命令输出匹配脚本问题：含双指定字符串时误判失败求助

Let's break down why your original script was failing when both keywords are present, and fix it properly.

The Root of the Problem #

Your original condition uses && (logical AND) between two negations:

if [ "$STATUS" != "Ok" ] && [ "$STATUS" != "Non-Critical" ]

When STATUS contains both "Ok" and "Non-Critical" (like "Ok Non-Critical"), both $STATUS != "Ok" and $STATUS != "Non-Critical" are true (since the full string doesn't match either exact value). The && makes the whole condition true, so it triggers the failure message—exactly the opposite of what you want.

Correct Logic Approach #

We need to check if the status contains at least one of the allowed keywords ("Ok" or "Non-Critical"). If it does, we pass; if it contains neither, we fail.

Fixed Script Options #

Here are two reliable ways to implement this:

Option 1: Using grep (simple and readable)

This uses extended regex to check for either keyword in the status string:

if echo "$STATUS" | grep -qE 'Ok|Non-Critical'; then
    echo "Status check good."
else
    echo "FAILED: STATUS does NOT contain 'Ok' OR 'Non-Critical'"
fi

• The -q flag makes grep quiet (no output, just exit code)
• -E enables extended regex for the | (OR) operator

Option 2: Using Bash's native string matching (no external tools)

Bash's [[ construct supports wildcard matching with *, so we can check if the string contains either keyword directly:

if [[ "$STATUS" == *Ok* || "$STATUS" == *Non-Critical* ]]; then
    echo "Status check good."
else
    echo "FAILED: STATUS does NOT contain 'Ok' OR 'Non-Critical'"
fi

• *Ok* matches any string that has "Ok" anywhere in it
• || is logical OR, so if either condition is true, the check passes

Verification of All Pass Scenarios #

Both scripts will correctly pass for:

• STATUS="Ok"
• STATUS="Non-Critical"
• STATUS="Ok Non-Critical"
• STATUS="Non-Critical Ok"

And fail for any string that doesn't include either keyword (like "Critical", "Error", or an empty string).

内容的提问来源于stack exchange，提问作者Boab