Question

我有多模块maven项目，我想要包含IntelliJ为我的.gitignore所有模块创建的.iml和.idea文件。我的app结构如下：

exports.relations = function (req, res) {

    var collection = [];

    async.parallel([
        function () {
            var firstqueryarray= [];
            connection.query(query1, function (err, rows, fields) {
                if (err) throw err;
                for (var i in rows) {
                    firstqueryarray.push({
                        name: rows[i].name,
                        email: rows[i].email
                    });
                }
                res.json({
                    firstqueryarray: firstqueryarray
                });
            });

        }

    , function () {
            var secondqueryarray= [];
            connection.query(query2, function (err, rows, fields) {
                if (err) throw err;

                for (var i in rows) {
                    secondqueryarray.push({
                        name: rows[i].name,
                        email: rows[i].email
                    });
                }
                res.json({
                    secondqueryarray: secondqueryarray
                });
            });
        }
      ]
    )
};

并且.gitignore的内容像傻瓜一样：

root/  

.gitignore

/module1

/module2

然而git仍然将.idea文件夹中的所有文件显示为未跟踪

你可以帮忙吗？

Answer 1

尝试将Cursor c=db.rawQuery("SELECT * FROM your_table_name WHERE Tname='r_c', null); if(c.getCount()==0) { showMessage("Error", "No records found"); return; } StringBuffer buffer=new StringBuffer(); while(c.moveToNext()) { buffer.append(c.getString(0)+"\n"); buffer.append("From: "+c.getString(2)+"\n"); buffer.append("Price: "+c.getString(1)+"\n"); buffer.append("Nots: "+c.getString(3)+"\n\n"); } // Displaying all records showMessage("Student Details", buffer.toString()); public void showMessage(String title,String message) { Builder builder=new Builder(this); builder.setCancelable(true); builder.setTitle(title); builder.setMessage(message); builder.show(); }更改为*/.idea/。

包括项目的intellij模块到gitignore

1 个答案: