Question

我是WS02的新手，但我很难找到一个url param并通过它。关注Sending a Simple Message example

我的资源是

<resource methods="GET" uri-template="/wholesales/{dlrcode}">
        <inSequence>
            <log level="custom">
                <property name="message" value="Whole sales request"/>
                <property expression="/default/expression" name="p_dlrCode"/>
            </log>
            <send>
                <endpoint key="WholeSales"/>
            </send>
        </inSequence>
        <outSequence>
            <send/>
        </outSequence>
        <faultSequence/>
    </resource>

我的END是

<?xml version="1.0" encoding="UTF-8"?>
<endpoint name="WholeSales" xmlns="http://ws.apache.org/ns/synapse">
    <http method="get" uri-template="http://xxx/RDRSvc//vehicle/wholesaleD/{url.var.dlrcode}?format=json"/>
</endpoint>

但是当我打电话给api时

http://xxx.xx.xx.xx:8280/rdr/wholesales/42103

我看不到dlrCode ???

TID [-1234] [ESB] [2017-03-07 11：08：29,464] INFO {org.apache.synapse.mediators.builtin.LogMediator} - message =＆＃34;整个销售请求＆＃34; ，p_dlrCode =

如何将42103放入属性p_dlrCode？

Answer 1

试试这个：

您的资源：

<resource methods="GET" uri-template="/wholesales/{dlrcode}">
        <inSequence>
            <log level="custom">
                <property name="message" value="Whole sales request"/>
                <property expression="get-property('uri.var.dlrcode')" name="p_dlrCode"/>
            </log>
            <send>
                <endpoint key="WholeSales"/>
            </send>
        </inSequence>
        <outSequence>
            <send/>
        </outSequence>
        <faultSequence/>
    </resource>

您的终端已修复：

<?xml version="1.0" encoding="UTF-8"?>
<endpoint name="WholeSales" xmlns="http://ws.apache.org/ns/synapse">
    <http method="get" uri-template="http://xxx/RDRSvc//vehicle/wholesaleD/{uri.var.dlrcode}?format=json"/>
</endpoint>

WSO2从请求传递值

1 个答案: