bash - 如何在 bash shell 脚本变量中查找子字符串

Question

我有一个像下面这样的变量。

variable = This script is not found

if [[ "$variable" = ~ "not found" ]];
then
echo "Not Found"
else
echo "Its there"
if

在执行即时通讯低于错误时，

line 4: syntax error in conditional expression
./test.sh: line 4: syntax error near `found"'
./test.sh: line 4: `if [[ "$variable" = ~ "not found" ]]; '

谁能指出我，我在这里缺少什么？

Answer 1

LIST="some string with a substring you want to match"
SOURCE="substring"

if echo "$LIST" | grep -q "$SOURCE"; then
    echo "matched";
else
    echo "no match";
fi

祝你好运 ;）

Answer 2

在指示点将其与您的版本进行比较：

variable="This script is not found"  # <--

if [[ "$variable" =~ "not found" ]]  # <--
then
    echo "Not Found"
else
    echo "Its there"
fi  # <--

您不能在赋值中的 = 周围放置空格，并且您需要引用包含空格的字符串文字。你不需要尾随 ; 如果你打算把 then 放在它自己的线上。if-then 以“fi”而不是“if”结尾。

Answer 3

这是您的 if 语句的正确构造

if [[ "$variable" =~ "not found" ]]; then
      echo "Not Found";
else
      echo "Its there";
fi

Answer 4

我尝试过以下代码，它们总是以真或假返回相同的结果

if [[ "$variable" =~ "not found" ]]; then
      echo "Not Found";
else
      echo "Its there";
fi