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getResource()找不到文件?(getResource() not finding file?)

当尝试在Eclipse中加载文件时,文件加载得很好,但是当我使用jar-splice将项目打包到.JAR文件时,似乎应用程序无法再找到其资源文件。

这是运行应用程序时抛出的错误

这是加载文件的方法:

public static File loadFile(String path) throws FileNotFoundException
{
    InputStream stream;

    stream = FileUtil.class.getClassLoader().getResourceAsStream(path);
    System.out.println("Stream = " + stream); //Debug purposes


    File file = new File(FileUtil.class.getClassLoader().getResource(path).getFile());
    if (!file.exists())
    {
        System.err.println("Path: " + FileUtil.class.getClassLoader().getResource(path).getPath()); //Also debug purposes
        throw new FileNotFoundException();
    }

    return file;
}

使用这两个System.out.printlns,很明显应用程序无法根据该路径找到该文件,但如果您查看该图片,那么该路径正是它所查找文件所在的位置。 我很困惑,因为之前从未发生过这种情况,而且它所说的无法找到文件的路径正是它的确切位置。 任何人的想法?


When trying to load the file in Eclipse the file loads just fine, however when I package the project into a .JAR file using jar-splice it seems that the application can no longer locate its resource files.

Here's the error thrown when the application is run

And here is the method that loads files:

public static File loadFile(String path) throws FileNotFoundException
{
    InputStream stream;

    stream = FileUtil.class.getClassLoader().getResourceAsStream(path);
    System.out.println("Stream = " + stream); //Debug purposes


    File file = new File(FileUtil.class.getClassLoader().getResource(path).getFile());
    if (!file.exists())
    {
        System.err.println("Path: " + FileUtil.class.getClassLoader().getResource(path).getPath()); //Also debug purposes
        throw new FileNotFoundException();
    }

    return file;
}

Using those two System.out.printlns it's clear that the application can't find the file based on that path, but if you look at the picture that path is exactly where the file it's looking for is located. I am so confused as this has never happened before, and the path it's saying it can't find the file at is exactly where it is. Any ideas anyone?


原文:https://stackoverflow.com/questions/36731493
更新时间:2023-06-14 08:06

最满意答案

首先在你的js中设置Bloodhound:

var dataSetBloodhound = new Bloodhound({
    datumTokenizer: Bloodhound.tokenizers.obj.whitespace('name'),
    queryTokenizer: Bloodhound.tokenizers.whitespace,
    local: data
});

其中data是数组中的建议列表。
我的例子是

data = [ { name: "Foo", url: "/bar.jpg" }, etc, etc ]

这就是我在Bloodhound.tokenizers.obj.whitespace('name')使用name的原因,因为我希望我的建议成为数据数组中的name

在我的HTML中我有我的输入:

<input id="quick-search-input" type="text" class="form-control" placeholder="Products" data-provide="typeahead"/>
//The important thing here is 'data-provide="typeahead"'

哪个是建议框将作用的输入。

然后在它后面设置js:

$('#quick-search-input').typeahead({
    hint: true,
    highlight: true,
    minLength: 1
},
{
    name : 'NameForFormInput',
    displayKey: 'name',
    templates:
    {
        suggestion: function(data)
        {
            return '<li class="list-group-item">
                    <p class="predictionText">'+data.name+'</p></li>';
        }
    },
    source : dataSetBloodhound
});

我认为这是你出错的地方,因为我在设置时遇到了类似的问题,但在我实现模板时修复了它。 此外,css将根据您的使用情况而有所不同。


First in your js set up the Bloodhound:

var dataSetBloodhound = new Bloodhound({
    datumTokenizer: Bloodhound.tokenizers.obj.whitespace('name'),
    queryTokenizer: Bloodhound.tokenizers.whitespace,
    local: data
});

Where data is the suggestion list in an array.
Mine for example is

data = [ { name: "Foo", url: "/bar.jpg" }, etc, etc ]

this is why I use name in Bloodhound.tokenizers.obj.whitespace('name') because I want my suggestions to be the name in the data array.

In my html I have my input:

<input id="quick-search-input" type="text" class="form-control" placeholder="Products" data-provide="typeahead"/>
//The important thing here is 'data-provide="typeahead"'

Which is the input that the suggestion box will act upon.

Then setting up the js behind it:

$('#quick-search-input').typeahead({
    hint: true,
    highlight: true,
    minLength: 1
},
{
    name : 'NameForFormInput',
    displayKey: 'name',
    templates:
    {
        suggestion: function(data)
        {
            return '<li class="list-group-item">
                    <p class="predictionText">'+data.name+'</p></li>';
        }
    },
    source : dataSetBloodhound
});

I think this is where you are going wrong as I had a similar issue when setting it up but fixed it when I implemented a template. Also the css will be different depending on what your using.

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