首页 \ 问答 \ Tomcat,MySQL - >无法为连接URL''创建类''的JDBC驱动程序(Tomcat, MySQL -> Cannot create JDBC driver of class '' for connect URL 'null')

Tomcat,MySQL - >无法为连接URL''创建类''的JDBC驱动程序(Tomcat, MySQL -> Cannot create JDBC driver of class '' for connect URL 'null')

我知道这看起来像一个重复的问题,但已尝试stackoverflow和其他论坛中的所有帖子但无济于事。

Apache Tomcat 8.0.32 Amazon RDS运行MySQL 5.6在EC2 Windows服务器上运行tomcat

我的web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.1" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd">
...
  <resource-ref>
    <description>datasource</description>
    <res-ref-name>jdbc/bbDataSource</res-ref-name>
    <res-type>javax.sql.DataSource</res-type>
    <res-auth>Container</res-auth>
    <res-sharing-scope>Shareable</res-sharing-scope>
  </resource-ref>
...
</web-app>

Context.xml位于我的应用程序的META-INF文件夹中:

<?xml version="1.0" encoding="UTF-8"?>
<Context path="/ch">
   <Resource name="jdbc/bbDataSource"
        auth="Container"
        type="javax.sql.DataSource"
        username="********"
        password="********"
        driverClassName="com.mysql.jdbc.Driver"
        url="jdbc:mysql://*url*:3306/x?zeroDateTimeBehavior=convertToNull"/>
 </Context>

在ServletContextListener中创建数据源

try {
        dataSource = (DataSource) new InitialContext().lookup("java:comp/env/jdbc/bbDataSource");
        mLogger.log(Level.CONFIG, "Startup - AppContextServletListener - contextInitialized - Created dataSource object: {0}", dataSource);
        sce.getServletContext().setAttribute("dataSource", dataSource);
}

启动期间没有问题。 访问dataSource tomcat时会抛出上述错误。 并且jar文件位于tomcat的lib文件夹中(我怀疑这是问题,因为我在同一个应用程序中也使用了hibernate,它能够访问数据库而没有任何问题)。

另外我注意到以下内容:$ CATALINA_HOME / conf / Catalina / localhost没有'ch.xml' - >尝试复制context.xml并重命名为'ch.xml',但无济于事。

server.xml - >在主机节点下

 <Host name="localhost"  appBase="webapps"
        unpackWARs="true" autoDeploy="true">
        <Context path="/ch" docBase="ch"></Context>
        <Context path="/ch/asdasd" docBase="C:\\xasdasd\\asdasd">   </Context>

I know this looks like a repeat question but have tried all posts in stackoverflow and other forums but to no avail.

Apache Tomcat 8.0.32 Amazon RDS running MySQL 5.6 Running tomcat on EC2 windows server

My web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.1" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd">
...
  <resource-ref>
    <description>datasource</description>
    <res-ref-name>jdbc/bbDataSource</res-ref-name>
    <res-type>javax.sql.DataSource</res-type>
    <res-auth>Container</res-auth>
    <res-sharing-scope>Shareable</res-sharing-scope>
  </resource-ref>
...
</web-app>

Context.xml sitting in the META-INF folder of my app:

<?xml version="1.0" encoding="UTF-8"?>
<Context path="/ch">
   <Resource name="jdbc/bbDataSource"
        auth="Container"
        type="javax.sql.DataSource"
        username="********"
        password="********"
        driverClassName="com.mysql.jdbc.Driver"
        url="jdbc:mysql://*url*:3306/x?zeroDateTimeBehavior=convertToNull"/>
 </Context>

Creating the datasource in a ServletContextListener

try {
        dataSource = (DataSource) new InitialContext().lookup("java:comp/env/jdbc/bbDataSource");
        mLogger.log(Level.CONFIG, "Startup - AppContextServletListener - contextInitialized - Created dataSource object: {0}", dataSource);
        sce.getServletContext().setAttribute("dataSource", dataSource);
}

No issues during startup. While accessing the dataSource tomcat throws the above error. And the jar file is sitting in the lib folder of tomcat (I doubt this is the problem since I'm also using hibernate in the same application and it is able to access the DB without any issues).

Also I noted the following: $CATALINA_HOME/conf/Catalina/localhost does not have the 'ch.xml' -> Tried copying the context.xml and renaming to 'ch.xml' there but to no avail.

server.xml -> Under the host node

 <Host name="localhost"  appBase="webapps"
        unpackWARs="true" autoDeploy="true">
        <Context path="/ch" docBase="ch"></Context>
        <Context path="/ch/asdasd" docBase="C:\\xasdasd\\asdasd">   </Context>

原文:https://stackoverflow.com/questions/36232894
更新时间:2022-01-10 08:01

最满意答案

更新:

用类似的东西循环遍历数组并不困难

foreach($messages as $k => $m){
    if(strtotime($m['Message']['updatedate']) > strtotime($m['Message']['createdate'])){
        $messages[$k]['Message']['NewDate'] = $m['Message']['updatedate'];
    }else{
        $messages[$k]['Message']['NewDate'] = $m['Message']['createdate'];
    }
}

原始答案:

我认为你的Message.updatedate总是最新的日期,所以你可以选择它。 但假设由于某种原因不是这种情况,您可以在模型中创建一个虚拟字段:

public $virtualFields = array(
    'NewDate' => "IF(Message.updatedate > Message.createdate, Message.updatedate, Message.createdate)"
);

这使用MySQL IF()函数。 如果您不使用MySQL,则必须弄清楚如何使用与数据库类似的操作。

http://dev.mysql.com/doc/refman/5.0/en/control-flow-functions.html#function_if


UPDATE:

It's not hard to loop over the array with something like

foreach($messages as $k => $m){
    if(strtotime($m['Message']['updatedate']) > strtotime($m['Message']['createdate'])){
        $messages[$k]['Message']['NewDate'] = $m['Message']['updatedate'];
    }else{
        $messages[$k]['Message']['NewDate'] = $m['Message']['createdate'];
    }
}

ORIGINAL ANSWER:

I would think that your Message.updatedate would always be the newest date, so you could just select that. But assuming that's not the case for some reason, you can create a virtual field in your model:

public $virtualFields = array(
    'NewDate' => "IF(Message.updatedate > Message.createdate, Message.updatedate, Message.createdate)"
);

This uses the MySQL IF() function. If you're not using MySQL you'd have to figure out how to do something similar with your database.

http://dev.mysql.com/doc/refman/5.0/en/control-flow-functions.html#function_if

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