json.decoder.JSONDecodeError: Expecting ',' delimiter: line 1 column 22 (char 21)
时间: 2023-12-20 15:31:45 浏览: 25
JSONDecodeError是JSON解码器遇到错误时引发的异常。在这种情况下,JSON解码器期望在第1行第22个字符(即第21个字符后面)处找到逗号分隔符,但是它找到了其他字符。这通常是由于JSON格式不正确引起的。要解决此问题,您可以按照以下步骤进行操作:
1.检查JSON字符串是否正确格式化,例如是否缺少逗号分隔符或引号等。
2.检查JSON字符串中的特殊字符是否正确转义。
3.检查JSON字符串中的键和值是否用双引号括起来。
以下是一个例子,演示了如何解决JSONDecodeError异常:
```python
import json
# JSON字符串格式不正确,缺少逗号分隔符
json_str = '{"name": "John" "age": 30}'
try:
data = json.loads(json_str)
print(data)
except json.JSONDecodeError as e:
print("JSONDecodeError: ", e)
# 输出:JSONDecodeError: Expecting ',' delimiter: line 1 column 22 (char 21)
```
相关问题
json.decoder.JSONDecodeError: Expecting ':' delimiter: line 1 column 91 (char 90)
json.decoder.JSONDecodeError: Expecting ':' delimiter: line 1 column 91 (char 90)错误的原因是在第一行的第91个字符处缺少了冒号(:)分隔符。
为了解决这个问题,需要在正确的位置添加冒号(:)分隔符。在给定的示例中,冒号(:)应该在 "uid" 键和其对应的值之间添加。正确的JSON字符串应该是:
{ "uid": "1", "data": [ ["数据1", 86, 81], ["曼联", 74, 66], ["利物浦", 69, 99] ] } <span class="em">1</span><span class="em">2</span><span class="em">3</span>
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json.decoder.JSONDecodeError: Expecting ',' delimiter: line 75 column 2
JSONDecodeError异常通常发生在解析JSON数据时,Python解析器期望在JSON对象或数组的元素之间看到逗号(','),但未找到逗号或者逗号的位置不正确。根据提供的引用内容,错误信息是"json.decoder.JSONDecodeError: Expecting ',' delimiter: line 75 column 2"。这意味着在JSON数据的第75行第2列位置上缺少了逗号。<span class="em">1</span><span class="em">2</span><span class="em">3</span>
#### 引用[.reference_title]
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