小编Mal*_*mar的帖子

我试图在同一个Ubuntu 16.10 32位框中创建MySQL 5.7的多个实例.我已将原始数据目录复制到新数据目录.为新实例创建单独的日志文件夹(日志正在运行).将所有者更改为mysql:mysql为新数据dir/var/lib/mysql2和log dir/var/log/mysql2 /.我删除了app-armor.我只是在经过其他SO答案之后发布了这个问题......其中没有一个解决了多实例问题.

这是我的my2.cnf:

[mysqld]
#
# * Basic Settings
#
user            = mysql
pid-file        = /var/run/mysqld/mysqld2.pid
socket          = /var/run/mysqld/mysqld2.sock
port            = 3307
basedir         = /usr
datadir         = /var/lib/mysql2
tmpdir          = /tmp
lc-messages-dir = /usr/share/mysql
skip-external-locking
#
# Instead of skip-networking the default is now to listen only on
# localhost which is more compatible and is not less secure.
bind-address            = 127.0.0.1
#
# * Fine Tuning
#
key_buffer_size         = 16M
max_allowed_packet      = 16M
thread_stack            = …

我想在我的网站上放置一个Google+登录按钮.按钮完美运作.我也有一个退出按钮.我在用户登录时隐藏登录按钮.我的代码在这里:

<span id="signinButton">
    <span
        class="g-signin"
        data-callback="signinCallback"
        data-clientid="******************"
        data-cookiepolicy="single_host_origin"
        data-requestvisibleactions="http://schema.org/AddAction"
        data-scope="https://www.googleapis.com/auth/plus.login https://www.googleapis.com/auth/userinfo.email">
    </span>
</span> &nbsp;&nbsp;&nbsp;&nbsp;
<button id="revokeButton" onclick="gapi.auth.signOut()">Sign Out</button>

<script>
    function signinCallback(authResult) {
        if (authResult['status']['signed_in']) {
            document.getElementById('signinButton').setAttribute('style', 'display: none');
            console.log("User successfully logged in!!");
        } else {
            console.log('Sign-in state: ' + authResult['error']);
        }
    }
</script>
<script>
    function disconnectUser(access_token) {
        var revokeUrl = 'https://accounts.google.com/o/oauth2/revoke?token=' +
                access_token;

        // Perform an asynchronous GET request.
        $.ajax({
            type: 'GET',
            url: revokeUrl,
            async: false,
            contentType: "application/json",
            dataType: 'jsonp',
            success: function(nullResponse) {
                document.getElementById('signinButton').setAttribute('style', 'display: display');
                // The response is …

设A是我的核心jar maven为基础的项目类.在另一个项目中,W是我的战争.我已经添加了A作为我对W的依赖.我已将prop.properties文件存储在资源文件夹中的A中.如何在战争中使用类路径从jar访问属性文件.使用Spring 4.我已经使用以下内容更新了jar的spring.xml:

 <context:property-placeholder 
 location="classpath:prop.properties" ignore-unresolvable="true"/>