Я использовал эту реализацию -> http://wiki.phonegap.com/w/page/16494756/Adding%20SQL%20Database%20support%20to%20your%20iPhone%20App
Вот мой код
var mydb=false;
// initialise the database
initDB = function() {
try {
if (!window.openDatabase) {
alert('not supported');
} else {
var shortName = 'test';
var version = '1.0';
var displayName = 'PhoneGap Test Database';
var maxSize = 65536; // in bytes
mydb = openDatabase(shortName, version, displayName, maxSize);
alert("Done");
}
} catch(e) {
// Error handling code goes here.
if (e == INVALID_STATE_ERR) {
// Version number mismatch.
alert("Invalid database version.");
} else {
alert("Unknown error "+e+".");
}
return;
}
}
celebsDataHandler=function(transaction, results) {
// Handle the results
var html = "<ul>";
alert("Length = "+results.rows.length);
for (var i=0; i<results.rows.length; i++) {
var row = results.rows.item(i);
html += '<li>'+row['id']+'</li>\n'
}
html +='</ul>';
alert(html);
}
// load the currently selected icons
loadCelebs = function() {
try {
mydb.transaction(
function(transaction) {
transaction.executeSql('SELECT * FROM call_details ORDER BY id',[], celebsDataHandler, errorHandler);
});
} catch(e) {
alert("Here"+e.message);
}
// db error handler - prevents the rest of the transaction going ahead on failure
errorHandler = function (transaction, error) {
alert(error);
return true;
}
}
После инициализации базы данных.Я вызываю loadCelebs () и затем получаю ошибку.Как это решить?