chmod | 易学教程

CakePHP chmod question

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问题 I downloaded CakePHP and put it in my www directory. I enabled mod-rewrite and all that and changed my document root. But I still get permission denied errors in my apache error log because Cake can't require the files from other directories. It's almost as if I need to chmod every single directory to 777 in order to make this work. I know that's not right but can someone tell me if I did not install it correctly or what to CHMOD. Because so far it's literally every folder I'm going through

How can I audit all chmod and chgrp commands?

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问题 Corporate security policies are starting to require low level event logging. For example, file access permission changes. One solution is to use SELinux but our knowledge of this is sparse at best. Another is to replace the command with a proxy which performs auditing (this sucks though). Any ideas? 回答1: I think you can look at auditd: http://linux.die.net/man/8/auditd Also check this thread please: http://www.linuxforums.org/forum/linux-security/109864-auditing-logging-all-commands-arguments

Linux terminal octal form of chmod

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问题 I want to set a directory to -rwx r-x r-x using the octal form of chmod what should i type in the terminal after chmod? and how do i calculate the digits in octal form? 回答1: It goes like this: rwx rwx rwx 421 421 421 so what you want is rwx r-x r-x 421 4-1 4-1 7 5 5 so you type chmod 755 nameofdirectory 来源： https://stackoverflow.com/questions/19085653/linux-terminal-octal-form-of-chmod

Phoenix file copying on Heroku

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问题 I'm trying to upload images to my Phoenix app on Heroku. I have a simple app that follows the instructions for file uploading from Phoenix's docs. I have a simple form and the controller uses File.cp_r() to copy the file from the temp directory. def create(conn, %{"user" => user_params}) do if upload = user_params["photo"] do File.cp_r(upload.path, "priv/static/images/foo.jpg") #non-dynamic name, doens't matter end ... end Works just file on my local. However when I upload this to Heroku,

maven-war-plugin chmod exit code was:127

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问题 I am running a shell to call a ant, which further call a maven in Redhat. Could anyone point out the context of this error? [ERROR] Failed to execute goal org.apache.maven.plugins:maven-war-plugin:2.3:wa r (default-war) on project sites: Error unpacking file [/scratch/mingtzha/12.1/.../sites-dojo/target/sites-dojo-1.6.1-12.1.4.0-SNAPSHOT.war]to [/scratch /mingtzha/12.1/.../sites-dojo]: chmod exit code was: 127 -> [Help 1] I found this "useJvmChmod" property maybe relevant. http://maven.apache

How to understand rwx permissions for directories? [closed]

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问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 6 years ago . What does it mean to read, write, and execute a directory? It's intuitive on files, but I don't understand it for directories. 回答1: If you have write permission for a directory, you can create new entries (files/folders/etc). If you have read permission for a directory, you may list ls the directories contents.

How run chmod in OSX with C#

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问题 I have a C# Mono program (Unity3d game) that unzips a zip archive using SharpZLib. My problem is that when the archive is unzipped on Mac OSX, the file permissions are different. Specifically they are set to 644 while I want them to be 755. I don't know why the file permissions are not kept the same after unzipping, but in any case I need to set permissions for all extracted files to 755 after unzipping. How can I do this with C#? I also need this to work on Windows and not only Mac OSX but

PHP files does not execute in windows machine

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问题 i created the application in PHP on my mac osx. and after the development i zipped and transferred to the windows 7 machine. and when i tried opening it i get this error. Warning: Unknown: failed to open stream: Permission denied in Unknown on line 0 Fatal error: Unknown: Failed opening required 'C:/wamp/www/app/index.php' (include_path='.;C:\php\pear') in Unknown on line 0 the above error suggest, the error is caused because of the file permission issue, is there any possible way i can fix

Unable to run .py file from putty, Syntax error: word unexpected (expecting “)”)

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问题 I am new to both Python and Linux and as such request simple explanations with minimal assumed knowledge where possible please, however I am more than willing to invest time and effort to learn. I have a Raspberry Pi 2 (Model B V1.1) that is running Linux. I interact with this pi via putty. I am trying to create a simple competitive reflex game, consisting of 2 buttons and a single LED. My goal is to have the LED light up after a short interval, and the first player to press their button wins

备份和恢复

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第五节课数据备份与恢复一数据备份相关概念 1.1 数据备份的目的？数据被误删除或设备损害导致数据丢失，是备份文件恢复数据。 1.2 数据备份方式？物理备份：指定备份库和表对应的文件 51 cp -r /var/lib/mysql/opt/mysql.bak cp -r /var/lib/mysql/bbsdb/opt/bbsdb.bak rm -rf /var/lib/mysql/bbsdb cp -r /opt/bbsdb.bak/var/lib/mysql/bbsdb chown -R mysql:mysql/var/lib/mysql/bbsdb systemctl restart mysqld 51 scp /opt/mysql.bak 192.168.4.51:/root/ 52 rm -rf /var/lib/mysql cp -r /root/mysql.bak /var/lib/mysql chown -R mysql:mysql /var/lib/mysql systemctl restart mysqld 逻辑备份：在执行备份命令时，根据备份的库表及数据生成对应的 sql 命令，把 sql 存储到指定的文件里。 1.3 数据备份策略？完全备份备份所有数据（一张表的所有数据一个库的所有数据一台数据库的所有数据）备份新产生数据