尝试使用Python中的字符串格式替换Python字符串中的值时出错

Question

我有这根绳子

S="191042709832779540946_1513246254239&source=%7B%22query%22%3A%7B%22match_all%22%3A%7B%7D%7D%2C%22from%22%3A%s%2C%22size%22%3A20000%2C%22facets%22%3A%7B%22_type%22%3A%7B%22terms%22%3A%7B%22field%22%3A%22_type%22%2C%22size%22%3A102%2C%22order%22%3A%22reverse_term%22%7D%7D%2C%22index.classification.exact%22%3A%7B%22terms%22%3A%7B%22field%22%3A%22index.classification.exact%22%2C%22size%22%3A110%2C%22order%22%3A%22count%22%7D%7D%2C%22index.has_seal.exact%22%3A%7B%22terms%22%3A%7B%22field%22%3A%22index.has_seal.exact%22%2C%22size%22%3A110%2C%22order%22%3A%22count%22%7D%7D%2C%22index.license.exact%22%3A%7B%22terms%22%3A%7B%22field%22%3A%22index.license.exact%22%2C%22size%22%3A110%2C%22order%22%3A%22count%22%7D%7D%2C%22index.publisher.exact%22%3A%7B%22terms%22%3A%7B%22field%22%3A%22index.publisher.exact%22%2C%22size%22%3A110%2C%22order%22%3A%22count%22%7D%7D%2C%22index.language.exact%22%3A%7B%22terms%22%3A%7B%22field%22%3A%22index.language.exact%22%2C%22size%22%3A110%2C%22order%22%3A%22count%22%7D%7D%7D%7D&_=1513246254241"

在上面的字符串中，我必须在某个地方替换一个整数。为此，我使用了%s

s = S % 100

然后我得到了下面的错误。你知道吗

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: unsupported format character 'B' (0x42) at index 51

注意：您可以搜索（ctrl+f）以在上面的字符串中看到%s。我试着突出它。但这是不可能的。你知道吗

如何替换上面字符串中的值。提前谢谢。你知道吗

Answer 1

您需要避免使用%s进行替换，而是将{}与Python的format()函数一起使用，如下所示：

text = "191042709832779540946_1513246254239&source=%7B%22query%22%3A%7B%22match_all%22%3A%7B%7D%7D%2C%22from%22%3A{}%2C%22size%22%3A20000%2C%22facets%22%3A%7B%22_type%22%3A%7B%22terms%22%3A%7B%22field%22%3A%22_type%22%2C%22size%22%3A102%2C%22order%22%3A%22reverse_term%22%7D%7D%2C%22index.classification.exact%22%3A%7B%22terms%22%3A%7B%22field%22%3A%22index.classification.exact%22%2C%22size%22%3A110%2C%22order%22%3A%22count%22%7D%7D%2C%22index.has_seal.exact%22%3A%7B%22terms%22%3A%7B%22field%22%3A%22index.has_seal.exact%22%2C%22size%22%3A110%2C%22order%22%3A%22count%22%7D%7D%2C%22index.license.exact%22%3A%7B%22terms%22%3A%7B%22field%22%3A%22index.license.exact%22%2C%22size%22%3A110%2C%22order%22%3A%22count%22%7D%7D%2C%22index.publisher.exact%22%3A%7B%22terms%22%3A%7B%22field%22%3A%22index.publisher.exact%22%2C%22size%22%3A110%2C%22order%22%3A%22count%22%7D%7D%2C%22index.language.exact%22%3A%7B%22terms%22%3A%7B%22field%22%3A%22index.language.exact%22%2C%22size%22%3A110%2C%22order%22%3A%22count%22%7D%7D%7D%7D&_=1513246254241"
replace = 100

print text.format(replace)

这会给你：

191042709832779540946_1513246254239&source=%7B%22query%22%3A%7B%22match_all%22%3A%7B%7D%7D%2C%22from%22%3A100%2C%22size%22%3A20000%2C%22facets%22%3A%7B%22_type%22%3A%7B%22terms%22%3A%7B%22field%22%3A%22_type%22%2C%22size%22%3A102%2C%22order%22%3A%22reverse_term%22%7D%7D%2C%22index.classification.exact%22%3A%7B%22terms%22%3A%7B%22field%22%3A%22index.classification.exact%22%2C%22size%22%3A110%2C%22order%22%3A%22count%22%7D%7D%2C%22index.has_seal.exact%22%3A%7B%22terms%22%3A%7B%22field%22%3A%22index.has_seal.exact%22%2C%22size%22%3A110%2C%22order%22%3A%22count%22%7D%7D%2C%22index.license.exact%22%3A%7B%22terms%22%3A%7B%22field%22%3A%22index.license.exact%22%2C%22size%22%3A110%2C%22order%22%3A%22count%22%7D%7D%2C%22index.publisher.exact%22%3A%7B%22terms%22%3A%7B%22field%22%3A%22index.publisher.exact%22%2C%22size%22%3A110%2C%22order%22%3A%22count%22%7D%7D%2C%22index.language.exact%22%3A%7B%22terms%22%3A%7B%22field%22%3A%22index.language.exact%22%2C%22size%22%3A110%2C%22order%22%3A%22count%22%7D%7D%7D%7D&_=1513246254241

这是必要的，因为您的text已经有其他%字符，并且这将与%s替换不兼容。通过使用此方法，replace也可以是字符串（如果需要）。你知道吗

Answer 2

您可以简单地使用replace

newS = S.replace("%s", "100")

在这里，通常的%语法很难应用，因为在您的字符串中有许多python没有按字面理解的'%x'（它认为它们是一些其他替换的占位符，但是在您的caes中只有转义的HTML字符）