PhoneGap将图片上传到服务器表单提交 [英] PhoneGap upload Image to server on form submit
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问题描述
我在这里面临的问题，因为在phonegap图像上传到服务器一旦你选择一个图片。我不想在提交表单之前上传图片。图像自动上传到服务器，这是我不想要的东西。我想上传图像与窗体，其中窗体包含更多的字段，需要与图像一起发送。提交表单的可能方式是什么？
I am facing problem here as in phonegap image is uploaded to the server once u select a picture.I don't want to upload image before submitting form. Image is uploaded automatically to server which is something i don't want.I want to upload image with the form, where form contains many more fields which is required to send along with image. What are the possible ways to submit with form?
<!DOCTYPE HTML >
<html>
<head>
<title>Registration Form</title>
<script type="text/javascript" charset="utf-8" src="phonegap-1.2.0.js"></script>
<script type="text/javascript" charset="utf-8">
// Wait for PhoneGap to load
document.addEventListener("deviceready", onDeviceReady, false);
// PhoneGap is ready
function onDeviceReady() {
// Do cool things here...
}
function getImage() {
// Retrieve image file location from specified source
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 = new Object();
params.value1 = "test";
params.value2 = "param";
options.params = params;
options.chunkedMode = false;
var ft = new FileTransfer();
ft.upload(imageURI, "http://yourdomain.com/upload.php", win, fail, options);
}
function win(r) {
console.log("Code = " + r.responseCode);
console.log("Response = " + r.response);
console.log("Sent = " + r.bytesSent);
alert(r.response);
}
function fail(error) {
alert("An error has occurred: Code = " = error.code);
}
</script>
</head>
<body>
<form id="regform">
<button onclick="getImage();">select Avatar<button>
<input type="text" id="firstname" name="firstname" />
<input type="text" id="lastname" name="lastname" />
<input type="text" id="workPlace" name="workPlace" class="" />
<input type="submit" id="btnSubmit" value="Submit" />
</form>
</body>
</html>
推荐答案
创建两个可以单独调用的函数。一个功能只是获取图像，另一个功能来上传图像。
Create two functions you can call separately. One function for just getting the image, and another function to upload the image.
您可以执行以下操作。
<!DOCTYPE html>
<html>
<head>
<title>Submit form</title>
<script type="text/javascript" charset="utf-8" src="cordova.js"></script>
<script type="text/javascript" charset="utf-8">
var pictureSource; // picture source
var destinationType; // sets the format of returned value
// Wait for device API libraries to load
//
document.addEventListener("deviceready",onDeviceReady,false);
// device APIs are available
//
function onDeviceReady() {
pictureSource = navigator.camera.PictureSourceType;
destinationType = navigator.camera.DestinationType;
}
// Called when a photo is successfully retrieved
//
function onPhotoURISuccess(imageURI) {
// Show the selected image
var smallImage = document.getElementById('smallImage');
smallImage.style.display = 'block';
smallImage.src = imageURI;
}
// A button will call this function
//
function getPhoto(source) {
// Retrieve image file location from specified source
navigator.camera.getPicture(onPhotoURISuccess, onFail, { quality: 50,
destinationType: destinationType.FILE_URI,
sourceType: source });
}
function uploadPhoto() {
//selected photo URI is in the src attribute (we set this on getPhoto)
var imageURI = document.getElementById('smallImage').getAttribute("src");
if (!imageURI) {
alert('Please select an image first.');
return;
}
//set upload options
var options = new FileUploadOptions();
options.fileKey = "file";
options.fileName = imageURI.substr(imageURI.lastIndexOf('/')+1);
options.mimeType = "image/jpeg";
options.params = {
firstname: document.getElementById("firstname").value,
lastname: document.getElementById("lastname").value,
workplace: document.getElementById("workplace").value
}
var ft = new FileTransfer();
ft.upload(imageURI, encodeURI("http://some.server.com/upload.php"), win, fail, options);
}
// Called if something bad happens.
//
function onFail(message) {
console.log('Failed because: ' + message);
}
function win(r) {
console.log("Code = " + r.responseCode);
console.log("Response = " + r.response);
//alert("Response =" + r.response);
console.log("Sent = " + r.bytesSent);
}
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);
}
</script>
</head>
<body>
<form id="regform">
<button onclick="getPhoto(pictureSource.PHOTOLIBRARY);">Select Photo:</button><br>
<img style="display:none;width:60px;height:60px;" id="smallImage" src="" />
First Name: <input type="text" id="firstname" name="firstname"><br>
Last Name: <input type="text" id="lastname" name="lastname"><br>
Work Place: <input type="text" id="workplace" name="workPlace"><br>
<input type="button" id="btnSubmit" value="Submit" onclick="uploadPhoto();">
</form>
</body>
</html>
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