2013-04-27 20:24
浏览 27


I am trying to pass image that I took in my mobile device to be saved in my server. the server side is coded in php. here is the javascript code:

var options = new FileUploadOptions();
                        options.fileKey = "image";
                        options.fileName =;
                        var params = new Object();
                        params.value1 = "test";
                        params.value2 = "param";

                        options.params = params;
                        options.chunkedMode = false;
                        var ft = new FileTransfer();
                        ft.upload(cameraImg, "", imageUploaded, imageUploadedError, options);

The server side code is:

    if(isset($_FILE['image'])) {
        echo "good";
            $ourFileName = "good.txt";
        $ourFileHandle = fopen($ourFileName, 'w') or die("can't open file");
        $file_name = $_FILE['image']['name'];
        $file_tmp = $_FILE['image']['tmp_name'];
        move_uploaded_file($file_tmp, 'images/'.$file_name);

    else {
        echo "not good";

I know I am connecting to the server and the php code is running but it is not catching the `if(isset($_FILE['image'])). I know this because I am getting an alert that says "response: not good". Where is my mistake? Thanks in advance!

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