How to check whether a string has at least one alphabetic character?

Question

I want to check whether a string has at least one alphabetic character? a regex could be like:

"^.*[a-zA-Z].*$"

however, I want to judge whether a string has at least one alphabetic character? so I want to use, like

 if [ it contains at least one alphabetic character];then
 ...
 else
 ...
 fi

so I'm at a loss on how to use the regex

I tried

if [ "$x"=~[a-zA-Z]+ ];then echo "yes"; else echo "no" ;fi
or
if [ "$x"=~"^.*[a-zA-Z].*$" ];then echo "yes"; else echo "no" ;fi

and test with x="1234", both of the above script output result of "yes", so they are wrong

how to achieve my goal?thanks!

I think you're looking for [this](http://stackoverflow.com/questions/304864/how-do-i-use-regular-expressions-in-bash-scripts). — David, Jan 02 '13 at 23:41
There are two problems: you need to use `[[` and `]]` instead of `[` and `]`; and you need to have spaces around the `=~` operator, otherwise bash sees it as part of one single long string. — Anders Johansson, Jan 03 '13 at 00:14
@Anders Johansson, you are right, it is ok now, but why do I need double [], can you explain this a bit? thanks! — user1944267, Jan 03 '13 at 00:27
@user1944267, the `[` builtin (see `man [`) is much less powerful than `[[` (see `man bash`; for example `[[` supports more operators and natural constructs like `&&` and `||` with their usual meaning while `[` requires `-a` and `-o` with explicit parentheses to get the same result). The `=~` operator only exists in `[[`. — Anders Johansson, Jan 03 '13 at 00:38

abhi.gupta200297 · Accepted Answer · 2013-01-03T02:27:37.670

3

Try this:

 #!/bin/bash

x="1234"
y="a1234"

if [[ "$x" =~ [A-Za-z] ]]; then
        echo "$x has one alphabet"
fi

if [[ "$y" =~ [A-Za-z] ]]; then
        echo "Y is $y and has at least one alphabet"
fi

edited Jan 03 '13 at 02:27

answered Jan 02 '13 at 23:46

abhi.gupta200297

881
6
12

I deleted 0-9, coz I only check ALPHABETIC characters. but the output is not as expected – user1944267 Jan 03 '13 at 00:06
@user1944267, this solution should work, and is preferrable since it works within bash instead of calling a separate tool. If the output is not as expected, please post your input data and expected output so we can help. (@abhi.gupta200297, the `+` is not needed). – Anders Johansson Jan 03 '13 at 00:09
Actually this won't work, you need to use `[[` and `]]` for regex in bash. – Anders Johansson Jan 03 '13 at 00:13
Ok i edited the code for a better example. @Anders thanks for pointing out. I corrected the brackets for regex. – abhi.gupta200297 Jan 03 '13 at 02:28
@user1944267 I think your problem is no spaces between tokens. – abhi.gupta200297 Jan 03 '13 at 02:31

score 2 · Answer 2 · answered Jan 02 '13 at 23:48

2

If you want to be portable, I'd call /usr/bin/grep with [A-Za-z].

answered Jan 02 '13 at 23:48

hd1

33,938
5
80
91

score 0 · Answer 3 · answered Jan 03 '13 at 00:45

0

Use the [:alpha:] character class that respects your locale, with a regular expression

[[ $str =~ [[:alpha:]] ]] && echo has alphabetic char

or a glob-style pattern

[[ $str == *[[:alpha:]]* ]] && echo has alphabetic char

answered Jan 03 '13 at 00:45

glenn jackman

238,783
38
220
352

score 0 · Answer 4 · answered Jan 03 '13 at 05:41

0

It's quite common in sh scripts to use grep in an if clause. You can find many such examples in /etc/rc.d/.

if echo $theinputstring | grep -q '[a-zA-Z]' ; then
    echo yes
else
    echo no
fi

answered Jan 03 '13 at 05:41

Wu Yongzheng

1,707
17
23

How to check whether a string has at least one alphabetic character?

4 Answers4