Question

我一直在编写一个bash脚本，我无法弄清楚为什么！=运算符无效。

#/bin/bash
vips=()
vips+=("        Ltm::HTTP Profile: Default_HTTP_Profile")
vips+=("        Ltm::Virtual Address: 10.206.16.76")
for i in "${vips[@]}";
do
    if [[ $i != *"TCP Profile"* ]] || [[ $i != *"OneConnect"* ]] || [[ $i != *"HTTP Profile"* ]]; then
        echo "test"
    fi
done
for i in "${vips[@]}";
do
    echo "$i"
done

结果是

    test
    test
        Ltm::HTTP Profile: Default_HTTP_Profile
        Ltm::Virtual Address: 10.206.16.76

如您所见，第二个数组元素不应与if逻辑匹配。

Answer 1

让我们跟踪它的执行方式：

i="        Ltm::HTTP Profile: Default_HTTP_Profile"
if [[ $i != *"TCP Profile"* ]] || [[ $i != *"OneConnect"* ]] || [[ $i != *"HTTP Profile"* ]]; then

首先，它运行[[ $i != *"TCP Profile"* ]]。此测试返回true，因为该字符串不包含TCP Profile。因此，if作为一个整体是正确的，并且它不需要运行任何其他测试。

您可能想要的内容如下：

case $i in
   *"TCP Profile"*|*"OneConnect"*|*"HTTP Profile"*) : ;; # do nothing
   *) echo "test" ;;
esac

......或者：

if ! [[ $i = *"TCP Profile"* || $i = *"OneConnect"* || $i = *"HTTP Profile"* ]]; then
  echo "test"
fi

Answer 2

我修改了我的逻辑如下，它的工作原理。谢谢您的帮助。

With Phone as (SELECT phone_id, patient_id, phone_type, PD.Number, row_number() over (partition by Patient_ID order by PD.Number) as RN

           FROM Phone_type PT
           INNER JOIN Phone_Details PD
              ON PT.Phone_ID = PD.Phone_ID
           WHERE Phone_Type = 1),
  Fax as (SELECT phone_id, patient_id, phone_type, PD.Number, row_number() over (partition by Patient_ID order by PD.Number) as RN
           FROM Phone_type PT
           INNER JOIN Phone_Details PD
              ON PT.Phone_ID = PD.Phone_ID
           WHERE Phone_Type = 1)
SELECT *
FROM PHONE P
FULL OUTER JOIN FAX F
 on P.Patent_Id =F.PatentID
and P.RN=F.RN

bash不等于操作员不工作

2 个答案: