String Bash：计算一个句子中字符串的出现次数

我只想计算一个特定字符串的出现次数。这是我的代码：

count=0
output="hai how are you ? my-code is incorrect. Do you have a new my-code ?"
echo $output

for word in $output
do
    if ["$word" == "my-code"];then
        count=$((count + 1))
    fi
done

echo $count

不幸的是，这是我的输出：

hai how are you ? my-code is incorrect. Do you have a new my-code ?
./test.sh: line 7: [hai: command not found
./test.sh: line 7: [how: command not found
./test.sh: line 7: [are: command not found
./test.sh: line 7: [you: command not found
./test.sh: line 7: [?: command not found
./test.sh: line 7: [my-code: command not found
./test.sh: line 7: [is: command not found
./test.sh: line 7: [incorrect.: command not found
./test.sh: line 7: [Do: command not found
./test.sh: line 7: [you: command not found
./test.sh: line 7: [have: command not found
./test.sh: line 7: [a: command not found
./test.sh: line 7: [new: command not found
./test.sh: line 7: [my-code: command not found
./test.sh: line 7: [?: command not found
0

---

我在字符串比较步骤中出错。我只是搜索了一下，发现了。我相信我也这样做了。这里的错误是什么？

[

是一个普通的shell命令。与所有命令一样，您需要在命令名和参数之间留出一个空格：

if [ "$word" == my-code ]; then

---

它还要求

]

是一个单独的参数，以便它在结尾处忽略它。

您有许多语法错误，如

[

之后和

]

之前以及

==

周围的空格等

但是，您可以通过使用

tr

完成工作来避免循环：

tr ' ' '\n' <<< "$output" | grep -c 'my-code'
2

---

tr''\n'
grep-o'my code'input.txt | wc-l