[英]Android upload file to website threw html and php
I'm getting a error when trying to upload a file to my website through android. 尝试通过android将文件上传到我的网站时出现错误。
Once I click on the photo i get this: 单击照片后,我得到以下信息:
this is my Main.java 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");
}
}
This is my HTML file Where everything happens etc etc etc 这是我的HTML文件,凡事都发生,等等,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>
And here is my PHP file loacated on my website 这是我在网站上放置的PHP文件
<?php
print_r($_FILES);
$new_image_name = "namethisimage.jpg";
move_uploaded_file($_FILES["file"]["tmp_name"], "/htdocs/".$new_image_name);
?>
Can anyone please give me a solution to this problem? 谁能给我解决这个问题的方法吗? Thanks! 谢谢!
remove this line from your PHP code 从您的PHP代码中删除此行
print_r($_FILES); print_r($ _ FILES);
If you wanted to return if the file upload was successful or not you could do something like 如果您想返回文件上传是否成功,则可以执行以下操作
$fileUploaded = move_uploaded_file($_FILES["file"]["tmp_name"], "/htdocs/".$new_image_name);
echo ($fileUploaded) ? 'File uploaded' : 'Upload failed';
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