Question

Phonegap和相机插件出了问题。我正在尝试在我的视图中显示拍摄的照片，但它不起作用。我可以使用DATA_URL，但我需要它与FILE_URI一起工作（因为他们建议！）

这是我的代码，但我想它不在代码中，因为使用DATA_URL它可以工作，但是FILE_URI却没有，尽管我的代码与文档完全相同。

$('#uploadBusinessCard').on('click', function() {
    console.log("capture this foto!");
    navigator.camera.getPicture(onSuccess, onFail, {
        quality: 10,
        destinationType: Camera.DestinationType.FILE_URI
    });
});

function onSuccess(imageURI) {
    console.log("on success!");
    var image = document.getElementById('myImageIdee');
    image.src = imageURI;
}

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

有效的代码几乎相同：

$('#uploadBusinessCard').on('click', function() {
    console.log("capture this foto!");
    navigator.camera.getPicture(onSuccess, onFail, {
        quality: 10,
        destinationType: Camera.DestinationType.DATA_URL
    });
});

function onSuccess(imageData) {
    var image = document.getElementById('myImageIdee');
    image.src = "data:image/jpeg;base64," + imageData;
}

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

Answer 1

试试这个。

$('#uploadBusinessCard').on('click', function() {
    console.log("capture this foto!");
    navigator.camera.getPicture(onSuccess, onFail, {
        quality: 10,
        destinationType: Camera.DestinationType.FILE_URI,
        saveToPhotoAlbum: true
    });
});

function onSuccess(imageURI) {
    console.log("on success!");
    var image = document.getElementById('myImageIdee');
    image.style.display = 'block';
    image.src = imageURI;
}

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

Answer 2

如果你在ipad上使用phonegap桌面应用程序和phonegap开发者应用程序，那么如果你使用FILE_URI就会出现一个没有显示图像的knonw问题

https://github.com/phonegap/phonegap-app-developer/issues/174

https://github.com/phonegap/phonegap-app-developer/issues/248

Phonegap FILE_URI，图片来源

2 个答案: