Question

我在Google Gears（Sqlite）中创建了一个表：

db.execute('create table if not exists SPALINKS (link_id int PRIMARY KEY, sid1 int, sid2 int, label text, user text, Timestamp int');

使用Google Gears数据库查询工具（在FF中）时，使用'last-insert-rowid'的INSERT可以正常工作。来自Javascript的插入工作正常，只要我不使用'last-insert-rowid'。但是当使用'last-insert-rowid'时它不再起作用了：

    var sql_stmt = 'INSERT INTO SPALINKS (link_id, sid1, sid2, label, user, Timestamp) VALUES (last_insert_rowid(),?,?,?,?,?)';
var arg_array = [sid1, sid2, label, user, creation_time];
db.execute(sql_stmt, arg_array);

Error: Database operation failed. ERROR: constraint failed DETAILS: constraint failed

为什么SQLite的'last-insert-rowid'可以正常使用数据库工具，但不能在Javascript函数内部创建和执行的SQL语句中运行？

Answer 1

似乎问题是相应的列未定义为AUTOINCREMENT。

Answer 2

我认为在这种情况下你不必使用last_insert_rowid()。为什么只做INTEGER NOT NULL PRIMARY KEY AUTOINCREMENT。

// if you writing for android and you have something like this:
//document.addEventListener("deviceready", onDeviceReady, false);
//function onDeviceReady(){
// make sure you comment it if you debugging in browser

// connection to the databse
var db = window.openDatabase("demo.db", "1.0", "Demo Database", 200000);

// There is problem with SQLite recognize new Date() format so I include http://code.google.com/p/datejs/ library in my project and format the date to be recognizable by SQLite

function nowTime(){
        return new Date().toString("yyyy-MM-dd HH:mm:ss");
    }

function errorCallBack(err) {
            alert("Error processing SQL: " + err.code);
        }
function callBack(){

    function readTransactionError(err){
        alert('Read Transaction Error: ' + err);
    }

// you can make your vars global if you want to use them out side of function!!!

    function querySuccess(tx, results) {
       var len = results.rows.length;
        for(var i = 0; i < len; i++) {
           var sid1Var = results.rows.item(i).sid1;
           var sid2Var = results.rows.item(i).sid2;
           var labelVar = results.rows.item(i).label;
           var userNameVar = results.rows.item(i).userName;
           var timeStampVar = results.rows.item(i).timeStamp;
    }
  }

tx.readTransaction('select * from SPALINKS;', [], querySuccess, readTransactionError);
}
function functionName(sid1, sid2, label, userName) {
    db.transaction(function(tx) {
        db.executeSql('create table if not exists SPALINKS (link_id int INTEGER NOT NULL PRIMARY KEY AUTOINCREMENT, sid1, sid2, label, userName, timeStamp');

        db.executeSql('insert into SPALINKS (sid1, sid2, label, userName, timeStamp) VALUES (?, ?, ?, ?, ?)', [sid1, sid2, label, userName, nowTime()], callBack, errorCallBack);
    });
}
// assuming you use jQuery
$('#selectorID').click(function(){
functionName(sid1, sid2, label, userName);
});

您可以在last_insert_rowid()：

等情况下使用UPDATE

function updateFunction() {
db.transaction(function(tx) {
    tx.executeSql('update SPALINKS set timeStamp = "' + nowTime() + '" WHERE link_id = last_insert_rowid()');
});
}

我希望这会起作用。

在Javascript函数中使用last_insert_row（）与在DB查询工具中使用时的区别？

2 个答案: