Question

我目前正在阅读O＆＃39; Reilly的书“Python编程”。下面是读取货架的代码，并创建一个Web界面，允许您从货架上访问这些值。您可以获取并更新值

'''
Implement a web-based interface for viewing and updating class instances
stored in a shelve; the shelve lives on server (same machine if localhost)
'''

import cgi, sys, os
import shelve, html
shelvename = 'class-shelve'
fieldnames = ('name', 'age', 'job', 'pay')

form = cgi.FieldStorage()
print('Content-type: text/html')
sys.path.insert(0, os.getcwd())

# main html template
replyhtml = """
<html>
<title>People Input Form</title>
<body>
<form method=POST action=peoplecgi.py>
    <table>
    <tr><th>key<td><input type=text name=key value="%(key)">
    $ROWS$
    </table>
    <p>
    <input type=submit value="Fetch", name=action>
    <input type=submit value="Update", name=action>
</form>
</body></html>
"""
# insert html for data rows at $ROWS$
rowhtml = '<tr><th>%s<td><input type=text name=%s value="%%(%s)s">\n'
rowshtml = ""
for fieldname in fieldnames:
    rowshtml += (rowhtml % ((fieldname,)*3))
replyhtml = replyhtml.replace('$ROWS$', rowshtml)


def htmlize(adict):
    new = adict.copy()
    for field in fieldnames:
        value = new[field]
        new[field] = html.escape(repr(value))
    return new

def fetchRecord(db, form):
    try:
        key = form['key'].value
        record = db[key]
        fields = record.__dict__
        fields['key'] = key
    except:
        fields = dict.fromkeys(fieldnames, '?')
        fields['key'] = 'Missing or invalid key!'
    return fields

def updateRecord(db, form):
    if not 'key' in form:
        fields = dict.fromkeys(fieldnames, '?')
        fields['key'] = 'Missing key input!'
    else:
        key = form['key'].value
        if key in db:
            record = db[key]
        else:
            from person_start import Person
            record = Person(name='?', age='?')
        for field in fieldnames:
            setattr(record, field, eval(form[field].value))
        db[key] = record
        fields = record.__dict__
        fields['key'] = key
    return fields

db = shelve.open(shelvename)
action = form['action'].value if 'action' in form else None
if action == 'Fetch':
    fields = fetchRecord(db, form)
elif action == 'Update':
    fields = updateRecord(db, form)
else:
    fields = dict.fromkeys(fieldnames, '?')
    fields['key'] = 'Missing or invalid action!'
db.close()
print(replyhtml % htmlize(fields))

然而，由于某种原因，打印不断失败。我已多次尝试删除＆＃34;＆＃34;错误是陈述，但无济于事。

有谁知道为什么这不能打印表格？

Answer 1

检查完整代码后，我认为问题出在replyhtml，在行 -

    <tr><th>key<td><input type=text name=key value="%(key)">

问题的格式为 - "%(key)"。您需要指定s或d等元素的类型，我相信您可能需要s（对于字符串）。示例 -

    <tr><th>key<td><input type=text name=key value="%(key)s">

ValueError：不支持的格式字符\＆＃39;＆＃34; \＆＃39; （0x22）用dict格式化时

1 个答案: