如何让PHP和AJAX上传的图像路径​​？ [英] how to get the uploaded image path in php and ajax?

问题描述

我有一个简单的形式：

 ＆LT;输入类型=隐藏名称=MAX_FILE_SIZE值=2097152ID =MAX_FILE_SIZE＆GT;
＆LT;输入类型=文件名称=file_uploadID =file_upload级=picture_main＆GT;
＆LT;输入类型=提交名称=upload_pictureID =upload_pictureVALUE =提交＆GT;
 

我试图以一个ajax上传：

  submit.on（点击，函数（）{
        var文件= $（'＃file_upload）VAL（）。
        uploadImageAjax（文件）;
        返回false;
}）;

VAR uploadmageAjax =功能（文件）
{
    $阿贾克斯（{
        键入：POST，
        网址：/测试/索引/ imageupload
        数据： {
            文件：文件
        }，
        数据类型：JSON，
        成功：功能（数据）{
            的console.log（数据）;
        }
    }）;
}
 

我得到的回复是，例如，文件：C：\ fakepath \ weirdan003-10.jpeg

但林不知道这是什么 fakepath 是！？

如果我要做到这一点在PHP只有我会得到这样的形象：

 如果（使用isset（$ _ POST ['upload_picture']））{
            $数据= $ formImageUpload-＆GT;的GetValues​​（）;
            $ pictureName = $数据['画面'];
....
 

，然后上传。

所以我想弄清楚的是，如果Ajax调用POST的该行动的权利文件，这样我就可以把它上传到磁盘。

将 $（'＃file_upload）VAL（）; 持有$ _FILE ??

什么想法？

感谢

解决方案

  var文件= $（'＃file_upload）VAL（）。
 

这将只返回一个路径，客户端计算机上的文件。对于SECURIT原因，它就像回到C：\ fakepath \ file_name.ext。如果我没有记错，在一些旧的浏览器有可能得到一个真实路径。但尽管如此，它不会帮助您获得服务器上的文件。

有关AJAX风格的上传，你可以使用一些插件，你有建议。或者只是使用 jQuery的形式插件。它会工作非常类似于 $。阿贾克斯。

现在，当文件正确上传，你会发现在 $这一切需要的信息_文件

在你的情况下，它会像 $ _ FILES ['file_upload'] 其中file_upload'是你的文件输入一个名称。

现在你可以从临时存储移动与 move_uploaded_file 。做任何你想要的文件

编辑： 我看你是使用Zend。看看<一href="http://ahsangill.word$p$pss.com/2009/02/17/zend-framework-file-upload-using-zend_form_element_file/"相对=nofollow>这个有关如何使用的FileUpload元素在服务器端运行。与Zend您可以使用文件上传方法而不是move_uploaded_file

i have a simple form:

<input type="hidden" name="MAX_FILE_SIZE" value="2097152" id="MAX_FILE_SIZE">
<input type="file" name="file_upload" id="file_upload" class="picture_main">
<input type="submit" name="upload_picture" id="upload_picture" value="Submit">

i am trying to to an ajax upload:

submit.on("click", function(){
        var file = $('#file_upload').val();
        uploadImageAjax(file);
        return false;
});

var uploadmageAjax = function(file)
{
    $.ajax({
        type: "POST",
        url: "/test/index/imageupload",
        data: {
            'file': file
        },
        dataType: "json",
        success: function (data) {
            console.log(data);
        }
    });
}

what i get back is, for example, file: "C:\fakepath\weirdan003-10.jpeg"

but im not sure what that fakepath is !?

if i were to do it in php only i would get the image like this:

if (isset($_POST['upload_picture']) ) {
            $data = $formImageUpload->getValues();
            $pictureName = $data['picture'];
....

and then upload it.

So what i want to figure out is if the ajax call POST's to that action the right file so i can then upload it to the disk.

will $('#file_upload').val(); hold the $_FILE??

any ideas?

thanks

解决方案

var file = $('#file_upload').val();

This will return only a path to file on client machine. for securit reason it is returned like c:\fakepath\file_name.ext. If I remember correctly, in some older browsers it was possible to get a real path. But still, it does not helps you to get a file on server.

For ajax style upload you can use some of plugins you got recommended. Or just use jQuery Forms plugin. It will work very similar to $.ajax.

Now, when file is uploaded correctly, you will find all required info about it in $_FILES

In your case it will be something like $_FILES['file_upload'] where 'file_upload' is a name of your file input.

Now you can move it from temporary storage with move_uploaded_file. And do whatever you want with that file

EDIT: And I see you are using Zend. Take a look at this about how to work with FileUpload element on the server side. With Zend you may use FileUpload methods instead of move_uploaded_file

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