我试图写一个bash脚本测试,需要一个参数,并通过curl发送到网站。我需要url编码的值,以确保特殊字符被正确处理。最好的方法是什么?

这是我到目前为止的基本脚本:

#!/bin/bash
host=${1:?'bad host'}
value=$2
shift
shift
curl -v -d "param=${value}" http://${host}/somepath $@

当前回答

使用curl——data-urlencode;来自man curl:

它发布数据,类似于其他——data选项,不同的是它执行url编码。为了与cgi兼容,<data>部分应该以名称开头,后面跟着分隔符和内容规范。

使用示例:

curl \
    --data-urlencode "paramName=value" \
    --data-urlencode "secondParam=value" \
    http://example.com

有关更多信息,请参阅手册页。

这需要curl 7.18.0或更新版本(发布于2008年1月)。使用curl -V来检查您拥有的版本。

你也可以对查询字符串进行编码:

curl --get \
    --data-urlencode "p1=value 1" \
    --data-urlencode "p2=value 2" \
    http://example.com
    # http://example.com?p1=value%201&p2=value%202

其他回答

awk版本的直接链接:http://www.shelldorado.com/scripts/cmds/urlencode 我用了很多年了,效果很好

:
##########################################################################
# Title      :  urlencode - encode URL data
# Author     :  Heiner Steven (heiner.steven@odn.de)
# Date       :  2000-03-15
# Requires   :  awk
# Categories :  File Conversion, WWW, CGI
# SCCS-Id.   :  @(#) urlencode  1.4 06/10/29
##########################################################################
# Description
#   Encode data according to
#       RFC 1738: "Uniform Resource Locators (URL)" and
#       RFC 1866: "Hypertext Markup Language - 2.0" (HTML)
#
#   This encoding is used i.e. for the MIME type
#   "application/x-www-form-urlencoded"
#
# Notes
#    o  The default behaviour is not to encode the line endings. This
#   may not be what was intended, because the result will be
#   multiple lines of output (which cannot be used in an URL or a
#   HTTP "POST" request). If the desired output should be one
#   line, use the "-l" option.
#
#    o  The "-l" option assumes, that the end-of-line is denoted by
#   the character LF (ASCII 10). This is not true for Windows or
#   Mac systems, where the end of a line is denoted by the two
#   characters CR LF (ASCII 13 10).
#   We use this for symmetry; data processed in the following way:
#       cat | urlencode -l | urldecode -l
#   should (and will) result in the original data
#
#    o  Large lines (or binary files) will break many AWK
#       implementations. If you get the message
#       awk: record `...' too long
#        record number xxx
#   consider using GNU AWK (gawk).
#
#    o  urlencode will always terminate it's output with an EOL
#       character
#
# Thanks to Stefan Brozinski for pointing out a bug related to non-standard
# locales.
#
# See also
#   urldecode
##########################################################################

PN=`basename "$0"`          # Program name
VER='1.4'

: ${AWK=awk}

Usage () {
    echo >&2 "$PN - encode URL data, $VER
usage: $PN [-l] [file ...]
    -l:  encode line endings (result will be one line of output)

The default is to encode each input line on its own."
    exit 1
}

Msg () {
    for MsgLine
    do echo "$PN: $MsgLine" >&2
    done
}

Fatal () { Msg "$@"; exit 1; }

set -- `getopt hl "$@" 2>/dev/null` || Usage
[ $# -lt 1 ] && Usage           # "getopt" detected an error

EncodeEOL=no
while [ $# -gt 0 ]
do
    case "$1" in
        -l) EncodeEOL=yes;;
    --) shift; break;;
    -h) Usage;;
    -*) Usage;;
    *)  break;;         # First file name
    esac
    shift
done

LANG=C  export LANG
$AWK '
    BEGIN {
    # We assume an awk implementation that is just plain dumb.
    # We will convert an character to its ASCII value with the
    # table ord[], and produce two-digit hexadecimal output
    # without the printf("%02X") feature.

    EOL = "%0A"     # "end of line" string (encoded)
    split ("1 2 3 4 5 6 7 8 9 A B C D E F", hextab, " ")
    hextab [0] = 0
    for ( i=1; i<=255; ++i ) ord [ sprintf ("%c", i) "" ] = i + 0
    if ("'"$EncodeEOL"'" == "yes") EncodeEOL = 1; else EncodeEOL = 0
    }
    {
    encoded = ""
    for ( i=1; i<=length ($0); ++i ) {
        c = substr ($0, i, 1)
        if ( c ~ /[a-zA-Z0-9.-]/ ) {
        encoded = encoded c     # safe character
        } else if ( c == " " ) {
        encoded = encoded "+"   # special handling
        } else {
        # unsafe character, encode it as a two-digit hex-number
        lo = ord [c] % 16
        hi = int (ord [c] / 16);
        encoded = encoded "%" hextab [hi] hextab [lo]
        }
    }
    if ( EncodeEOL ) {
        printf ("%s", encoded EOL)
    } else {
        print encoded
    }
    }
    END {
        #if ( EncodeEOL ) print ""
    }
' "$@"

我发现在python中可读性更好:

encoded_value=$(python3 -c "import urllib.parse; print urllib.parse.quote('''$value''')")

三重'确保单引号的值不会有伤害。Urllib在标准库中。它的工作,例如这个疯狂的(现实世界)url:

"http://www.rai.it/dl/audio/" "1264165523944Ho servito il re d'Inghilterra - Puntata 7

以下是基于奥威尔的回答,但解决了多字节 通过设置LC_ALL=C (vte.sh的一个技巧)在评论中提到的错误。 我把它写成了函数PROMPT_COMMAND的形式,因为 我就是这么用的。

print_path_url() {
  local LC_ALL=C
  local string="$PWD"
  local strlen=${#string}
  local encoded=""
  local pos c o

  for (( pos=0 ; pos<strlen ; pos++ )); do
     c=${string:$pos:1}
     case "$c" in
        [-_.~a-zA-Z0-9/] ) o="${c}" ;;
        * )               printf -v o '%%%02x' "'$c"
     esac
     encoded+="${o}"
  done
  printf "\033]7;file://%s%s\007" "${HOSTNAME:-}" "${encoded}"
}

问题是关于在bash中这样做,不需要python或perl,因为实际上有一个命令,它完全是你想要的-“urlencode”。

value=$(urlencode "${2}")

这也更好,因为上面的perl答案,例如,没有正确编码所有字符。尝试使用从Word中获得的长破折号,你会得到错误的编码。

注意,你需要安装"gridsite-clients"来提供这个命令:

sudo apt install gridsite-clients

另一种php方法:

echo "encode me" | php -r "echo urlencode(file_get_contents('php://stdin'));"