Question

我现在正在制作一个混合应用程序。然后，实现在设备库中上传图像并不顺利。

使用通用FTP进行上传，错误消除了PHP端。

[剧本来源]

＆＃13;

var pictureSource;  
var destinationType;  
document.addEventListener("deviceready",onDeviceReady,false);

function onDeviceReady() {
    pictureSource=navigator.camera.PictureSourceType;
    destinationType=navigator.camera.DestinationType;
}
function onPhotoDataSuccess(imageData) {
  var smallImage = document.getElementById('smallImage');
  smallImage.style.display = 'block';
  smallImage.src = "data:image/jpeg;base64," + imageData;
}
function onPhotoURISuccess(imageURI) {
  var largeImage = document.getElementById('largeImage');
  largeImage.style.display = 'block';
  largeImage.src = imageURI;
}
function capturePhoto() {
  navigator.camera.getPicture(onPhotoDataSuccess, onFail, { quality: 90,
    correctOrientation : true,
    saveToPhotoAlbum : true,
    destinationType: destinationType.DATA_URL });
}
function capturePhotoEdit() {
   navigator.camera.getPicture(onPhotoDataSuccess, onFail, { quality: 90, 
    allowEdit: true,
    correctOrientation : true,
    saveToPhotoAlbum : true,
    destinationType: destinationType.DATA_URL });
}
function getPhoto(source) {
//  navigator.camera.getPicture(onPhotoURISuccess, onFail, { quality: 90,
	navigator.camera.getPicture(uploadPhoto, onFail, { quality: 90,
    correctOrientation : true, 
    destinationType: destinationType.FILE_URI,
    sourceType: source });
}
function onFail(message) {
  alert('Failed because: ' + message);
}

function uploadPhoto(imageURI) {
    var options = new FileUploadOptions();
    options.chunkedMode = false;
    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;

    var fileTransfer = new FileTransfer();
    fileTransfer.upload(
		imageURI, 
		"http://192.168.0.26/android/upload.php", 
		win, 
		fail, 
		options
	);
}

function win(r){
	console.log("Code = " + r.responseCode.toString()+"\n");
    console.log("Response = " + r.response.toString()+"\n");
    console.log("Sent = " + r.bytesSent.toString()+"\n");
}

function fail(error){
	alert("error :Code = " + error.code);
}

＆＃13;

[PHP来源]

＆＃13;

<?php
	print_r($_FILES);
	move_uploaded_file($_FILES["file"]["tmp_name"], 'C:/Users/MyPC/Desktop/');
?>

＆＃13;

当它使用路径的原因时，希望保存在桌面中。如果在图库中选择了照片并且实施了上传，则会出现一个无法形成的错误，并且不会发生某些错误。

Logcat Description 1

Logcat Description 2

明确将类型放入选项并发送。但Type值为空白值。

只有文件名（=图像URI）似乎正确传递。

如果与PHP端问题相同，你如何修复，因为我看到了？

供您参考，upload.php的路径是＆＃34; C：\ Apache24 \ htdocs \ android＆＃34;。

我希望能解决这个问题。感谢。

如何将照片从cordova上传到服务器？

0 个答案: