Bash 意外标记“elif'；”附近出现语法错误已经读过其他帖子了

首先，我已经在这里列出的其他评论中查看了前面的评论，但不幸的是，提供的任何帮助都没有解决我的问题

我在CentOS 7中作为我的环境工作，我正在将一些错误处理编码到我的添加用户脚本中

#! /bin/bash
echo -n "Enter username: "
read -r username
/bin/egrep -i "^${username}:" /etc/passwd
if [ $? -eq 0 ]
echo "User $username already exists. Please check the username and try again."
elif [ $? eq 1 ]
echo "User $username does not exist. Please proceed with account creation."
then
adduser "$username"
echo -n "Enter password: "
read -r -s password
echo $username:$password | chpasswd
touch /etc/sudoers.d/sugroup
chmod 0440 /etc/sudoers.d/sugroup
echo "$username ALL=(ALL) NOPASSWD: ALL" >> /etc/sudoers.d/sugroup
else
echo "Error encountered."
fi

当我去测试它时，我得到以下错误消息：

./testscript-error.sh line 7: syntax error near unexpected token 'elif'
./testscript-error.sh line 7: elif [ $? eq 1 ]

我试过：

elif [ $? eq 1 ]**;**
echo "User $username does not exist. Please proceed with account creation."
then
adduser "$username"
echo -n "Enter password: "
read -r -s password
echo $username:$password | chpasswd
touch /etc/sudoers.d/sugroup
chmod 0440 /etc/sudoers.d/sugroup
echo "$username ALL=(ALL) NOPASSWD: ALL" >> /etc/sudoers.d/sugroup**;**

我也试过：

elif [ $? eq 1 ]
then
echo "User $username does not exist. Please proceed with account creation."
then
adduser "$username"
echo -n "Enter password: "
read -r -s password
echo $username:$password | chpasswd
touch /etc/sudoers.d/sugroup
chmod 0440 /etc/sudoers.d/sugroup
echo "$username ALL=(ALL) NOPASSWD: ALL" >> /etc/sudoers.d/sugroup

---

同样的结果。不确定我遗漏了什么，可以用另一双眼睛来观察。

给你。。。我希望您能更好地理解语法和用法：

#!/bin/bash

while true; do
    echo -n "Enter username: "
    read -r username
    /bin/egrep -i "^${username}:" /etc/passwd
    if [ $? -eq 0 ]; then
        echo "User $username already exists. Please check the username and try again."
    else
        echo "User $username does not exist. Proceed with account creation."
        break
    fi
done

adduser "$username"
if [ $? -gt 0 ]; then
    echo "Error encountered."
    exit 1
fi

echo -n "Enter password: "
read -r -s password
echo "$username:$password" | chpasswd
if [ $? -gt 0 ]; then
    echo "Error encountered."
    exit 1
fi

touch /etc/sudoers.d/sugroup
chmod 0440 /etc/sudoers.d/sugroup
echo "$username ALL=(ALL) NOPASSWD: ALL" >> /etc/sudoers.d/sugroup
if [ $? -gt 0 ]; then
    echo "Error encountered."
    exit 1
fi

---

这是完成的工作代码

#!/bin/bash
#========================================================================================================
# This script allows for account creation on a server                                                   |
# It also performs error handling to ensure that the user doesn't currently exist on the system.        |
# Also provides feedback from the input to verify the entries are correct.                              |
#========================================================================================================
while true; do
    echo -n "Enter username: "
    read -r username
    /bin/egrep -i "^${username}:" /etc/passwd
    if [ $? -eq 0 ]; then
        echo "User $username already exists. Please check the username and try again."
    else
        echo "User $username does not exist. Proceed with account creation."
        break
    fi
done

adduser "$username"
if [ $? -gt 0 ]; then
    echo "Error encountered."
    exit 1
fi

echo -n "Enter password: "
read -r -s password
echo "$username:$password" | chpasswd
echo "Password was succesfully set for $username."
if [ $? -gt 0 ]; then
    echo "Error encountered. There was a problem with your entry. Please re-run the script and try again."
    exit 1
fi

usermod -a -G wheel "$username"
echo "User was succesfully added to the group wheel."
if [ $? -gt 0 ]; then
    echo "Error encountered."
    exit 1
fi
echo "Successfully added $username to the system."

---

请看一看：如果

<0
是一个命令，但
[0
不是。
[1
也不是。当您编写
[$？
时，您正试图运行名为
[0
或
[1
的命令（或
$？
的任何值）。您需要一个空格：
，如果[$？…
@williampersell，我检查了我在这里输入的代码并添加了空格。它在我运行它的实际框上有空格。我已经通过shellchecker运行了它，甚至从帮助页面也无法理解为什么它会给我一个解析错误。@ladycoder2098没有关系。这仍然是问题所在。旁白：
如果[$？-gt 0]
比仅仅将
if
s合并到您想要测试其成功或失败的语句中要糟糕得多。
if！adduser“$username”；然后…
，或者更好的是，一行代码：
adduser“$username”|{echo”遇到错误“>&2；exit 1；
…或者一行代码带有函数：
die（）{echo“$*”>&2；退出1；}
，然后稍后：
useradd“$username”| | die“无法添加用户”
编写此类脚本的其他方法大约有100万种。只要完成了这项工作，就没有好坏之分。基于这个问题，很明显，一个初学者根本看不到自己的错误，只需要一个简单的例子来纠正他简单的逻辑。没有复杂的语法可以做到这一点。当然有一些措施其中之一是健壮性：如果代码在修改时不太可能被破坏，那么这比它更容易被破坏要好。（例如，添加新的日志记录可能会破坏
$？
，使对其值的依赖变得脆弱）。另一个措施是重复：遵循DRY的代码（不要重复自己的代码）更容易在一个地方修改（即，在我上面给出的函数中，您可以通过仅在一个地方进行更改来更改是否将错误记录到stdout或stderr）。