json - 在没有尾随分隔符的情况下分隔 AWK 中的输出记录

Question

我有以下记录:

31 Stockholm
42 Talin
34 Helsinki
24 Moscow
15 Tokyo

我想用 AWK 将它转换为 JSON。使用此代码:

#!/usr/bin/awk
BEGIN {
    print "{";
    FS=" ";
    ORS=",\n";
    OFS=":";
};

{    
    if ( !a[city]++ && NR > 1 ) {
        key = $2;
        value = $1;
        print "\"" key "\"", value;
    }
};

END {
    ORS="\n";
    OFS=" ";
    print "\b\b}";
};

给我这个:

{
"Stockholm":31,
"Talin":42,
"Helsinki":34,
"Moscow":24,
"Tokyo":15, <--- I don't want this comma
}

问题是最后一个数据行的尾随逗号。它使 JSON 输出 Not Acceptable 。我怎样才能得到这个输出:

{
"Stockholm":31,
"Talin":42,
"Helsinki":34,
"Moscow":24,
"Tokyo":15
}

Answer 1

介意对您发布的脚本的一些反馈吗？

#!/usr/bin/awk        # Just be aware that on Solaris this will be old, broken awk which you must never use
BEGIN {
    print "{";        # On this and every other line, the trailing semi-colon is a pointless null-statement, remove all of these.
    FS=" ";           # This is setting FS to the value it already has so remove it.
    ORS=",\n";
    OFS=":";
};

{
    if ( !a[city]++ && NR > 1 ) {      # awk consists of <condition>{<action} segments so move this condition out to the condition part
                                       # also, you never populate a variable named "city" so `!a[city]++` won't behave sensibly.
        key = $2;
        value = $1;
        print "\"" key "\"", value;
    }
};

END {
    ORS="\n";                          # no need to set ORS and OFS when the script will no longer use them.
    OFS=" ";
    print "\b\b}";                     # why would you want to print a backspace???
};

所以你的原始脚本应该写成:

#!/usr/bin/awk
BEGIN {
    print "{"
    ORS=",\n"
    OFS=":"
}

!a[city]++ && (NR > 1) {    
    key = $2
    value = $1
    print "\"" key "\"", value
}

END {
    print "}"
}

以下是我如何真正编写脚本以将您发布的输入转换为您发布的输出:

$ cat file
31 Stockholm
42 Talin
34 Helsinki
24 Moscow
15 Tokyo
$
$ awk 'BEGIN{print "{"} {printf "%s\"%s\":%s",sep,$2,$1; sep=",\n"} END{print "\n}"}' file
{
"Stockholm":31,
"Talin":42,
"Helsinki":34,
"Moscow":24,
"Tokyo":15
}