摘要:
我不知道如何重命名其中有破折号的钥匙。例如
obj.Fast-Italian = obj.Fast-Car1;下面的代码适用于不包含破折号的键"-":
var json = '[{"_id":"5078c3a803ff4197dc81fbfb","email":"user1@gmail.com","image":"some_image_url","name":"Name 1"}]';
var obj = JSON.parse(json)[0];
obj.id = obj._id;
delete obj._id;
json = JSON.stringify([obj]);
fs.writeFileSync('output1.json', json);我不能用上面的这个JSON:
var json = '[{"Fast-Car1":"Ferrari F40 Cabrio","Fast-Car2":"Audi R8 Black Edition","Fast-Car3":"Nissan GTR Blue"},{"Fast-Car1":"Lambo Diablo Fire Colors","Fast-Car2":"Skoda RS 4 doors","Fast-Car3":"Honda NSX red paint"}]'
// what I need to go here is change Fast-Car1 Fast-Italian, Fast-Car2 = Fast-German, Fast-Car3 = Fast-Japanese问题是我不知道该怎么做:
obj.Fast-Italian = obj.Fast-Car1;是因为键名中的"-“。
//so final JSON would look like this:
var json = '[{"Fast-Italian":"Ferrari F40 Cabrio"},{"Fast-German":"Audi R8 Black Edition"},{"Fast-Japanese":"Nissan GTR Blue"},,{"Fast-Italian":"Lambo Diablo Fire Colors","Fast-German":"VW Golf RS silver","Fast-Japanese":"Honda NSX red paint"}]'
// JSON has big amount of those, so I will loop anyway, but I have no idea how to 我试过这个:
var obj = JSON.parse(json)[0];
obj.[Fast-Italian] = obj.['Fast-Car1'];
delete obj._id;但后来出错了:
// obj.id = obj.['Fast-Car1'];
// SyntaxError: Unexpected token '['额外注意: JSON来自Excel,其中每个列都有快速-。
发布于 2020-06-19 22:25:48
你在正确的轨道上:
var obj = JSON.parse(json)[0];
obj.[Fast-Italian] = obj.['Fast-Car1'];
delete obj._id;有两个问题..。
第一,你不混合和匹配点表示法和括号符号。所以不要做obj.keyname,你只需做objkeyname。
第二,据我所见,快速意大利语不是一个已声明的变量名.它应该是字符串键名..。所以你需要把它用引号括起来
var obj = JSON.parse(json)[0];
obj['Fast-Italian'] = obj['Fast-Car1'];
delete obj['Fast-Car1'];https://stackoverflow.com/questions/62479309
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