使用Spring2.5将字符串或HTML返回到ajax请求

我正在使用Spring2.5（基于XML的配置）。在我的Spring控制器中，如何将字符串返回到AJAX请求？ 我的ajax代码如下所示：

 jq("#editAdTextSave").click(function() {
        alert( "Handler for .click() called." );
        jq.ajax({
              url: 'bookingDetail.html',
              type: 'POST',
              data: 'action=updateAdText',
              success: function(data) {
                //called when successful
                alert(model);
                //$('#ajaxphp-results').html(data);
              },
              error: function(e) {
                //called when there is an error
                //console.log(e.message);
  }
});
    });

---

Submit上的控制器需要ModelAndView对象返回值。但是我只需要返回字符串。我如何才能这样做。请建议。

您需要在映射方法中使用

@ResponseBody

。使用

@ResponseBody

会将返回的值写入

HttpResponseBody

，并且不会作为视图进行计算。下面是快速演示：

@Controller
public class DemoController {

    @RequestMapping(value="/getString",method = RequestMethod.POST)
    @ResponseBody
    public String getStringData() {
        return "data";
    }    
}

---

如问题中所述，使用了Spring 2.5

@ResponseBody

是在Spring3.0中引入的

@看

---

回到问题上来

在我的Spring控制器中，如何返回字符串

您需要将

OutputStream

作为参数添加到控制器方法中

基于XML的配置应该如下所示

<?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:context="http://www.springframework.org/schema/context"
   xsi:schemaLocation="
        http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
        http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context.xsd">

    <context:component-scan base-package="your.controllers"/>

</beans>

package your.controllers;

@Controller
public class YourController {

    @RequestMapping(value = "/yourController")
    protected void handleRequest(OutputStream outputStream) throws IOException {
        outputStream.write(
            "YOUR DESIRED STRING VALUE".getBytes()
        );
    }
}

<?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:context="http://www.springframework.org/schema/context"
       xsi:schemaLocation="
            http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
            http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context.xsd">

    <bean class="org.springframework.web.servlet.handler.BeanNameUrlHandlerMapping"/>

    <bean name="/yourController" class="your.controllers.YourController"/>

</beans>

package your.controllers;

public class YourController extends AbstractController {

    @Override
    protected ModelAndView handleRequestInternal(HttpServletRequest request, HttpServletResponse response) throws Exception {
        response.getOutputStream().write(("YOUR DESIRED STRING VALUE").getBytes());
        return null;
    }
}

---

最后，如果你不想做任何注释，那么

基于XML的配置应该如下所示

<?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:context="http://www.springframework.org/schema/context"
   xsi:schemaLocation="
        http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
        http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context.xsd">

    <context:component-scan base-package="your.controllers"/>

</beans>

package your.controllers;

@Controller
public class YourController {

    @RequestMapping(value = "/yourController")
    protected void handleRequest(OutputStream outputStream) throws IOException {
        outputStream.write(
            "YOUR DESIRED STRING VALUE".getBytes()
        );
    }
}

<?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:context="http://www.springframework.org/schema/context"
       xsi:schemaLocation="
            http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
            http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context.xsd">

    <bean class="org.springframework.web.servlet.handler.BeanNameUrlHandlerMapping"/>

    <bean name="/yourController" class="your.controllers.YourController"/>

</beans>

package your.controllers;

public class YourController extends AbstractController {

    @Override
    protected ModelAndView handleRequestInternal(HttpServletRequest request, HttpServletResponse response) throws Exception {
        response.getOutputStream().write(("YOUR DESIRED STRING VALUE").getBytes());
        return null;
    }
}

---

我没有使用注释我实现SimpleFormController并使用受保护的ModelAndView onSubmit（HttpServletRequest请求、HttpServletResponse响应、Object命令、BindException错误）抛出异常{method

@ResponseBody

是在spring 3.0中引入的，这里的问题是关于spring 2.5的。