Question

目前，在我的应用程序中，我有一个屏幕，用户点击“发送”并将一些数据发送到服务器（wcf服务）。

现在，我想添加一项功能：除了实际发送的数据外，用户还必须拍摄照片并将其与其他参数一起发送。

我添加了插件（相机）并创建了模板。

function capturePhoto() {
    alert("1");
    navigator.camera.getPicture(onPhotoDataSuccess, onFail, {
        quality: 50,
        destinationType: navigator.camera.DestinationType.FILE_URI,
        //destinationType: navigator.camera.DestinationType.NATIVE_URI,
        //sourceType: sourceType.CAMERA
    });
}

function onPhotoDataSuccess(imageData) {
    alert("S");
    var smallImage = document.getElementById('myImage');
    smallImage.style.display = 'block';
    smallImage.src = "data:image/jpeg;base64," + imageData;
    image = "data:image/jpeg;base64," + imageData;
    alert("Image = " + image);
}

function onFail(message) {
    alert('Failed because: ' + message);
}

我确定我必须在onSuccess函数中管理一些东西，但我怎样才能达到目标？拍摄照片后，我必须将照片添加到此处作为参数：

$.ajax({
    type: "POST",
    contentType: "application/json",
    dataType: "json",
    url: url,
    data: JSON.stringify({ "data": date_str, "PICTUREHERE":pictureHere }),
    success: function (output) {
          //my stuff
            },
    error: function (XMLHttpRequest, textStatus, errorThrown) {
          //my error stuff

     }
 });

我的deviceReady功能是：

$(document).on("deviceready", function () {
    navigator.splashscreen.hide();
});

我认为在服务器端，“pictureHere”参数必须是byte []，但我不知道如何管理app端。

更新：我的config.xml

<widget id="com.devexpress.apptemplate" version="1.0" versionCode="1">
  <name>ApplicationTemplate</name>
  <description>Template</description>
  <author email="info@info.com" href="http://www.info.com/">info</author>
  <preference name="permissions" value="none" />
  <preference name="prerendered-icon" value="true" />
  <preference name="android-windowSoftInputMode" value="adjustPan" />
  <!--<preference name="phonegap-version" value="cli-7.0.1" />-->

  <!--<preference name="SplashScreen" value="splash" />-->
  <preference name="SplashScreenDelay" value="60000" />
  <preference name="AutoHideSplashScreen" value="false" />
  <preference name="SplashShowOnlyFirstTime" value="false" />
  <preference name="FadeSplashScreen" value="false" />
  <preference name="ShowSplashScreenSpinner" value="false" />
  <preference name="DisallowOverscroll" value="true" />
  <preference name="StatusBarOverlaysWebView" value="false" />
  <preference name="StatusBarBackgroundColor" value="#000000" />
  <preference name="android-minSdkVersion" value="14" />
  <preference name="android-targetSdkVersion" value="22" />
  <!--<plugin name="cordova-plugin-file" />-->
  <!--<plugin name="cordova-plugin-camera" />-->
  <plugin name="cordova-plugin-camera" spec="^2.4.1">
    <variable name="CAMERA_USAGE_DESCRIPTION" value=" " />
    <variable name="PHOTOLIBRARY_USAGE_DESCRIPTION" value=" " />
  </plugin>
  <plugin name="cordova-plugin-splashscreen" onload="true" />
  <plugin name="cordova-plugin-whitelist" />
  <plugin name="cordova-plugin-ios-longpress-fix" />
  <plugin name="cordova-plugin-statusbar" onload="true" />
  <access origin="*" />
</widget>

在我的html页面中，我添加了

<img id="myImage" />

Answer 1

function getImage() {
  // Retrieve image file location from specified source
  navigator.camera.getPicture(function(pictureHere){
    $.ajax({
      type: "POST",
      contentType: "application/json",
      dataType: "json",
      url: url,
      data: JSON.stringify({ "data": date_str, "PICTUREHERE":pictureHere}),
      success: function (output) {
        //my stuff
      },
      error: function (XMLHttpRequest, textStatus, errorThrown) {
        //my error stuff
      }
      });
    }, function (message) {
      Alert("Error", "error");
    }, {
      quality: 50,
      destinationType: navigator.camera.DestinationType.FILE_URI,
      sourceType: navigator.camera.PictureSourceType.PHOTOLIBRARY
    }
  );
}

pictureHere 必须是base64字符串

Cordova / phonegap：拍照并将其发送到我的服务器

1 个答案: