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Home/ Questions/Q 7931523
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Editorial Team
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Editorial Team
Asked: June 3, 20262026-06-03T20:44:43+00:00 2026-06-03T20:44:43+00:00

I have an AJAX request that is posting to the URL http://localhost:8080/spellchecker/google-spellchecker.do which is

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I have an AJAX request that is posting to the URL http://localhost:8080/spellchecker/google-spellchecker.do which is mapped below. Any idea why this is throwing a 404 error? I have a System.out.println("=========="); in GoogleSpellChekerServlet.service() and it doesn’t get called.

<!-- Setup Spring ServletWrappers -->
<bean id="urlMapping" class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
    <property name="mappings">
        <props>
            <prop key="/spellchecker/google-spellchecker.do">googleSpellChekerWrappingController</prop>
        </props>
    </property>
</bean>

<bean id="googleSpellChekerWrappingController" class="org.springframework.web.servlet.mvc.ServletWrappingController">
    <property name="servletClass">
        <value>org.tinymce.spellchecker.GoogleSpellChekerServlet</value>
    </property>
</bean>

Here’s a screenshot of the ajax request…

Screenshot of ajax request

web.xml

<servlet>
    <servlet-name>dispatcher</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <load-on-startup>1</load-on-startup>
</servlet>

<servlet-mapping>
    <servlet-name>dispatcher</servlet-name>
    <url-pattern>*.html</url-pattern>
    <url-pattern>*.json</url-pattern>
</servlet-mapping>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-03T20:44:44+00:00Added an answer on June 3, 2026 at 8:44 pm

    Your web.xml isn’t mapping *.do to the servlet.

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