Question

我使用下面的代码填充一个非常大的表，表格按降序填充。有时，整个表格不会显示，新内容也不会显示。我希望这个表以升序填充，因为我需要的是最新的内容。谢谢你的帮助。

      <?xml version="1.0" encoding="UTF-8"?>
          <beans xmlns="http://www.springframework.org/schema/beans"
            xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
          xmlns:jaxws="http://cxf.apache.org/jaxws"
          xmlns:cxf="http://cxf.apache.org/core" 
         xmlns:soap="http://cxf.apache.org/bindings/soap"
          xsi:schemaLocation="http://cxf.apache.org/core 
          http://cxf.apache.org/schemas/core.xsd 
          http://www.springframework.org/schema/beans 
          http://www.springframework.org/schema/beans/spring-beans.xsd 
          http://cxf.apache.org/bindings/soap 
          http://cxf.apache.org/schemas/configuration/soap.xsd 
           http://cxf.apache.org/jaxws http://cxf.apache.org/schemas/jaxws.xsd">

        <jaxws:endpoint id="customerOrderid" address="/customerOrders"
            wsdlLocation="WEB-INF/wsdl/customerOrders.wsdl"  
            implementor="com.bharath.ws.train.imp.CustomersOrdersWS"
            serviceName="customersOrders:CustomerOrdersService" 
            xmlns:customersOrders="http://train.ws.bharath.com/" >
         </jaxws:endpoint>
    <cxf:bus>
        <cxf:features>
            <cxf:logging /> 
        </cxf:features>
    </cxf:bus>
   </beans>

Answer 1

更改

 $stmt = $conn->prepare("SELECT mydate, mball14, mball14a, mball15, mball15a, mball16, mball16a, mball17, mball17a, mball18, mball18a, mball19, mball19a, mball20, mball20a FROM manager");

要

 $stmt = $conn->prepare("SELECT mydate, mball14, mball14a, mball15, mball15a, mball16, mball16a, mball17, mball17a, mball18, mball18a, mball19, mball19a, mball20, mball20a FROM manager ORDER BY mydate DESC");

这将采用日期，并将最新的日期放在顶部

以升序填充表

1 个答案: