paragraph = [{content:["12","3","4"]}, {content:["12","3","4"]},{content:["12","3","4"]}] const result = paragraphs.map(paragraph => paragraph.content.join('\n')).join('\n'); 为什么会报错
paragraph = [{content:["12","3","4"]}, {content:["12","3","4"]},{content:["12","3","4"]}] const result = paragraphs.map(paragraph => paragraph.content.join('\n')).join('\n'); 为什么会报错
paragraph = [{content:["12","3","4"]}, {content:["12","3","4"]},{content:["12","3","4"]}]
const result = paragraphs.map(paragraph => paragraph.content.join('\n')).join('\n'); 为什么会报错
用户500598139
2024年03月21日
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浏览 117
这段代码会报错是因为在map方法的回调函数中使用了与外部变量同名的参数名。在这里，参数名 paragraph 与外部变量名相同，导致冲突。为了解决这个问题，可以将参数名改为其他名称，例如：
复制代码
const paragraphs = [{content:["12","3","4"]}, {content:["12","3","4"]},{content:["12","3","4"]}];
const result = paragraphs.map(para => para.content.join('\n')).join('\n');
这样就避免了参数名与外部变量名相同的冲突，代码应该可以正常运行了。
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