我从Google bigquery得到了这种格式,但我需要更改以下格式:
"Rows: [[{"insert_time":{"value":"2020-12-23T21:04:12.316Z"},"_id":null,"viewId":"106851427","userId":"9f84898d-7e13-4218-835a-6db82ade9931","AletName":"Conv between: video_screen - click_on_screen - true and camera - click - success_save_recording_settings","AletType":"Conversion has been change","date":{"value":"2020-12-05"},"Hour":"04","EventsConv":4,"ConvMedianHourly":0.3666666666666667,"Actual_VS_expected":"Got: 4 but expected: 0.37"},{"insert_time":{"value":"2020-12-23T21:04:12.316Z"},"_id":null,"viewId":"104831427","userId":"9f84898d-7e13-4218-835a-6db72ace9931","AletName":"Conv between: video - select - youtube and category - select - New - Karaoke","AletType":"Conversion has been change","date":{"value":"2020-12-05"},"Hour":"03","EventsConv":21,"ConvMedianHourly":3,"Actual_VS_expected":"Got: 21 but expected: 3"}]]到这个
"[{"viewId":"62437650","source":"Google},{"viewId":"6166150","source":"Google}]"我试着用
x.rows[0]但它不起作用。
发布于 2020-12-29 05:32:49
如果我猜对了,并且您的查询结果是一个字符串,那么您可以使用:
let newX = x.substring(8,x.length-1);//如果需要object:
JSON.parse(newX);//如果不是字符串而是对象,可以这样做:
JSON.stringify(x) //之前
发布于 2020-12-29 05:21:07
如果您愿意与viewId和source一起维护其他属性,则可以使用x[0]。
https://stackoverflow.com/questions/65484189
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