Question

没有显示任何错误，但重新发现仍然没有更新.. 在我的快递函数下面删除。

waitingModel.delete = function(id,callback) {
    db.getConnection(function(err, connection) {
        var sql = 'UPDATE customers_ariving SET deleted = 1 WHERE id='+id;
        console.log(sql);
        connection.query(sql,function(err,data) {
            var result = {};
            if (err) {
                connection.release();
                callback(err, null);
            } else {
                connection.release();
                result = data;
                callback(null, result);
            }
       });
   });
};

Answer 1

我找到了来自ref link的解决方案：Node.js mysql query syntax issues UPDATE WHERE

我想更新列deleted = 1，所以我直接添加了值。

waitingModel.delete = function(id,callback) {
    db.getConnection(function(err, connection) {
        var sql = 'UPDATE customers_ariving SET deleted = ? WHERE id = ?';
        connection.query(sql, ['1',id],function(err,data){
            var result = {};
            if (err) {
                connection.release();
                callback(err, null);
            } else {
                connection.release();
                result = data;
                callback(null, result);
            }
        });
    });
};

更新查询在快速js中不起作用

1 个答案: