我正在使用phonegap上传图像文件。我想读取PHP文件上的选项参数,我该怎么做?
这是我的移动应用程序代码。
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;
var ft = new FileTransfer();
ft.upload(imageURI, encodeURI("http://some.server.com/upload.php"), win, fail, options);
}
这是我的上传.php文件
<?php
print_r($_FILES);
$new_image_name = uniqid().uniqid ().".jpg";
move_uploaded_file($_FILES["file"]["tmp_name"], "image/".$new_image_name);
?>
如何在php文件上读取以下值
params.value1 = "test";
params.value2 = "param";
options.params
应被读取为POST
,这应显示它们:
print_r($_POST);