我来帮你排查下这个脚本里的语法问题，你遇到的报错主要是几个常见的Shell脚本规范没做好，尤其是elif的结构和条件语句的格式问题，咱们一步步来修正：

核心语法错误点 #

• if/elif后必须分隔then：Shell要求条件表达式结束后，要么用分号;和then隔开，要么把then单独放一行。比如你写的if [ "$1" = "-a" -a $# -lt 3 ] then，正确写法是if [ "$1" = "-a" -a $# -lt 3 ]; then，你所有的if/elif都没做这个处理，这是最主要的语法问题。
• elif结构不完整：elif的格式是elif [ 条件 ]; then，你写的elif [ "$1" = "-s" -a $# -lt 4 ]后面直接接了嵌套if，没有加then，导致Shell无法识别分支结构，-c、-d、-n分支都有这个问题。
• 注释符号用错了：Shell里单行注释是#，你用了// syntax error，//会被当作命令执行，这直接触发了语法报错，得改成# syntax error或者删掉。
• 嵌套if的fi闭合问题：每个if都需要对应的fi来收尾，你有些分支的fi位置不对，导致整个脚本逻辑结构混乱。

修正后的完整脚本 #

#!/bin/bash

if [ "$1" = "-a" -a $# -lt 3 ]; then
    echo "Add a new line in katalogos!"
    read -p "Give me a name... " name
    read -p "Give me a surname... " surname
    read -p "Give me a city name... " cityName
    read -p "Give me a phone number... " num
    echo "$name $surname $cityName $num" > katalogos
elif [ "$1" = "-l" -a $# -lt 3 ]; then
    echo "Content of katalogos will be sorted numerically and blank lines will be excluded!"
    sort -b -n katalogos
elif [ "$1" = "-s" -a $# -lt 4 ]; then
    if [[ $2 != *[!0-9]* ]]; then
        echo "Content of katalogos will be sorted according to the second argument!"
        sort +$3 katalogos
    fi
elif [ "$1" = "-c" -a $# -lt 4 ]; then  # 修正了注释符号
    if [[ $2 = *[!0-9]* ]]; then
        echo "Content of katalogos will be sorted according to the keyword!"
        if [ $(grep -e "$2" katalogos | wc -l) -eq 0 ]; then
            echo "String is not matched."
        else
            grep -e "$2" katalogos
        fi
    fi
elif [ "$1" = "-d" -a ( "$3" = "-b" -o "$3" = "-r" ) ]; then
    if [[ $2 = *[!0-9]* ]]; then
        echo "Katalogos's string matching lines will be deleted and blank lines will be in their place, assuming that the third argument equals -b, else just the lines will be deleted!"
        if [ $(grep -e "$2" katalogos | wc -l) -eq 0 ]; then
            echo "String is not matched."
        else
            if [ "$3" = "-b" ]; then
                sed -i "/$2/d" katalogos && sed -i '$a ' katalogos  # 修正了sed的语法错误
                echo "A blank line inserted in place of the deleted one."
            else
                sed -i "/$2/d" katalogos  # 修正了sed的语法错误
                echo "Line deleted."
            fi
        fi
    fi
elif [ "$1" = "-n" ]; then
    echo "katalogos's number of blank lines will be shown with the ability to delete them!"
    grep -cvP '\S' katalogos
    read -p "Do you want to delete them? Type 1 for yes or 0 for no... " ans
    if [ $ans -eq 1 ]; then
        sed -i '/^$/d' katalogos  # 修正了删除空行的逻辑
        echo "Lines deleted."
    fi
else
    echo "Help centre!"
    echo "-Type ./telcat -a to insert a new line to katalogos."
    echo "-Type ./telcat -l to see the contents of katalogos sorted numerically (excluding blank lines)."
    echo "-Type ./telcat -s plus a number to see the contents of katalogos sorted by the data that the number points to."
    echo "-Type ./telcat -c plus a keyword to see only the lines that match with the word given."
    echo "-Type ./telcat -d plus a keyword and -b or -r to delete the lines that contain the word given. Specifically if the third argument is -b it will automatically add a blank line to the deleted one and if it is -r it will not."
    echo "-Type ./telcat -n to see the number of blank lines of katalogos."
    echo "End of help centre!"
fi

额外的优化建议 #

• 尽量用[[ ]]替代[ ]，它支持更丰富的字符串匹配和逻辑表达式，可读性更强，比如可以把[ "$1" = "-a" -a $# -lt 3 ]改成[[ "$1" == "-a" && $# -lt 3 ]]。
• 参数引用尽量都加双引号，避免空格或特殊字符导致的错误，比如grep -e "$2" katalogos这种写法是对的，不要漏了引号。
• sort +$3是旧版语法，新版sort推荐用-k $3，比如sort -k $3 katalogos兼容性更好。

内容的提问来源于stack exchange，提问作者Marios Koni