Cordova 无法使用Phonegap在SQLite数据库中创建表

Cordova 无法使用Phonegap在SQLite数据库中创建表,cordova,Cordova,我尝试使用phonegap连接到数据库。看起来数据库被打开了,但我无法在数据库中创建表。数据库创建后,执行似乎停止了。使用db.transaction的下一步不起作用 代码- function createDB() { alert("in createDB..."); var db = window.openDatabase("Database60", "1.0", "Phonegap Demo60", 200000); alert("after openDatabase

我尝试使用phonegap连接到数据库。看起来数据库被打开了,但我无法在数据库中创建表。数据库创建后,执行似乎停止了。使用db.transaction的下一步不起作用

代码-

function createDB() {
    alert("in createDB...");
    var db = window.openDatabase("Database60", "1.0", "Phonegap Demo60", 200000);
    alert("after openDatabase...");

    db.transaction(
            function(tx)
            {
                    alert("in function(tx)...")
                    tx.executeSql('DROP TABLE IF EXISTS DEMO');
                    tx.executeSql('CREATE TABLE IF NOT EXISTS DEMO (id unique, data)');
                    tx.executeSql('INSERT INTO DEMO (id, data) VALUES (1, "First row")');
                    tx.executeSql('INSERT INTO DEMO (id, data) VALUES (2, "Second row")');                  
                 },
            function(tx, err)
            {
                    alert("Error processing SQL: "+err);
            }
            )

    alert("after table creation...");
} 

尝试使用lawnchair。最方便的包装

//this creates the database 

db.transaction(populateDB, transaction_error, populateDB_success);

function populateDB(tx) {
tx.executeSql('DROP TABLE IF EXISTS Registration');
var sql = "CREATE TABLE IF NOT EXISTS Registration (" + " id INTEGER PRIMARY KEY AUTOINCREMENT," + "name VARCHAR(50), " + "address VARCHAR(50),"+" age INTEGER(2), "+" birthdate DATE, "+" gender VARCHAR(10), "+" hobbies VARCHAR(100), "+" email VARCHAR(50),"+" uname VARCHAR(50),"+" pwd VARCHAR(50))";

tx.executeSql(sql);

tx.executeSql("INSERT INTO Registration  (name,address,age,birthdate,gender,hobbies,email,uname,pwd) VALUES ('"+ v1 +"','"+ v2 +"','"+ v3 +"','"+ v4 +"','"+ v5 +"','"+ v6 +"','"+ v7 +"','"+ v8 +"','"+ v9 +"')");}

function transaction_error(tx, error) {
alert("Database Error: " + error);
}

function populateDB_success() {
dbCreated = true;
alert("Successfully inserted");
db.transaction(queryDB, errorCB);
}


This works for me.Try this. May be helpful to you.

谢谢。

你能提供比一行更好的答案吗?请看这里的答案