需要使用awk匹配在文件中查找字符串

Question

Ii have javascript file which contains following

    ..............
    templateUrl: 'views/product.html'
    url: '/categories',
    templateUrl: 'views/categories.html'
    url: '/report',
    templateUrl: 'views/report.html'
    url: '/publisher',
    templateUrl: 'views/publisher.html'
    url: '/manageoutlet',
    templateUrl: 'views/outlet.html'
    url: '/order',
    templateUrl: 'views/order.html'
    url: '/mobileapp',
    templateUrl: 'views/mobileapp.html'
    url: '/cartitem/:id',
    .....
    .....
    'url': 'http://fhdjhjd/fkdf',
    ........
    ........

I need to get the line 'url': 'http://fhdjhjd/fkdf', from the file using awk , for that am using the following code

sudo awk '{
    if(match($0,/'url'\s*:\s*'.*'/)){
        pattern = substr($0,RSTART,RLENGTH);
        printf "%s\n",pattern;
    }
}' "app.js"

but the output is

url: '/categories',
url: '/report',
url: '/publisher',
url: '/manageoutlet',
url: '/order',
url: '/mobileapp',
url: '/cartitem/:id',

I don't know why it's happen . I need to get the line

'url': 'http://fhdjhjd/fkdf',

Answer 1

Using \\047 to match "'"

sudo awk '{
    if(match($0,/\047url\047\s*:\s*'.*'/)){
        pattern = substr($0,RSTART,RLENGTH);
        printf "%s\n",pattern;
    }
}' "app.js"

Answer 2

尝试

awk -F":" '$2~/fhdjhjd/{ print }' file

Answer 3

Try this:

awk '($0~/'\''url'\'':/) {print}' File

In your case, it will be:

sudo awk '($0~/'\''url'\'':/) {print}' "app.js"

Answer 4

You should escape the single quotes ' using '\\''

the regex must be

'\''url'\''\s*:\s*'.*'

The code can be

sudo awk '{
    if(match($0,/'\''url'\''\s*:\s*'.*'/)){
        pattern = substr($0,RSTART,RLENGTH);
        printf "%s\n",pattern;
    }
}' "app.js"

But I would use a simpler awk code as

$ awk ' /'\''url'\''\s*:\s*'.*'/ ' "app.js"
    'url': 'http://fhdjhjd/fkdf',