, false);
function onDeviceReady()
{
var db = window.openDatabase("shia.sqlite", "1.0", "shia", 100000);
}
function queryDB(tx) {
tx.executeSql('SELECT * FROM posts', [], querySuccess, errorCB);
}
function querySuccess(tx, results) {
console.log("Returned rows = " + results.rows.length);
// this will be true since it was a select statement and so rowsAffected was 0
if (!results.rowsAffected) {
console.log('No rows affected!');
return false;
}
// for an insert statement, this property will return the ID of the last inserted row
console.log("Last inserted row ID = " + results.insertId);
}
function errorCB(err) {
alert("Error processing SQL: "+err.code);
}
</script>
This error displayed into console :
Uncaught TypeError: Cannot read property 'executeSql' of undefined
Please help me .

Answer 1

#Behzad,
your code has the function
function queryDB(tx) {
tx.executeSql('SELECT * FROM posts', [], querySuccess, errorCB);
}
However, when you call queryDB, you did not pass the handle. So in this function the parameter tx is null because you did not pass the handle.