Android的PhoneGap的相机和图片上传图片上传、相机、Android、PhoneGap

2023-09-06 13:55:57 作者:旧劫

我一直在努力使这项工作,谷歌搜索,并在这里自上周五以来。 我的最终目标是要能够采取一个标题和描述的每个多张照片,并将其上传到服务器,然后显示在网页上。

I have been trying to make this work, searched google and here since Friday. My ultimate goal is to be able take multiple pictures with a title and description for each and upload them to a server, then display on a web page.

我至今是:能够给一个图像的标题和描述,浏览画廊,发现一个图像,然后选择它。但是当我做了图像与形式上传沿,立即执行。我希望能够利用这一个提交按钮做的。

What I have so far is: the ability to give one image a title and description, browse the gallery, find an image and select it. BUT when I do the image is uploaded along with the form, immediately. I would like to be able to do this using a submit button.

我也有一个按钮来拍摄图像代替,并在网页上显示的图像的preVIEW。但是当我做使用相机拍摄的图像,我不知道如何上传我的表格。我能够打印到屏幕上使用DIV和innerHTML的通话将图像数据......但老实说,我很迷失,甚至不知道从哪里开始张贴code的特定代码片段。因为它目前存在的,现在我会后整个页面....

I also have a button to take an image instead, and a preview of the image on the page appears. BUT when I do take an image with the camera I do not know how to upload my form. I was able to print to the screen the image data using a div and innerHTML call... but honestly i'm so lost and do not even know where to start posting specific snippets of code. I will post the entire page as it currently exists right now....

<html>
<head>
    <title>File Transfer Example</title>

    <script type="text/javascript" charset="utf-8" src="cordova-2.3.0.js"></script>
    <script type="text/javascript" charset="utf-8">

        var pictureSource;   // picture source
        var destinationType; // sets the format of returned value

        // Wait for Cordova to load
        //
        document.addEventListener("deviceready", onDeviceReady, false);

        // Cordova is ready
        //
function onDeviceReady() {
    pictureSource=navigator.camera.PictureSourceType;
    destinationType=navigator.camera.DestinationType;
}


        function browse(){
            navigator.camera.getPicture(uploadPhoto,
                                        function(message) { alert('get picture failed'); },
                                        { quality: 50,
                                        destinationType: navigator.camera.DestinationType.FILE_URI,
                                        sourceType: navigator.camera.PictureSourceType.PHOTOLIBRARY }
                                        );
        }

        function uploadPhoto(imageURI) {
            var options = new FileUploadOptions();
            options.fileKey="file";
            options.fileName=imageURI.substr(imageURI.lastIndexOf('/')+1);
            options.mimeType="image/jpeg";

            var params = {};
            params.value1 = "test";
            params.value2 = document.getElementById('file_name').value + "";
            params.value3 = document.getElementById('file_description').value + "";
            options.params = params;

            var ft = new FileTransfer();
            ft.upload(imageURI, encodeURI("http://site.com/pages/upload.php"), win, fail, options);
        }

        function win(r) {
            console.log("Code = " + r.responseCode);
            console.log("Response = " + r.response);
            console.log("Sent = " + r.bytesSent);
        }

        function onFileSystemSuccess(fileSystem) {
            console.log(fileSystem.name);
        }

        function onResolveSuccess(fileEntry) {
            console.log(fileEntry.name);
        }

        function fail(evt) {
            console.log(evt.target.error.code);
        }


        function fail(error) {
            alert("An error has occurred: Code = " + error.code);
            console.log("upload error source " + error.source);
            console.log("upload error target " + error.target);
        }






function capturePhoto() {
  // Take picture using device camera, allow edit, and retrieve image as base64-encoded string  
  navigator.camera.getPicture(onPhotoDataSuccess, onFail, { quality: 20, allowEdit: true,
    destinationType: destinationType.DATA_URL });
}

function onPhotoDataSuccess(imageData) {
  // console.log(imageData);

  var smallImage = document.getElementById('smallImage');
  smallImage.style.display = 'block';
  smallImage.src = "data:image/jpeg;base64," + imageData;

  var smallTEXT = document.getElementById('smallTEXT');
  smallTEXT.style.display = 'block';
  smallTEXT.innerHTML = "data:image/jpeg;base64," + imageData;
}

function onPhotoURISuccess(imageURI) {
  // Uncomment to view the image file URI 
  // console.log(imageURI);

  // Get image handle
  //
  var largeImage = document.getElementById('largeImage');

  // Unhide image elements
  //
  largeImage.style.display = 'block';

  largeImage.src = imageURI;
}
function capturePhoto() {
  // Take picture using device camera and retrieve image as base64-encoded string
  navigator.camera.getPicture(onPhotoDataSuccess, onFail, { quality: 50,
    destinationType: destinationType.DATA_URL });
}



// Called if something bad happens.
// 
function onFail(message) {
  alert('Failed because: ' + message);
}







        </script>
</head>
<body>

    <h1>Example</h1>
    <p>Upload File</p>

    <form name ="filename" id="file_name_form" action="#">
    Title <br><input type="text" name="name" id="file_name" /><br>
    Description <br><textarea type="text" name="description" id="file_description" /></textarea>
    </form>
    <button onclick="capturePhoto();">Use Camera</button> <br>
    <button onclick="browse();">browse gallery</button><br>
    <img style="display:none;width:160px;" id="smallImage" src="" />
    <hr>
    <div id="smallTEXT">ggg</div>
    <button onclick"uploadPhoto();">submit</button>


</body>

推荐答案

根据这样的回答:enter这里链接描述 不能直接使用URI ....

According to this answer: enter link description here You cannot use the URI directly....

不过,这似乎URI可以直接使用...(见我的$ C $低于C)

But, it seems the uri can be used directly... (see my code below)

修改25-7-2013 我得到这个工作有: 称这样的:

Edit 25-7-2013 I got this working with: call like this:

navigator.camera.getPicture(onPhotoUriSuccess, onFailCamera, { quality: 50,
        destinationType: pictDestinationType.FILE_URI });

和更迭:

function onPhotoUriSuccess(imageUriToUpload){
    var url=encodeURI("http://your_url_for_the_post/");

    var username='your_user';
    var password='your_pwd';

    var params = new Object();
    params.your_param_name = "something";  //you can send additional info with the file

    var options = new FileUploadOptions();
    options.fileKey = "the_name_of_the_image_field"; //depends on the api
    options.fileName = imageUriToUpload.substr(imageUriToUpload.lastIndexOf('/')+1);
    options.mimeType = "image/jpeg";
    options.params = params;
    options.chunkedMode = true; //this is important to send both data and files

    var headers={'Authorization':"Basic " + Base64.encode(username + ":" + password)};
    options.headers = headers;

    var ft = new FileTransfer();
    ft.upload(imageUriToUpload, url, succesFileTransfer, errorFileTransfer, options);

}

顺便说一句,我使用的Apache网络服务器上的API的网站,我在这里看到,nginx的可能与分块模式的问题: PhoneGap的chunckedMode真正的上传错误

By the way, I use an apache webserver on the api site, I saw here, nginx could have a problem with the chunked mode: PhoneGap chunckedMode true upload error

 
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