春季文件上传混合形式 [英] Spring file upload in a mixed form

问题描述

我已经有了下面的实体：

  @Entity 
 @RooJavaBean 
 @RooToString 
 @RooEntity 
公共类SelniumFile {
 
 @ManyToOne（targetEntity = ShowCase.class）
 @JoinColumn 
私人ShowCase showcase; 
 
 @Lob 
 @Basic（fetch = FetchType.LAZY）
 private byte [] file; 
 
 @NotNull 
私人字符串名称; 
 
 
 
 
 
 $ b 
但我不确定如何在视图/控制器端实现它。我可以自由地将< form：input> 等spring-form标签与< input type = file ...> / code>？
 
 
 
我已经看到了漂亮的，但仍然需要一些帮助，以将其应用于我的具体情况。
 
 




解决方案

更新：我想我的问题是严重的公式化。我想要做的是创建一个spring



我在老的spring文档中找到了一个很好的解释，并将其应用到Spring 3.0的新MVC中。基本上这意味着你需要在你的控制器@InitBinder方法中注册一个PropertyEditor。之后，所有内容都将按照预期运行（只要您已将MultiPartResolver添加到上下文并设置正确的表单编码）。
这里是我的示例：

$ $ $ $ $ $ $ $ c $ @ $ @RequestMapping（/ scriptfile / **）
@Controller
public class ScriptFileController {

//我们需要一个特殊的属性编辑器，知道如何将数据
//从请求绑定到一个byte []
@InitBinder
public void initBinder（WebDataBinder binder）
{
binder.registerCustomEditor（byte []。class，new ByteArrayMultipartFileEditor（））;

$ b @RequestMapping（value =/ scriptfile，method = RequestMethod.POST）
public String create（@Valid ScriptFile scriptFile，BindingResult result，ModelMap modelMap）{
if（scriptFile == null）抛出新的IllegalArgumentException（需要scriptFile）;
if（result.hasErrors（））{
modelMap.addAttribute（scriptFile，scriptFile）;
modelMap.addAttribute（showcases，ShowCase.findAllShowCases（））;
返回scriptfile / create;
}
scriptFile.persist（）;
返回redirect：/ scriptfile /+ scriptFile.getId（）;
}
}





I want to upload a file to my spring 3.0 applicatoin (created with roo).

I already have the following entity:

@Entity
@RooJavaBean
@RooToString
@RooEntity
public class SelniumFile {

    @ManyToOne(targetEntity = ShowCase.class)
    @JoinColumn
    private ShowCase showcase;

    @Lob
    @Basic(fetch = FetchType.LAZY)
    private byte[] file;

    @NotNull
    private String name;
}

But I am not sure how to implement it on the view/controller side. Can I freely mix spring-form tags like <form:input> with normal tags like <input type=file ...>?

I have seen the nice multipart upload section in the MVC-Documentation but still need a little help to apply it to my specific case.

解决方案

Update: I think my question was badly formulated. What I wanted to do is create a spring

I found a very good explanation on how to do it in the older spring documentation and applied it to the new Spring 3.0 MVC. Basically this means you need to register a PropertyEditor in your controllers @InitBinder method. Afterwards everything will behave as expected (provided you have added MultiPartResolver to the context and set the correct form encoding). Here is my sample:

@RequestMapping("/scriptfile/**")
@Controller
public class ScriptFileController {

    //we need a special property-editor that knows how to bind the data
    //from the request to a byte[]
    @InitBinder
    public void initBinder(WebDataBinder binder)
    {
        binder.registerCustomEditor(byte[].class, new ByteArrayMultipartFileEditor());
    }

    @RequestMapping(value = "/scriptfile", method = RequestMethod.POST)    
    public String create(@Valid ScriptFile scriptFile, BindingResult result, ModelMap modelMap) {    
        if (scriptFile == null) throw new IllegalArgumentException("A scriptFile is required");        
        if (result.hasErrors()) {        
            modelMap.addAttribute("scriptFile", scriptFile);            
            modelMap.addAttribute("showcases", ShowCase.findAllShowCases());            
            return "scriptfile/create";            
        }        
        scriptFile.persist();        
        return "redirect:/scriptfile/" + scriptFile.getId();        
    }    
}

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