bash shell脚本中的整数表达式预期错误
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Integer expression expected error in shell script
提问by user2904832
I'm a newbie to shell scripts so I have a question. What Im doing wrong in this code?
我是 shell 脚本的新手，所以我有一个问题。我在这段代码中做错了什么？
#!/bin/bash
echo " Write in your age: "
read age
if [ "$age" -le "7"] -o [ "$age" -ge " 65" ]
then
echo " You can walk in for free "
elif [ "$age" -gt "7"] -a [ "$age" -lt "65"]
then
echo " You have to pay for ticket "
fi
When I'm trying to open this script it asks me for my age and then it says
当我试图打开这个脚本时，它会询问我的年龄，然后它说
./bilet.sh: line 6: [: 7]: integer expression expected
./bilet.sh: line 9: [: missing `]'
I don't have any idea what I'm doing wrong. If someone could tell me how to fix it I would be thankful, sorry for my poor English I hope you guys can understand me.
我不知道我做错了什么。如果有人能告诉我如何解决它，我将不胜感激，对不起，我的英语不好我希望你们能理解我。
回答by kamituel
You can use this syntax:
您可以使用以下语法：
#!/bin/bash
echo " Write in your age: "
read age
if [[ "$age" -le 7 || "$age" -ge 65 ]] ; then
echo " You can walk in for free "
elif [[ "$age" -gt 7 && "$age" -lt 65 ]] ; then
echo " You have to pay for ticket "
fi
回答by chepner
If you are using -o(or -a), it needs to be inside the brackets of the testcommand:
如果您使用-o(或-a)，它需要在test命令的括号内：
if [ "$age" -le "7" -o "$age" -ge " 65" ]
However, their use is deprecated, and you should use separate testcommands joined by ||(or &&) instead:
但是，不推荐使用它们，您应该使用test由||(或&&)连接的单独命令：
if [ "$age" -le "7" ] || [ "$age" -ge " 65" ]
Make sure the closing brackets are preceded with whitespace, as they are technically arguments to [, not simply syntax.
确保右括号前面有空格，因为它们是 的技术参数[，而不仅仅是语法。
In bashand some other shells, you can use the superior [[expression as shown in kamituel's answer. The above will work in any POSIX-compliant shell.
在bash和其他一些 shell 中，您可以使用高级[[表达式，如kamituel's answer所示。以上将在任何符合 POSIX 的 shell 中工作。
回答by Mr-IDE
This error can also happen if the variable you are comparing has hidden characters that are not numbers/digits.
如果您比较的变量具有不是数字/数字的隐藏字符，也会发生此错误。
For example, if you are retrieving an integer from a third-party script, you must ensure that the returned string does not contain hidden characters, like "\n"or "\r".
例如，如果您从第三方脚本中检索整数，则必须确保返回的字符串不包含隐藏字符，例如"\n"或"\r"。
For example:
例如：
#!/bin/bash
# Simulate an invalid number string returned
# from a script, which is "1234\n"
a='1234
'
if [ "$a" -gt 1233 ] ; then
echo "number is bigger"
else
echo "number is smaller"
fi
This will result in a script error : integer expression expectedbecause $acontains a non-digit newline character "\n". You have to remove this character using the instructions here: How to remove carriage return from a string in Bash
这将导致脚本错误，: integer expression expected因为$a包含非数字换行符"\n"。您必须使用此处的说明删除此字符：如何从 Bash 中的字符串中删除回车符
So use something like this:
所以使用这样的东西：
#!/bin/bash
# Simulate an invalid number string returned
# from a script, which is "1234\n"
a='1234
'
# Remove all new line, carriage return, tab characters
# from the string, to allow integer comparison
a="${a//[$'\t\r\n ']}"
if [ "$a" -gt 1233 ] ; then
echo "number is bigger"
else
echo "number is smaller"
fi
You can also use set -xvto debug your bash script and reveal these hidden characters. See https://www.linuxquestions.org/questions/linux-newbie-8/bash-script-error-integer-expression-expected-934465/
您还可以set -xv用来调试 bash 脚本并显示这些隐藏字符。见https://www.linuxquestions.org/questions/linux-newbie-8/bash-script-error-integer-expression-expected-934465/
回答by muha
./bilet.sh: line 6: [: 7]: integer expression expected
Be careful with " "
小心 " "
./bilet.sh: line 9: [: missing `]'
This is because you need to have space between brackets like:
这是因为您需要在括号之间留有空格，例如：
if [ "$age" -le 7 ] -o [ "$age" -ge 65 ]
look: added space, and no " "
看：增加了空间，没有 " "
回答by Harry1992
Try this:
尝试这个：
If [ $a -lt 4 ] || [ $a -gt 64 ] ; then \n
Something something \n
elif [ $a -gt 4 ] || [ $a -lt 64 ] ; then \n
Something something \n
else \n
Yes it works for me :) \n
回答by markfree
If you are just comparing numbers, I think there's no need to change syntax, just correct those lines, lines 6 and 9 brackets.
如果您只是比较数字，我认为无需更改语法，只需更正那些行、第 6 行和第 9 行括号即可。
Line 6 before: if [ "$age" -le "7"]-o [ "$age" -ge " 65"]
第 6 行之前： if [ "$age" -le "7"]-o [ "$age" -ge " 65"]
After: if [ "$age" -le "7" -o "$age" -ge "65" ]
后： if [ "$age" -le "7" -o "$age" -ge "65" ]
Line 9 before: elif [ "$age" -gt "7"] -a [ "$age" -lt "65"]
第 9 行之前： elif [ "$age" -gt "7"] -a [ "$age" -lt "65"]
After: elif [ "$age" -gt "7" -a "$age" -lt "65" ]
后： elif [ "$age" -gt "7" -a "$age" -lt "65" ]
