加入收藏 | 设为首页 | 会员中心 | 我要投稿 李大同 (https://www.lidatong.com.cn/)- 科技、建站、经验、云计算、5G、大数据,站长网!
当前位置: 首页 > 编程开发 > Python > 正文

python – TypeError:强制转换为Unicode:需要字符串或缓冲区,

发布时间:2020-12-20 11:41:08 所属栏目:Python 来源:网络整理
导读:我是编程特别是 Python的新手.我正试图用 this tutorial 学习它. 我现在正在练习19,其中一个学习练习是创建我自己的设计功能,并运行十次不同的时间.我认为到目前为止我做得很好,直到第10次跑.这是我第一次尝试创建并写入文件,然后再次打开它以取出数字并将其
我是编程特别是 Python的新手.我正试图用 this tutorial 学习它.
我现在正在练习19,其中一个学习练习是创建我自己的设计功能,并运行十次不同的时间.我认为到目前为止我做得很好,直到第10次跑.这是我第一次尝试创建并写入文件,然后再次打开它以取出数字并将其放入bread_and_butter变量中.

错误如下所示:

Traceback (most recent call last):
  File "ex19b.py",line 95,in <module>
    j2 = open(j1)
TypeError: coercing to Unicode: need string or buffer,file found
from sys import argv

def bread_and_butter(bread,butter):
    print "I have %d bread(s) and %d grams buttern" % (bread,butter)


print "1. I can give it the numbers directly:"
bread_and_butter(10,15)

print "2. Or I could use variables from my script:"
amount_of_bread = 10
amount_of_butter = 15

bread_and_butter(amount_of_bread,amount_of_butter)

print "3. We can even do math inside it:"
bread_and_butter(10 + 5,7 + 9)

print "4. And we can combine both,math and variables:"
bread_and_butter(amount_of_bread + 7,amount_of_butter + 9)

print "5. We can let the user decide how much we have:"
x = raw_input("How much bread do we have?")
i = raw_input("And how much butter do we have?")

y = int(x)
z = int(i)

bread_and_butter(y,z)

print "6. Also we can take arguments for that:"

script,how_much_bread,how_much_butter = argv

q = int(how_much_bread)
w = int(how_much_butter)

bread_and_butter(q,w)

print "7. Another mathematic way:n"
x1 = raw_input("How much breads do we have left?")
x2 = raw_input("How much breads do we need?")
y1 = raw_input("How much grams butter was left again?")
y2 = raw_input("How much grams butter do we still need?")

a1 = int(x1)
a2 = int(x2)
b1 = int(y1)
b2 = int(y2)

c,d = a2 - a1,b2 - b1

print "We still need %d bread(s) and %d gram butter butn" % (c,d)

bread_and_butter(a1,b1)

print "8. Another way:n"
bread = 10
butter = 300

y = bread - 8
z = butter - 1000

bread_and_butter(y,z)

print "9. Second last way:n"

script,bread,butters = argv

f = int(bread)
g = int(butters)

l = f - 30
k = g - 500

bread_and_butter(l,k)

print "10. The last way:n"

j = raw_input("I'll create a file with numbers in it,what's it's name?")

j1 = open(j,'w')

number1 = raw_input("How many breads do we have?")
number2 = raw_input("And how much butter do we have?")

print "I'm going to write these to the file..."

j1.write("%sn%sn" % (number1,number2))

j1.close()

print "The filename is: %s." % j1.name

j2 = open(j1)
j3 = open(j1)

j4 = j2.read()
j5 = j3.read()

j6 = int(j4)
j7 = int(j5)

bread_and_butter(j6,j7)

当有人能在这里帮助我时,我会很高兴.我想这不会是我不会理解的最后一件事但是现在让我感到困惑.当你看到它时,我确定你的“专业人士”已经解决了这个问题.

无论如何,提前谢谢,

Wumpel

解决方法

您收到此错误是因为open正在查找文件名,并且您将文件对象本身传递给它j1.例如,如果您打开j,则不会再出现错误.

编辑:
如果你想打开一个文件并读取特定的行,我会使用这样的东西:

j1 = open(j,'w')
for i,line in enumerate(j1):
    if i == 12:
        # this is the 13th line
    elif i == 312:
        # this is the 313th line
    elif i > 1000:
        break
j1.close()

(编辑:李大同)

【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容!

    推荐文章
      热点阅读