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我想通過形象,我參加了我的移動設備要保存在我的服務器傳輸圖像時。服務器端用php編碼。 這裏是javascript代碼:得到錯誤與PhoneGap的到PHP
var options = new FileUploadOptions();
options.fileKey = "image";
options.fileName = newPlace.id;
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(cameraImg, "http://www.myserver.com/upload.php", imageUploaded, imageUploadedError, options);
服務器端代碼:
<?php
if(isset($_FILE['image'])) {
echo "good";
$ourFileName = "good.txt";
$ourFileHandle = fopen($ourFileName, 'w') or die("can't open file");
fclose($ourFileHandle);
$file_name = $_FILE['image']['name'];
$file_tmp = $_FILE['image']['tmp_name'];
move_uploaded_file($file_tmp, 'images/'.$file_name);
}
else {
echo "not good";
}
?>
我知道我連接到服務器和PHP代碼在運行,但它沒有趕上了`如果( isset($ _ FILE [ '圖像']))。我知道這是因爲我得到的是說「迴應:不好」的警報。我的錯誤在哪裏?提前致謝!