0

因此，我将其设置为通过 Phonegap 本机 api 从设备图像/相机上传图像，并在图像名称上附加随机数。我试图弄清楚如何访问服务器上图像位置的最终 url，以便我可以在制作 Facebook 帖子时利用该图像的 url，因为它们需要外部上传的图像。

代码：.php

<?php
$dir = "upload"; 
$randomImg = rand(1, 9999)."image".rand(1, 999999).".jpg";
if ($_FILES) {
    print_r($_FILES);
    mkdir ($dir, 0777, true); 
    move_uploaded_file($_FILES["file"]["tmp_name"],$dir."/".$randomImg);
}
else if (isset($_GET['image'])) {
    $image = $dir."/".$_GET['image'];
    header('Content-type: image/jpeg');
    list($width, $height) = getimagesize($image); 
    $newWidth = 120.0; 
    $size = $newWidth / $width;
    $newHeight = $height * $size; 
    $resizedImage = imagecreatetruecolor($newWidth, $newHeight); 
    $image = imagecreatefromjpeg($image); 
    imagecopyresampled($resizedImage, $image, 0, 0, 0, 0, $newWidth, $newHeight, $width, $height); 
    imagejpeg($resizedImage, null, 80); 
}
else {
    $images = scandir($dir);
    $ignore = Array(".", "..");
    $baseURI = "http://".$_SERVER['SERVER_NAME'].':'.$_SERVER['SERVER_PORT'].$_SERVER['REQUEST_URI'];
    if ($images) {
        foreach($images as $curimg){ 
            if (!in_array($curimg, $ignore)) {
                echo "Image: ".$curimg."<br>";
                echo "<img src='".$baseURI."?image=".$curimg."&rnd=".uniqid()."'><br>"; 
            }
        }
    }
    else {
        echo "We apologize, but there are no images found on our server at this time.";
    }
}
?>

.html

<section>
                            <input value="Upload Photo" id="uploadBtn" type="button" onclick="uploadImg();"  />
                        </section>

.js

function uploadImg() {
    var img = document.getElementById('camImg');
    var imageURI = img.src;
    if (!imageURI || (img.style.display === "none")) {
        $('#status').html("Snap a photo or select one from your gallery by clicking the 'Attach Photo...' button").css('color', 'red');
        return;
    }
    paramount = "http://www.param0unt.com/app/upload.php";
    if (paramount) {
        var opts = new FileUploadOptions();
        opts.fileKey = "file";
        opts.fileName = imageURI.substr(imageURI.lastIndexOf('/') + 1);
        opts.mimeType = "image/jpeg";
        opts.chunkedMode = false;
        var media = new FileTransfer();
        media.upload(imageURI, paramount, function(r) {
            $('#status').html("Photo successfully stored on our server! (<font color='red'>NOTE:</font> This is alpha version, you will be able to utilize these photos via facebook group ad posts in an upcoming beta release!)").css('color', 'green');
        }, function(error) {
             $('#status').html("Photo Upload Failed, please try again later.").css('color', 'red');
            //console.log(error.code);          
        }, opts);
    }

}

4

1 回答 1

0

使用 上传文件后move_uploaded_file，生成 URL 并将其作为 JSON 编码的响应回显：

$url = $baseURI . "?image=" . $randomImg; // http://example.com/app/upload.php?image=1234image5678.jpg
echo json_encode(array("url" => $url));
exit;

然后在你的成功回调 JS 函数中，解析响应：

var data = JSON.parse(r);
console.log(data.url); // http://example.com/app/upload.php?image=1234image5678.jpg
于 2013-03-21T17:35:22.697 回答