javascript mysql内部函数

当我运行这个函数时，当它返回时，我得到null，因为我在mysql查询中有返回，我如何解决这个问题

function CheckStorage(Host, Slots, SSLocation) {
    connection.query("SELECT * FROM `machines` WHERE `shost` LIKE '" + Host + "' AND `location` LIKE '" + SSLocation + "' AND `slotsa` >= " + Slots + "", function(err, rows, fields) {
        if(rows[0]) {
            console.log(rows[0]["ip"]);
            return rows[0]["ip"];
        } else {
            return false;
        }   
    });
}

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connection.query是一个异步函数，因此CheckStorage在connection.query完成之前返回。您需要将处理查询结果的回调函数传递给CheckStorage，然后在查询完成时调用该函数

function CheckStorage(Host, Slots, SSLocation, callback) {
    connection.query("SELECT * FROM `machines` WHERE `shost` LIKE '" + Host + "' AND `location` LIKE '" + SSLocation + "' AND `slotsa` >= " + Slots + "", function(err, rows, fields) {
        if(rows[0]) {
            console.log(rows[0]["ip"]);
            if (typeof callback == "function") {
               callback(rows[0]["ip"]); 
            }
        }  
    });
}

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该函数是异步的。。回调就是答案。可能重复的