在对象中上传文件

本文介绍了在对象中上传文件的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我做错了什么？

fileUpload.cfm

<cfcomponent name="fileAttachment" hint="This is the File Attachment Object">

    <cffunction name="uploadFile" access="public" output="no" returntype="string">
        <cfargument name="fileToUpload" type="string" required="no">
        <cfargument name="pDsn" required="no" type="string">
        <cfset var cffile = "">
        <cffile action="upload" destination="D:\apache\htdocs\abc\uploads" filefield="#ARGUMENTS.fileToUpload#" nameconflict="makeunique">
        <cfreturn cffile.clientFile />
    </cffunction>

</cfcomponent>

test_fileUpload.cfm

<form action="fileUpload.cfm" enctype="multipart/form-data" method="post">
    <input type="file" name="fileToUpload"><br/>
    <input type="submit">
</form>

推荐答案

此行：

<cffile action="upload" destination="D:\apache\htdocs\abc\uploads" filefield="#ARGUMENTS.fileToUpload#" nameconflict="makeunique">

filefield属性想要表单字段的名称上传文件。你是在正确的轨道，但不幸的是，这不是＃ARGUMENTS.fileToUpload＃的价值，目前 - 根据你的构造，它持有一个引用

The filefield attribute wants the name of the form field that will hold the uploaded file. You're on the right track, but unfortunately, that's not what the value of #ARGUMENTS.fileToUpload# is, presently--based on your construction, it holds a reference to the actual file itself.

在您的表单中添加一个新的隐藏字段：

Add a new hidden field to your form:

<input type="hidden" name="nameOfField" value="fileToUpload">

然后，将 FORM.nameOfField uploadFile（）方法作为第一个参数。 CFFILE将负责其余的工作。

Then, pass FORM.nameOfField to your uploadFile() method as the first parameter. CFFILE will take care of the rest.

这篇关于在对象中上传文件的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！

uploadfile