$ grep "console.log" * -R
account/db.js: console.log(err);
account/db.js: console.log(info);
account/db.js: console.log(err);
account/db.js: console.log(err2);
account/controller.js: console.log(covers);
account/controller.js: console.log(req.api_user);
account/controller.js: console.log(code);
account/controller.js: console.log(user);
account/helper.js: console.log(err);
messages/db.js: console.log("Error " + err);
messages/helper.js: console.log('No email notify.');
messages/helper.js: console.log(msg_body);
messages/helper.js: console.log(message.sid);
messages/helper.js: console.log(message.dateCreated);
messages/helper.js: console.log(error);
products/controller.js: console.log(product);
products/controller.js: console.log(product);
products/helper.js: console.log(data)
products/helper.js: console.log('removing index....');
profile/db.js: console.log(err);
profile/db.js: console.log(info);
profile/db.js: console.log(err);
profile/controller.js: console.log("sending phone confirmation text...");
profile/helper.js: console.log(message.sid);
profile/helper.js: console.log(message.dateCreated);
profile/helper.js: console.log(error);
receiver/controller.js: console.log(from);
receiver/controller.js: console.log(body);
receiver/controller.js: console.log(from_email);
receiver/controller.js: console.log(to_id_gen);
receiver/controller.js: console.log(finalbody);
receiver/controller.js: console.log(result);
reviews/db.js: console.log(err);
reviews/db.js: console.log(results);
reviews/controller.js: console.log(review);
reviews/controller.js: console.log(review_id);
search/controller.js: console.log(JSON.stringify(data.hits.hits, null, 4));
如您所见,当我编写代码时,我到处都在做 console.log。
现在,我想删除所有这些行。我不想手动进入每个文件来删除它们。相反,我想通过命令来做到这一点。
与 类似grep "console.log" * -R
,我怎样才能做同样的事情,但递归地删除这些行?(从我当前目录一直向下查看每个文件)