Question

我需要计算一个帖子的所有赞成票和赞成票，

并且只应显示用户是否是其他用户的朋友。

我的问题是，upvotes和downvotes的计数不准确。

示例：

post_id 5和4只有1个upvote，但检索到的数据表示他们有3个upvotes。

Here's我创建的sqlfiddle。

Answer 1

你可以尝试使用这个小改动它返回预期的输出：

import numpy as np
from matplotlib import pyplot as plt

###############################################################################
# The data (change all of this to your actual data, this is just a mockup)
variables = [
    'apple',
    'juice',
    'orange',
    'peach',
    'gum',
    'stones',
    'bags',
    'lamps',
]

base = 3000

lows = np.array([
    base - 246 / 2,
    base - 1633 / 2,
    base - 500 / 2,
    base - 150 / 2,
    base - 35 / 2,
    base - 36 / 2,
    base - 43 / 2,
    base - 37 / 2,
])

values = np.array([
    246,
    1633,
    500,
    150,
    35,
    36,
    43,
    37,
])

###############################################################################
# The actual drawing part

# The y position for each variable
ys = range(len(values))[::-1]  # top to bottom

# Plot the bars, one by one
for y, low, value in zip(ys, lows, values):
    # The width of the 'low' and 'high' pieces
    low_width = base - low
    high_width = low + value - base

    # Each bar is a "broken" horizontal bar chart
    plt.broken_barh(
        [(low, low_width), (base, high_width)],
        (y - 0.4, 0.8),
        facecolors=['white', 'white'],  # Try different colors if you like
        edgecolors=['black', 'black'],
        linewidth=1,
    )

    # Display the value as text. It should be positioned in the center of
    # the 'high' bar, except if there isn't any room there, then it should be
    # next to bar instead.
    x = base + high_width / 2
    if x <= base + 50:
        x = base + high_width + 50
    plt.text(x, y, str(value), va='center', ha='center')

# Draw a vertical line down the middle
plt.axvline(base, color='black')

# Position the x-axis on the top, hide all the other spines (=axis lines)
axes = plt.gca()  # (gca = get current axes)
axes.spines['left'].set_visible(False)
axes.spines['right'].set_visible(False)
axes.spines['bottom'].set_visible(False)
axes.xaxis.set_ticks_position('top')

# Make the y-axis display the variables
plt.yticks(ys, variables)

# Set the portion of the x- and y-axes to show
plt.xlim(base - 1000, base + 1000)
plt.ylim(-1, len(variables))

Answer 2

看看这是否会让你到处（在SQLite中测试，网站无法正常工作）。

SELECT vote_type, post_id, user_id, COUNT(*) FROM (
   SELECT * FROM post JOIN vote ON vote.post_id = post.post_id WHERE (post.user_id || "," || vote.user_id) IN (
      SELECT DISTINCT user_id || "," || friend_id FROM friend WHERE request = 'friend'
   )
) GROUP BY vote_type;

这个查询的一个大问题是连接（||），我相信它可以更加健壮。

Answer 3

出现问题是因为user_id = 8发出了3个好友请求，因此您的查询为每个请求生成一行，您只需要一行 - 如果您更改了这样的查询，则可以避免多次计数：

SELECT user_info.id,user_info.fb_id,user_info.profile_pic,
   user_info.fname,user_info.lname, post.post_id,post.message,post.time,post.img,post.user_id,
   COUNT(CASE WHEN vote.vote_type = 'upvote' then 1 ELSE NULL END) AS 'upvote', COUNT(CASE WHEN vote.vote_type = 'downvote' then 1 ELSE NULL END) AS 'downvote'

   FROM user_info

   JOIN post
   on post.user_id = user_info.id

    LEFT JOIN vote ON
    vote.post_id = post.post_id

     WHERE EXISTS (select * FROM friend
     WHERE (friend.user_id = user_info.id OR friend.friend_id = user_info.id)
     AND friend.request = 'friend' AND (friend.user_id = 8 OR friend.friend_id = 8))

     GROUP BY post.post_id
     ORDER BY post.post_id DESC

SQL - 计算其连接ID的列值

3 个答案: