尝试通过 android 将文件上传到我的网站时出现错误。
一旦我点击照片，我得到这个:
这是我的 Main.java
package test.com.phonegap.app;
import org.apache.cordova.DroidGap;
import android.os.Bundle;
public class Main extends DroidGap {
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
super.loadUrl("file:///android_asset/www/index.html");
}
}
这是我的 HTML 文件，一切都在这里发生等等
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html>
<head>
<title>File Transfer Example</title>
<script type="text/javascript" charset="utf-8" src="cordova-1.5.0.js"></script>
<script type="text/javascript" charset="utf-8">
// Wait for PhoneGap to load
//
document.addEventListener("deviceready", onDeviceReady, false);
// PhoneGap is ready
//
function onDeviceReady() {
}
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://monaiz.byethost7.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>
<button onclick="getImage();">Upload a Photo</button>
</body>
</html>
这是我网站上的 PHP 文件
<?php
print_r($_FILES);
$new_image_name = "namethisimage.jpg";
move_uploaded_file($_FILES["file"]["tmp_name"], "/htdocs/".$new_image_name);
?>
谁能给我一个解决这个问题的方法？谢谢!
最佳答案
从您的 PHP 代码中删除这一行
print_r($_FILES);
如果你想返回文件上传成功与否，你可以做类似的事情
$fileUploaded = move_uploaded_file($_FILES["file"]["tmp_name"], "/htdocs/".$new_image_name);
echo ($fileUploaded) ? 'File uploaded' : 'Upload failed';
关于java - Android通过html和php上传文件到网站，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/12289990/