如何使用 BeautifulSoup 和 Flask 从循环中打印所有字典值

Question

def search():
    query = "xxx"
    r = requests.get('website?q=' + query)
    soup = BeautifulSoup(r.content, 'lxml')
    results=[]
    for row in soup.findAll('li', attrs = {'class':'res'}):
        result = {}
        result['urltext'] = row.a.text
        result['url'] = row.cite.text
        results.append(result)

        return render_template('site.html', urll = result['url'], urltext = result['urltext'])
    else:
        return render_template('start.html')

When used like that i am getting single result in site.html, it's like the loop is stopping and getting first result and passing it to the site.html. And i want all results it finds to be passed.

def search():
    query = "xxx"
    r = requests.get('website?q=' + query)
    soup = BeautifulSoup(r.content, 'lxml')
    results=[]
    for row in soup.findAll('li', attrs = {'class':'res'}):
        result = {}
        result['urltext'] = row.a.text
        result['url'] = row.cite.text
        results.append(result)

        print (result['url'])

Using print gives me all results. But that is pointless, unless i will drop the render_template idea and print whole HTML with result['url'] and result['urltext'] in it somehow.

Answer 1

Your problem is return which always ends function at once so next loop will be never executed. So don't use return inside for -loop.

And using else with for is also wrong idea because else will be executed only when you use break in for -loo. for/else doesn't work like if/else .

Another problem is result['url'] and result['urltext'] in render_template - it will have only last value. You have all values in results , not in result['url'] and result['urltext'] . Maybe you should send it as

return render_template('site.html', results=results)

and inside template use

{% for row in results %}
   <a href="{{ row.url }}">{{ row.ulrtext }}</a>
{% endfor %}

---

def search():
    query = "xxx"
    r = requests.get('website?q=' + query)
    soup = BeautifulSoup(r.content, 'lxml')
    results = []
    for row in soup.findAll('li', attrs={'class': 'res'}):
        result = {}
        result['urltext'] = row.a.text
        result['url'] = row.cite.text
        results.append(result)

    if results:
        return render_template('site.html', results=results)
    else:
        return render_template('start.html')