jq将日期解析为时间戳答案

【问题标题】：jq parsing date to timestampjq将日期解析为时间戳
【发布时间】：2018-03-18 01:55:15
【问题描述】：

我有以下脚本：

curl -s -S 'https://bittrex.com/Api/v2.0/pub/market/GetTicks?marketName=BTC-NBT&tickInterval=thirtyMin&_=1521347400000' | jq -r '.result|.[] |[.T,.O,.H,.L,.C,.V,.BV] | @tsv | tostring | gsub("\t";",") | "(\(.))"'

这是输出：

(2018-03-17T18:30:00,0.00012575,0.00012643,0.00012563,0.00012643,383839.45768188,48.465051)
(2018-03-17T19:00:00,0.00012643,0.00012726,0.00012642,0.00012722,207757.18765437,26.30099514)
(2018-03-17T19:30:00,0.00012726,0.00012779,0.00012698,0.00012779,97387.01596624,12.4229077)
(2018-03-17T20:00:00,0.0001276,0.0001278,0.00012705,0.0001275,96850.15260027,12.33316229)

我想用时间戳替换日期。

我可以在 shell 中使用日期进行这种转换

date -d '2018-03-17T18:30:00' +%s%3N
1521325800000

我想要这个结果：

(1521325800000,0.00012575,0.00012643,0.00012563,0.00012643,383839.45768188,48.465051)
(1521327600000,0.00012643,0.00012726,0.00012642,0.00012722,207757.18765437,26.30099514)
(1521329400000,0.00012726,0.00012779,0.00012698,0.00012779,97387.01596624,12.4229077)
(1521331200000,0.0001276,0.0001278,0.00012705,0.0001275,96850.15260027,12.33316229)

此数据存储在 MySQL 中。

是否可以在单个命令行中使用 jq 或其他命令（如 awk、sed、perl）执行日期转换？

【问题讨论】：

标签： datetime awk sed jq

【解决方案1】：

这是一个假设“Z”（UTC+0）时区的全 jq 解决方案。

简而言之，只需将.T 替换为：

((.T + "Z") | fromdate | tostring + "000")

要验证这一点，请考虑：

timestamp.jq

[splits("[(),]")]
| .[1] |= ((. + "Z")|fromdate|tostring + "000")  # milliseconds
| .[1:length-1]
| "(" + join(",") + ")"

调用

jq -rR -f timestamp.jq  input.txt

输出

(1521311400000,0.00012575,0.00012643,0.00012563,0.00012643,383839.45768188,48.465051)
(1521313200000,0.00012643,0.00012726,0.00012642,0.00012722,207757.18765437,26.30099514)
(1521315000000,0.00012726,0.00012779,0.00012698,0.00012779,97387.01596624,12.4229077)
(1521316800000,0.0001276,0.0001278,0.00012705,0.0001275,96850.15260027,12.33316229)

【讨论】：

jq 获胜，只需替换 .T ，代码更短。再次感谢！

【解决方案2】：

这是一个不可移植的 awk 解决方案。它不可移植，因为它依赖于系统date 命令；在我正在使用的系统上，相关调用如下所示：date -j -f "%Y-%m-%eT%T" STRING "+%s"

awk -F, 'BEGIN{OFS=FS}
  NF==0 { next }
  { sub(/\(/,"",$1);
    cmd="date -j -f \"%Y-%m-%eT%T\" " $1 " +%s";
    cmd | getline $1;
    $1=$1 "000";     # milliseconds
    printf "%s", "(";
    print;
  }' input.txt

输出

(1521325800000,0.00012575,0.00012643,0.00012563,0.00012643,383839.45768188,48.465051)
(1521327600000,0.00012643,0.00012726,0.00012642,0.00012722,207757.18765437,26.30099514)
(1521329400000,0.00012726,0.00012779,0.00012698,0.00012779,97387.01596624,12.4229077)
(1521331200000,0.0001276,0.0001278,0.00012705,0.0001275,96850.15260027,12.33316229)

【讨论】：

我修改了命令日期：awk -F, 'BEGIN{OFS=FS} NF==0 { next } { sub(/(/,"",$1); cmd="date - d " $1" +%s%3N"; cmd | getline $1; printf "%s", "("; print;}'; 这可行，但它从系统中获取日期 gmt。谢谢

【解决方案3】：

使用 sed 的解决方案：

 sed -e 's/(\([^,]\+\)\(,.*\)/echo "(\$(date -d \1 +%s%3N),\2"/g' | ksh

测试：

  <commande_curl> | sed -e 's/(\([^,]\+\)\(,.*\)/echo "(\$(date -d \1 +%s%3N),\2"/g' | ksh

或：

  <commande_curl> > results_curl.txt
  cat results_curl.txt | sed -e 's/(\([^,]\+\)\(,.*\)/echo "(\$(date -d \1 +%s%3N),\2"/g' | ksh

【讨论】：