Question

我遇到以下代码的问题：

IFS=: read c1 c2 c3 c4 rest <<< "$line"

不要错误地认为这段代码运行良好，但它似乎并没有用于ksh。我基本上需要在没有＆＃34;＆lt;＆lt;＆lt;＆＃34;的情况下编写相同的代码。关于＆＃34;＆lt;＆lt;＆lt;＆＃34;的信息不多。线上。如果有人有任何想法，将非常感激。

修改

编程的整个部分的代码如下：

m|M)
            #Create Modify Message
            clear
            echo "                                 Modify Record                                 "
            echo -en '\n'
            echo -en '\n'
            while true
            do
                echo "What is the last name of the person you would like to modify:"
                read last_name
                if line=$(grep -i "^${last_name}:" "$2")
                then
                    oldIFS=$IFS 
                    IFS=:
                    set -- $line
                    IFS=$oldIFS
                    c1=$1
                    c2=$2
                    c3=$3
                    c4=$4
                    shift; shift; shift; shift 
                    rest="$*"
                    echo -e "Last Name: $1\nFirst Name: $2\nState: $4"
                    while true
                    do
                        echo "What would you like to change the state to?:"
                        read state
                            if echo $state | egrep -q '^[A-Z]{2}$'
                            then
                                echo "State: $state"
                                echo "This is a valid input"
                                break
                            else
                                echo "Not a valid input:"
                            fi
                    done
                    echo -e "Last Name: $c1\nFirst Name: $c2\nState: $state"
                    echo "State value changed"
                    break                       
                else
                    echo "ERROR: $last_name is not in database"
                    echo "Would you like to search again (y/n):"
                    read modify_choice
                    case $modify_choice in [Nn]) break;; esac
                fi  
            done

;;

好的，除了

外，一切正常

echo -e "Last Name: $c1\nFirst Name: $c2\nState: $state"

它只会显示：

姓氏：

名字：

州：

所以我可以看到它没有正确地将它添加到我的回声中。

最终编辑

CODE：

#Case statement for modifying an entry
        m|M)
            #Create Modify Message
            clear
            echo "                                 Modify Record                                 "
            echo -en '\n'
            echo -en '\n'
            while true
            do
                echo "What is the last name of the person you would like to modify:"
                read last_name
                if line=$(grep -i "^${last_name}:" "$2")
                then
                    echo "$line" |
                    while IFS=: read c1 c2 c3 c4 rest; do
                        echo -e "Last Name: $c1\nFirst Name: $c2\nState: $c4"
                        last=$c1
                        first=$c2
                    done
                    while true
                    do
                        echo "What would you like to change the state to?:"
                        read state
                            if echo $state | egrep -q '^[A-Z]{2}$'
                            then
                                echo "State: $state"
                                echo "This is a valid input"
                                break
                            else
                                echo "Not a valid input:"
                            fi
                    done
                    echo -e "Last Name: $last\nFirst Name: $first\nState: $state"
                    echo "State value changed"
                    break           
                else
                    echo "ERROR: $last_name is not in database"
                    echo "Would you like to search again (y/n):"
                    read modify_choice
                    case $modify_choice in [Nn]) break;; esac
                fi  
            done
 ;;

Answer 1

Bash中的here string

command <<<"string"

基本等同于

echo "string" | command

明显的例外是后者使用管道，这意味着你无法有意义地将它与read一起使用。常见的解决方法是使用set内置函数从字符串或外部命令中捕获令牌：

oldIFS=$IFS
IFS=:
set -- $line   # no quotes
IFS=$oldIFS
c1=$1
c2=$2
c3=$3
c4=$4
shift; shift; shift; shift
rest="$*"    # loses spacing / quoting

另一种解决方法是使用循环，只迭代一次;这看起来似乎很优雅，但如果伪循环的主体很长或很复杂，可能会导致相当笨重的代码。

echo "$line" |
while IFS=: read c1 c2 c3 c4 rest; do
    : stuff which uses those variables
done

这解决了echo stuff | read variable在子进程中运行read并因此立即忘记variable的值的问题 - while循环的正文是read发生的所有相同过程，因此初始化的变量值在循环内可见。

另一种类似的解决方法是将read和procesIn委托给一个函数;

process () {
    IFS=: read c1 c2 c3 c4 rest
    : stuff which uses those variables
}

echo "$line" | process

这是笨重还是优雅取决于功能中发生的事情。如果它整齐地封装，它可能是相当有吸引力的;但如果你最终传递了一堆不相关的变量（或者更糟糕的是，修改函数内部的全局变量！）它可能恰恰相反。

试着写“＆lt;＆lt;＆lt;＆lt;”在ksh

1 个答案: