Question

我正在尝试创建一个函数，该函数返回玩家基于本地数据库获得的胜利数量。然而，值'p1'和'p2'似乎每次退出循环时都会重置为0.

function getAmountOfWins(player) {
var p1 = 0, p2 = 0;
db.transaction(function (tx) {
    tx.executeSql("SELECT * FROM Filter WHERE player1='" + player + "'", [], function (tx, results) {
        for (var i = 0; i < results.rows.length; i++) {
            current = results.rows.item(i);
            p1 = p1 + current.p1wins;
        }
    });
});
db.transaction(function (tx) {
    tx.executeSql("SELECT * FROM Filter WHERE player2='" + player + "'", [], function (tx, results) {
        for (var i = 0; i < results.rows.length; i++) {
            current = results.rows.item(i);
            p2 = p2 + current.p2wins;
        }
    });
});
console.log("Total Wins before return: " + (p1 + p2));
return p1 + p2;
}

在这种情况下，当p1和p2退出'db.transaction'时，它们似乎被重置为0。我很遗憾如何保存数据，所以我可以实际返回它。

任何人都有一个可以防止这种情况的魔法修复？

Answer 1

他们没有重置。他们从未被设定过。 executeSQL和transaction方法是异步。这意味着你必须做一些这样的事情：

function getAmountOfWins(player, callback) {
    // Call function, callback is executed after the query calculations
    getPlayerValues(1, player, function(p1num){
        // Call function, callback is executed after the query calculations
        getPlayerValues(2, player, function(p2num){
            // Call the original callback function
            callback( p1num + p2num );
        });
    });
}

// Get the player values
function getPlayerValues(playernr, player, callback){
    db.transaction(function (tx) {
        tx.executeSql("SELECT * FROM Filter WHERE player" + playernr + "='" + player + "'", [], function (tx, results) {
            var p = 0;
            for (var i = 0; i < results.rows.length; i++) {
                current = results.rows.item(i);
                p = p + current.p2wins;
            }
            callback(p);
        });
    });
}

电话：

// Get the amount of wins, this is an async functionm with the callback
getAmountOfWins(player, function(value){
    console.log("Total Wins before return: " + (value));

});

Answer 2

我将假设executeSql是异步的，因为它接受回调函数。

您执行此功能的主体将在console.log运行很久之后的某个时间点执行。您需要重构代码，以便通过传递给console.log的回调运行executeSql（以及您想要包含的任何其他逻辑）。

Javascript - 在返回之前重置值

2 个答案: