Question

我有.dat个文件：

#id|firstName|lastName|gender|birthday|creationDate|locationIP|browserUsed
933|Mahinda|Perera|male|1989-12-03|2010-03-17T13:32:10.447+0000|192.248.2.123|Firefox
1129|Carmen|Lepland|female|1984-02-18|2010-02-28T04:39:58.781+0000|81.25.252.111|Internet Explorer
4194|Hồ Chí|Do|male|1988-10-14|2010-03-17T22:46:17.657+0000|103.10.89.118|Internet Explorer
8333|Chen|Wang|female|1980-02-02|2010-03-15T10:21:43.365+0000|1.4.16.148|Internet Explorer
8698|Chen|Liu|female|1982-05-29|2010-02-21T08:44:41.479+0000|14.103.81.196|Firefox
8853|Albin|Monteno|male|1986-04-09|2010-03-19T21:52:36.860+0000|178.209.14.40|Internet Explorer
10027|Ning|Chen|female|1982-12-08|2010-02-22T17:59:59.221+0000|1.2.9.86|Firefox

我希望firstName从特定行开始lastName，birthday和id。

示例：如果输入是933，我想提取（用空格分隔）：

Mahinda Perera 1989-12-03

Answer 1

这应该这样做：

#!/bin/sh
id="$1"
awk -F '|' -v ID="$id" '($1==ID){print $2, $3, $5}' infile

用作：

$ script.sh 933
Mahinda Perera 1989-12-03

Answer 2

awk -F'|' '$1 ~ /933/{print $2, $3, $5}' file 
Mahinda Perera 1989-12-03

如果字段1匹配933打印以下字段：2,3和5.

Answer 3

使用GNU sed：

 public void getImage(){
    //String id = editTextId.getText().toString().trim();
    session = new SessionManager(getApplicationContext());
    HashMap<String, String> user = session.getUserDetails();

    String user_id = user.get(SessionManager.KEY_ID);

    class GetImage extends AsyncTask<String,Void,Bitmap> {
        ProgressDialog loading;

        @Override
        protected void onPreExecute() {
            super.onPreExecute();
            loading = ProgressDialog.show(Profile.this, "Uploading...", null, true, true);
        }

        @Override
        protected void onPostExecute(Bitmap b) {
            super.onPostExecute(b);
            loading.dismiss();
            imageView.setImageBitmap(b);
        }

        @Override
        protected Bitmap doInBackground(String... params) {
            String user_id = params[0];
            String add = "http://www.example.com/folder/getProfile.php?user_id="+user_id;
            URL url = null;
            Bitmap image = null;
            try {
                url = new URL(add);
                image = BitmapFactory.decodeStream(url.openConnection().getInputStream());
            } catch (MalformedURLException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return image;
        }
    }

    GetImage gi = new GetImage();
    gi.execute(user_id);
}

$ id=933 $ sed -n "/^$id"'|/{s/^[^|]*|$[^|]*$|$[^|]*$|[^|]*|$[^|]*$|.*$/\1 \2 \3/p;q}' infile Mahinda Perera 1989-12-03阻止打印;其余的做到了这一点：

-n

要让它在BSD sed下运行，"/^$id"'|/ { # If a line starts with the ID... (notice quoting for parameter expansion) # Capture second, third and fifth field, discard rest, print s/^[^|]*|$[^|]*$|$[^|]*$|[^|]*|$[^|]*$|.*$/\1 \2 \3/p q # Quit to avoid processing the rest of the file for nothing }'和右括号之间必须有另一个分号。

表明awk更适合这个问题。

Answer 4

使用：

awk '{print $2 $3 $5}' infile.dat

从管道分隔文件中获取特定列和特定行

4 个答案: