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错误：“[：缺少`]'”，括号前有空格

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错误：“[：缺少`]'”，括号前有空格

当我运行以下代码时：

#!/bin/bash

PREVIOUS_COMMIT=e099d95d52b8fca99e47fd7cee5f782287178b27
SERVICE=service-web-prj1

if [ ! git diff "$PREVIOUS_COMMIT" HEAD --name-only | grep -qs "$SERVICE" ] || [ ! git diff "$PREVIOUS_COMMIT" HEAD --name-only | grep -qs 'service-web' ];
    then
        echo Didnt pass first
        exit 0;
fi

echo passed first with $SERVICE

if ( ! echo "$SERVICE" | grep -q "^service-web" );
    then
        echo Didnt pass second
        exit 0;
fi

echo passed second with $SERVICE

我正进入（状态：

scripts/getLastCommit.1.sh: line 9: [: missing `]'
scripts/getLastCommit.1.sh: line 9: [: missing `]'

我环顾四周，每个人都提到了最后一个“]”之前的空格，我已经检查了三遍，我确实有这个

有人知道我做错了什么吗？

谢谢，丹尼斯

发现这对我有用：

if ! echo $GITDIFF | grep -q -e 'service-web*' -e $SERVICE;

答案1

您无法测试 shelltest括号内的整个命令。您想要完全删除这些括号，这将导致它只测试命令的退出代码。

#!/bin/bash

PREVIOUS_COMMIT=e099d95d52b8fca99e47fd7cee5f782287178b27
SERVICE=service-web-prj1

if ! git diff "$PREVIOUS_COMMIT" HEAD --name-only | grep -qs "$SERVICE" || ! git diff "$PREVIOUS_COMMIT" HEAD --name-only | grep -qs 'service-web'
then
        echo "Didn't pass first"
        exit 0
fi

echo passed first with "$SERVICE"

if ( ! echo "$SERVICE" | grep -q "^service-web" )
then
        echo "Didn't pass second"
        exit 0
fi

echo passed second with "$SERVICE"

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