Chinaunix

标题: 如何将文件名拼接成一个字符串 [打印本页]

作者: yufeiluo    时间: 2017-03-07 16:28
标题: 如何将文件名拼接成一个字符串
ls /var/www/html/wp
index.php        wp-blog-header.php    wp-includes        wp-signup.php
license.txt      wp-comments-post.php  wp-links-opml.php  wp-trackback.php
readme.html      wp-config.php         wp-load.php        xmlrpc.php
wc               wp-config-sample.php  wp-login.php
wp-activate.php  wp-content            wp-mail.php
wp-admin         wp-cron.php           wp-settings.php

如何将这些输出,用一个空格连接起来,赋值给x.

echo $x  将输出

index.php wp-blog-header.php  等等


作者: haooooaaa    时间: 2017-03-07 16:31
  1. x=`ls | xargs`
复制代码

作者: moperyblue    时间: 2017-03-07 16:34

  1. ls /var/www/html/wp|paste -sd' '
复制代码

作者: yufeiluo    时间: 2017-03-07 17:57
$ x=$(ls /var/www/html/wp|paste -sd' ')
$ echo $x
index.php license.txt readme.html wc wp-activate.php wp-admin wp-blog-header.php wp-comments-post.php wp-config.php wp-config-sample.php wp-content wp-cron.php wp-includes wp-links-opml.php wp-load.php wp-login.php wp-mail.php wp-settings.php wp-signup.php wp-trackback.php xmlrpc.php





欢迎光临 Chinaunix (http://bbs.chinaunix.net/) Powered by Discuz! X3.2