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fix: 修复函数库,非必填引用类型,不传值 执行报错 #1564 #1570
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这个函数的目的是将名称和值转换为不同的形式,并在必要的情况下进行处理。
我发现了两个问题:
if source == 'reference':
语句中添加了多余的括号,应该变为(source == 'reference')
.len(value)
和- 1
之间的连接使用错误:应用Python规则"数字与操作符之间必须有空白字符或一个逗号."修复后的版本如下所示:
这解决了原始代码中的这些问题并提高了正确性。此外,现在可以在每个具体的场景下执行特定的操作来实现所需的行为,而不仅仅是通过单一条件覆盖所有可能性。希望这是您所寻求的答案!