Bash Shell脚本失败，错误为预期的二进制运算符

我在运行脚本时遇到群集\u网络\u操作。sh:line 44:[：0:应为二进制运算符错误

在我的脚本中，我调用ssh copy id命令，然后根据其输出设置返回代码

ssh拷贝id:output

WARNING: All keys were skipped because they already exist on the remote system.

Successful:

Number of key(s) added: 1

Now try logging into the machine, with "ssh 'bhupesh@10.236.61.133'", and check in:

2. All keys were skipped because they already exist on the remote system

我使用了一个函数，该函数包含检查字符串中的子字符串

contains() {
    # check if the substring exits in a string. 
    string="$1"
    substring="$2"
    if test "${string#*$substring}" != "$string"
    then
        return 0    # $substring is in $string
    else
        return 1    # $substring is not in $string
    fi
}

enable_auth(){
#   Enabling the authentication by sending the pub_key of all the nodes in
#   the cluster to the new gateway.
#   Currently, its a one way secure authentication.
    #pdsh -S -g all -N cat /root/.ssh/id_rsa.pub | ssh "$2"@"$1" 'cat >> .ssh/authorized_keys'
    ssh-copy-id -i ~/.ssh/id_rsa.pub "$2"@"$1"
    ret=$?
    #sub_str = "Now try logging into the machine, with"
    if [ contains "$ret" "Now try logging into the machine, with" ]
    then
        echo "Successfully Configured"
        ret=0
    elif [ contains "$ret" "All keys were skipped because they already exist on the remote system" ]
    then
        echo "System "$1" is already configured for the user: $2"
        ret=0
    else
        ret=1
    fi
    echo 0 
}

---

在shell中，返回代码0表示成功或true，0表示错误

使用退出状态的示例

function contains {
    local string substring
    string=$1
    substring=$2
    [[ ${string#*$substring} != ${string} ]]
    return $?
}

if contains abc b; then echo ok; else echo KO; fi

if ! contains xyz c; then echo ok; else echo KO; fi

contains abc b && echo success
contains abc b || echo failed

---

Remove

[

和

]

调用

contains

函数$？不是ssh复制id输出，而是返回代码，包含0“any string”对IMHO没有什么意义