Apache ant：从文件中读取值并将其分配给属性的值

Question

我想用文本文件中保存的内容设置替换过滤器的值，请帮助

Answer 1

ant property task可用于读取属性文件

例如

var uri = new Uri("https://stackoverflow.com/yo?oldKey=oldValue");

// Test with a string/string dictionary.
var q = uri.WithQuery(new Dictionary<string, string>
{
    ["k1"] = string.Empty,
    ["k2"] = null,
    ["k3"] = "v3"
});

Debug.Assert(q == new Uri(
    "https://stackoverflow.com/yo?k1&k2&k3=v3"));

// Test with a string/object dictionary.
q = uri.WithQuery(new Dictionary<string, object>
{
    ["k1"] = "v1",
    ["k2"] = new[] { "v2a", "v2b" },
    ["k3"] = null
});

Debug.Assert(q == new Uri(
    "https://stackoverflow.com/yo?k1=v1&k2=v2a&k2=v2b&k3"));

// Test with a name/value collection.
var nvc = new NameValueCollection()
{
    ["k1"] = string.Empty,
    ["k2"] = "v2a"
};

nvc.Add("k2", "v2b");

q = uri.WithQuery(nvc);
Debug.Assert(q == new Uri(
    "https://stackoverflow.com/yo?k1&k2=v2a&k2=v2b"));

// Test with any dictionary.
q = uri.WithQuery(new Dictionary<int, HashSet<string>>
{
    [1] = new HashSet<string> { "v1" },
    [2] = new HashSet<string> { "v2a", "v2b" },
    [3] = null
});

Debug.Assert(q == new Uri(
    "https://stackoverflow.com/yo?1=v1&2=v2a&2=v2b&3"));

// Test with an anonymous object.
q = uri.WithQuery(new
{
    k1 = "v1",
    k2 = new[] { "v2a", "v2b" },
    k3 = new List<string> { "v3" },
    k4 = true,
    k5 = null as Queue<string>
});

Debug.Assert(q == new Uri(
    "https://stackoverflow.com/yo?k1=v1&k2=v2a&k2=v2b&k3=v3&k4=True&k5"));

// Keep existing query using a name/value collection.
nvc = HttpUtility.ParseQueryString(uri.Query);
nvc.Add("newKey", "newValue");

q = uri.WithQuery(nvc);
Debug.Assert(q == new Uri(
    "https://stackoverflow.com/yo?newKey=newValue&oldKey=oldValue"));

// Merge two query objects using the RouteValueDictionary.
var an1 = new { k1 = "v1" };
var an2 = new { k2 = "v2" };

q = uri.WithQuery(
    new RouteValueDictionary(an1).Concat(
        new RouteValueDictionary(an2)));

Debug.Assert(q == new Uri(
    "https://stackoverflow.com/yo?k1=v1&k2=v2"));