Question

我无法在远程服务器图像文件夹中找到从phonegap相机api拍摄的上传照片。照片的名称如“23497678900.jpg”保存在我的mysql表中，但实际照片未保存在指定的服务器文件夹中。

以下是我用来捕获，上传和在服务器上的代码：

<!DOCTYPE html>
<html>
 <head>
<title>Submit form</title>

<script type="text/javascript" charset="utf-8" src="cordova.js"></script>
<script type="text/javascript" charset="utf-8">

var pictureSource;   // picture source
var destinationType; // sets the format of returned value

// Wait for device API libraries to load
//
document.addEventListener("deviceready",onDeviceReady,false);

// device APIs are available
//
function onDeviceReady() {
    pictureSource = navigator.camera.PictureSourceType;
    destinationType = navigator.camera.DestinationType;
}


// Called when a photo is successfully retrieved
//
function onPhotoURISuccess(imageURI) {

    // Show the selected image
    var smallImage = document.getElementById('smallImage');
    smallImage.style.display = 'block';
    smallImage.src = imageURI;
}


  // A button will call this function
  //
 function getPhoto() {
  // Retrieve image file location from specified source
  navigator.camera.getPicture(onPhotoURISuccess, onFail, { quality: 50,
    destinationType: destinationType.FILE_URI });
  }

  function uploadPhoto() {

    //selected photo URI is in the src attribute (we set this on getPhoto)
    var imageURI = document.getElementById('smallImage').getAttribute("src");
    if (!imageURI) {
        alert('Please select an image first.');
        return;
    }

    //set upload options
    var options = new FileUploadOptions();
    options.fileKey = "file";
    options.fileName = imageURI.substr(imageURI.lastIndexOf('/')+1);
    options.mimeType = "image/jpeg";
    options.chunkedMode = false;

    options.params = {
        firstname: document.getElementById("firstname").value,
        lastname: document.getElementById("lastname").value
    }


    options.headers = {
      Connection: "close"
    };

    var ft = new FileTransfer();
    ft.upload(imageURI, encodeURI("http://hostname.com/image/upload.php"), win, fail,

 options);
}

// Called if something bad happens.
//
function onFail(message) {
  console.log('Failed because: ' + message);
}

function win(r) {
    console.log("Code = " + r.responseCode);
    console.log("Response = " + r.response);
    alert("Response =" + r.response);
    console.log("Sent = " + r.bytesSent);
}

function fail(error) {
    alert("An error has occurred: Code = " + error.code);
    console.log("upload error source " + error.source);
    console.log("upload error target " + error.target);
}

</script>
</head>
<body>

    <button onclick="getPhoto();">Select Photo:</button><br>
    <img style="display:none;width:60px;height:60px;" id="smallImage" src="" /><br>

<form id="regform">
    First Name: <input type="text" id="firstname" name="firstname"><br>
    Last Name: <input type="text" id="lastname" name="lastname"><br>
    <input type="button" id="btnSubmit" value="Submit" onclick="uploadPhoto();">
</form>
 </body>
</html>

upload.php代码是：

<?php
$server = "localhost";
$username = "username";
$password = "pw";
$database = "mydbase";

 $con = mysqli_connect($server, $username, $password, $database) or die ("Could not

connect: " . mysqli_connect_error());

 $firstName2=$_POST["firstname"];
 $lastName2=$_POST["lastname"];

//storing file in filename variable
 $fileName = $_FILES['file']['name'];

//This is the directory where images will be saved
$target = "image/";
$target = $target . basename( $_FILES['file']['name']);

move_uploaded_file($_FILES['file']['tmp_name'],$target);

$sql = "INSERT INTO postinfo (firstname,lastname,imgfile) ";
$sql .= "VALUES ('$firstName2', '$lastName2', '".basename($_FILES['file']'name'])."')";

if($con->query($sql)){
echo "Your comment has been sent";
}
else{
echo "Error in sending your comment";
}
?>

关于如何将实际图片上传到我服务器的图片文件夹中，我已经没想过了。

Answer 1

这很可能是PHP脚本中带有$target变量的问题。如果您有权访问服务器的根目录，则可能会在/image中找到您的图像。

如果您希望图像与脚本位于同一目录中，则需要更多类似的内容：

$target = dirname(__FILE__) . '/image/';

我找不到从远程服务器图像文件夹中的phonegap相机api拍摄的上传照片

1 个答案: