小编puf*_*fin的帖子

django没有这样的表:

我在Django中有模型设置如下.

class Pupil(models.Model):
    forename = models.CharField(max_length=30)
    surname = models.CharField(max_length=30)
    dateofbirth = models.DateField()
    year = models.IntegerField()
    class_group = models.CharField(max_length=30)
    email = models.EmailField(blank=True)
    assignments = models.ManyToManyField('Assignment', verbose_name='related assignments')

    def __unicode__(self):
        return u'%s %s' % (self.forename, self.surname)

class Subject(models.Model):
    name = models.CharField(max_length=30)
    level = models.CharField(max_length=30)
    teachers = models.ManyToManyField('Teacher', verbose_name='related teachers')

    def __unicode__(self):
        return self.name

class Teacher(models.Model):
    forename = models.CharField(max_length=30)
    surname = models.CharField(max_length=30)
    email = models.EmailField(blank=True)

    def __unicode__(self):
        return u'%s %s' % (self.forename, self.surname)

class Assignment(models.Model):
    assignment_name = models.CharField(max_length=30)
    date_assigned = models.DateField()
    date_submitted = …
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django django-south

42
推荐指数
3
解决办法
7万
查看次数

python pip仍在寻找以前的安装

在经历了sqlite3的这个冲泡问题之后,我做到了

brew rm sqlite python python3
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然后

brew install python python3
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这安装python2.7.5作为默认解释器和brew安装pip以及python,我以为我能够

pip install virtualenv
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为新的python2.7.5安装virtualenv.但是,我得到了

-bash: /usr/local/share/python/pip: /usr/local/Cellar/python/2.7.3/bin/python: bad interpreter: No such file or directory
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我怎样才能解决这个问题?我应该在之间创建一个符号链接

/usr/local/share/python/pip --> /usr/local/Cellar/python/2.7.5/bin/pip-2.7
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python homebrew

15
推荐指数
1
解决办法
2万
查看次数

jQuery On Click无效

我无法启动我的事件处理程序.我正在将JSON调用到页面中,如下所示.然后,我希望能够单击我创建的一个按钮,该按钮将执行"hello there"警报.它适用于页面上的每个其他元素,但不适用于<button>.

有人可以帮忙吗?

test.js

$(document).ready(function() {
  $.getJSON('/api/v1/recipe/?format=json', function(data) {
    $.each(data.objects, function(i, recipe) {
      $('table#recipes').append('<tr><td>' + recipe.title + '</td><td>' + recipe.ingredients + '</td><td>' + recipe.method + '</td><td><button class="btn" type="button" id="rmv">remove</button></td></tr>');
      });
    });

    $('rmv').on('click', function() {
      alert('hello there!');
    });
  });
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recipe.html

{% extends 'base.html' %}

{% block content %}
<div class="container-fluid">
  <div class="row-fluid">
    <div class="span6">
      <form action='' method="post">{% csrf_token %}
        {{ form.as_p }}
        <input type="submit" value="go baby!">
      </form>
    </div>
    <div class="span6">
      <table class="table table-striped table-bordered" id="recipes">
        <tr>
          <th>Title</th>
          <th>Ingredients</th>
          <th>Method</th> …
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jquery

11
推荐指数
3
解决办法
4万
查看次数

用Scrapy和Selenium刮痧

我有一个scrapy蜘蛛爬行网站,通过页面上的JavaScript重新加载内容.为了进入下一页抓取,我一直在使用Selenium点击网站顶部的月份链接.

问题在于,即使我的代码按预期移动每个链接,蜘蛛也会抓住月份的第一个月(Sept)数据并返回此重复数据.

我怎么能绕过这个?

from selenium import webdriver

class GigsInScotlandMain(InitSpider):
        name = 'gigsinscotlandmain'
        allowed_domains = ["gigsinscotland.com"]
        start_urls = ["http://www.gigsinscotland.com"]


    def __init__(self):
        InitSpider.__init__(self)
        self.br = webdriver.Firefox()

    def parse(self, response):
        hxs = HtmlXPathSelector(response)
        self.br.get(response.url)
        time.sleep(2.5)
        # Get the string for each month on the page.
        months = hxs.select("//ul[@id='gigsMonths']/li/a/text()").extract()

        for month in months:
            link = self.br.find_element_by_link_text(month)
            link.click()
            time.sleep(5)

            # Get all the divs containing info to be scraped.
            listitems = hxs.select("//div[@class='listItem']")
            for listitem in listitems:
                item = GigsInScotlandMainItem()
                item['artist'] = listitem.select("div[contains(@class, 'artistBlock')]/div[@class='artistdiv']/span[@class='artistname']/a/text()").extract()
                #
                # …
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python selenium scrapy

6
推荐指数
1
解决办法
6002
查看次数

如何正确设置python3的路径?

$ echo $PATH
/usr/local/share/python:/usr/bin:/bin:/usr/sbin:/sbin:/usr/local/bin:/usr/X11/bin
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所以/ usr/local/bin就在我的路上.

当我做`哪个python3'

/usr/local/bin/python3
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当我然后尝试创建virtualenv

mkvirtualenv py3000 --python=python3

The executable /Users/misdirectedpuffin/python3 (from --python=/Users/misdirectedpuffin/python3) does not exist
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Virtualenv在使用默认python时工作 mkvirtualenv testenv' and when doing 'mkvirtualenv py3000 --python=/usr/local/bin/python3

它似乎在寻找$ HOME for python3.我怎么能纠正这个?

**编辑**

我还可以设置export python3=/usr/local/bin/python3bash_profile,然后调用$python3--python=$python3,但我真正想要的是--python=python3不美元符号.

python virtualenvwrapper

6
推荐指数
1
解决办法
1万
查看次数