如何查找具有特定数字作为属性的所有列?

时间:2015-01-18 07:45:20

标签: python python-2.7 flask flask-sqlalchemy

我在SQLAlchemy中有一个表的类:

class pinCodes(db.Model):
id = db.Column(db.Integer, primary_key=True)
officename = db.Column(db.String(60), nullable=False)
pincode = db.Column(db.Integer, nullable=False)
taluk = db.Column(db.String(50), nullable=False)
district = db.Column(db.String(50), nullable=False)
state = db.Column(db.String(50), nullable=False)
__tablename__ = "pincodes"

def __init__(self, officename, pincode, taluk, district, state):
    self.officename = officename
    self.pincode = pincode
    self.taluk = taluk
    self.district = district
    self.state = state 

def __repr__(self):
    return '<officename {}>'.format(self.officename)

现在我想用&#34; pincode&#34;返回所有列。作为特定值: 这是代码:

def fmt_pin(code):
    return code.replace(" ","").replace("-","").replace("_","").strip()

def int_pin(code):
    code = fmt_pin(code)
    return int(code)



@app.route('/find_pin_codes', methods=['POST','GET'])
def pincodes():
    if request.method == 'POST':  

        if request.form.get("pincode", "") != "":
            pincode = request.form['pincode']
            try:
                num = pinCodes.query.filter(pincode = int_pin(pincode)).all()
                return render_template('pincodes.html', 
                                        bypincode = num

                                        )
            except Exception as e:
                return render_template('pincodes.html',
                                        error = e)

但是当我搜索一个数字时,我收到以下错误: filter()得到了一个意想不到的关键字参数&#39; pincode&#39;。

编辑:我使用filter_by后,我没有得到任何结果。 这是我的Jinja模板代码:

  {% if pincode is defined %}

    <thead>
        <tr>
            <td><h4>Pincode</h4></td>
            <td><h4>Office Name</h4></td>
            <td><h4>Taluk</h4></td>
            <td><h4>District</h4></td>
            <td><h4>State</h4></td>
        </tr>
    </thead>

    <tbody>

    {% for o in bypincode %}
        <tr>
            <td><h5>{{ o.pincode }}</h5></td>
            <td><h5>{{ o.officename }}</h5></td>
            <td><h5>{{ o.taluk }}</h5></td>
            <td><h5>{{ o.district }}</h5></td>
             <td><h5>{{ o.state }}</h5></td>
        </tr>
    {% endfor %}  
    {% endif %}

1 个答案:

答案 0 :(得分:1)

尝试使用filter_by代替filter

filter_by(**kwargs)正如您预期的那样工作:

pinCodes.query.filter_by(pincode = int_pin(pincode))

但是filter(*criterion)没有关键字参数,你应该像这样使用它:

pinCodes.query.filter(pinCodes.pincode == int_pin(pincode))