awk:从bash传递变量 [英] awk: passing variables from bash

问题描述

我收到以下代码的语法错误.是否有不支持"-v"选项的awk版本，或者我缺少某些内容?谢谢.

I am getting syntax errors with the following code. Is there an awk version that does not support the "-v" option or am I missing something? Thanks.

#!/usr/local/bin/bash
f_name="crap.stat" 
S_Date="2012-02-10"
E_Date="2012-02-13"

awk -F "\t" -v s_date="$S_Date" -v e_date="$E_Date" 'BEGIN {print s_date,e_date}' $f_name

推荐答案

Solaris 10(也称为oawk)上的默认awk程序似乎不支持-v选项.替代nawk程序确实支持它.有些人将名称更改为awk，因此它是nawk的链接，因此您无法轻易预测将找到的名称为awk.

The default awk program on Solaris 10 (aka oawk) does not seem to support the -v option; the alternative nawk program does support it. Some people switch the name awk so it is a link to nawk, so you can't readily predict which you'll find as awk.

HP-UX 11.x，AIX 6.x和Mac OS X(10.7.x)上的awk程序都支持-v表示法，这并不奇怪，因为

The awk programs on HP-UX 11.x, AIX 6.x and Mac OS X (10.7.x) all support the -v notation, which isn't very surprising since POSIX expects support for -v.

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