我正进入(状态
开放失败:
EACCES (Permission denied)
在线上 OutputStream myOutput = new FileOutputStream(outFileName);
我检查了根,我试过了android.permission.WRITE_EXTERNAL_STORAGE.
我该如何解决这个问题？
try {
InputStream myInput;
myInput = getAssets().open("XXX.db");
// Path to the just created empty db
String outFileName = "/data/data/XX/databases/"
+ "XXX.db";
// Open the empty db as the output stream
OutputStream myOutput = new FileOutputStream(outFileName);
// Transfer bytes from the inputfile to the outputfile
byte[] buffer = new byte[1024];
int length;
while ((length = myInput.read(buffer)) > 0) {
myOutput.write(buffer, 0, length);
}
// Close the …android android-permissions android-5.0-lollipop android-storage
我试图在Android上使用ffmpeg.so我正在获取工作目录:null环境:null错误.
try {
Process p = Runtime.getRuntime().exec("/data/data/com.example.foo/files/ffmpeg -f image2 -i "
+ Environment.getExternalStorageDirectory().getAbsolutePath() + "/img/b%d.jpg "
+ Environment.getExternalStorageDirectory().getAbsolutePath() + "/DCIM/Camera/a444.mp4");
} catch (IOException e) {
e.printStackTrace();
}
错误日志
04-09 01:50:45.683: I/Adreno-EGL(18393): <qeglDrvAPI_eglInitialize:320>: EGL 1.4 QUALCOMM Build: I0404c4692afb8623f95c43aeb6d5e13ed4b30ddbDate: 11/06/13
04-09 01:50:45.713: D/OpenGLRenderer(18393): Enabling debug mode 0
04-09 01:53:56.487: D/dalvikvm(18551): Trying to load lib /data/app-lib/com.example.sodeneme-1/libffmpeg.so 0x423d7d00
04-09 01:53:56.487: D/dalvikvm(18551): Added shared lib /data/app-lib/com.example.sodeneme-1/libffmpeg.so 0x423d7d00
04-09 01:53:56.487: D/dalvikvm(18551): No JNI_OnLoad found in /data/app-lib/com.example.sodeneme-1/libffmpeg.so 0x423d7d00, skipping init
04-09 01:53:56.537: W/System.err(18551): java.io.IOException: Error running exec(). …我试图在OneToMany关联上映射相同类型的属性.我试图与描述区分,但有点卡在这里.
public class User
{
public virtual int UserId { get; set; }
public virtual string UserName { get; set; }
[Description("From")]
public virtual IList<Message> FromMessageList { get; set; }
[Description("To")]
public virtual IList<Message> ToMessageList { get; set; }
}
public class Message
{
public virtual int MessageId { get; set; }
public virtual string Text { get; set; }
[Description("From")]
public virtual User FromUser { get; set; }
[Description("To")]
public virtual User ToUser { get; set; }
}
public …这就是我想要的
(IList<Foo>)listPropertyInfo.GetValue(item)
这就是我获得Foo类型的方式
listPropertyInfo.GetValue(item).GetType().GenericTypeArguments[0]
这是我尝试但无法成功的原因
Convert.ChangeType(listPropertyInfo.GetValue(item), IList<listPropertyInfo.GetValue(item).GetType().GenericTypeArguments[0]>)
还有这个;
((typeof(IList<>).MakeGenericType(listPropertyInfo.GetValue(item).GetType().GenericTypeArguments.Single())))(listPropertyInfo.GetValue(item))
这是我试图实现的方法
public static void trigger(IList<T> result)
{
foreach (var item in result)
{
foreach (var listPropertyInfo in typeof(T).GetProperties().ToList().FindAll(x => x.PropertyType.Name == typeof(IList<>).Name))
{
trigger((IList<Foo>)listPropertyInfo.GetValue(item));
}
}
}
我有一个应用程序使用sqlite db,首先安装我检查如果数据库文件夹也不存在db文件如果没有调用updateDB函数.但有些情况例如在galaxy note 10.1它给我错误.
负载 ;
this.dhn = DataHelper.getDataHelper(this);
File directory = new File(Environment.getDataDirectory() + File.separator + "data" + File.separator + "XXX" + File.separator + "databases");
if(!directory.exists())
{
directory.mkdirs();
updateDB();
}
try {
androidCheckout = this.dhn.Guid();
if(this.dhn.getSettings("dbVersion") == null || Integer.parseInt(this.dhn.getSettings("dbVersion")) != Version || !this.dhn.isTableExists("UserInfo"))
{
updateDB();
}
}
catch (SQLiteException e)
{
try {
updateDB();
androidCheckout = this.dhn.Guid();
}
catch (SQLiteException e11)
{
ManuelYukle();
}
}
public void updateDB()
{
this.dhn.close();
try {
InputStream myInput;
myInput = getAssets().open("XXX.db");
// …我在使用 SAP .NET Connector 3.0 x64 时遇到此错误。我在 stackoverflow 上找不到任何解决方案，所以我想在这里与其他像我一样遇到同样问题的人分享。
我如何理解如果用户点击了admob广告？ontouch听众没有工作.
Display display = ((WindowManager) this.getSystemService(Context.WINDOW_SERVICE)).getDefaultDisplay();
int width = display.getWidth();
int height = display.getHeight();
Window window = getWindow();
adsLayout = new RelativeLayout(this);
RelativeLayout.LayoutParams lp2 = new RelativeLayout.LayoutParams(//width,height);
RelativeLayout.LayoutParams.WRAP_CONTENT, RelativeLayout.LayoutParams.FILL_PARENT);
adsLayout.setGravity(Gravity.BOTTOM);
adView = new AdView(this, AdSize.BANNER, "XXX");
com.google.ads.AdRequest adRequest1 = new com.google.ads.AdRequest();
adRequest1.addTestDevice(com.google.ads.AdRequest.TEST_EMULATOR); // Emulator
int adwidth = height;
adView.setPadding((width - adwidth) / 2, 0, 0, 0);
adView.loadAd(adRequest1);
adsLayout.addView(adView);
adView.setOnTouchListener( (android.view.View.OnTouchListener) mOnTouchListener );
window.addContentView(adsLayout,lp2);
第二个问题是如何管理广告如果他们在按钮的顶部或触摸的东西？android适用于许多手机,无法测试所有..只是阅读禁止..
我想在代码后面写一个帖子.我有简单的html帖子正在工作但是当我尝试使它成为WebRequest我无法使它工作.
谢谢你的提前时间.
这是工作>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" >
<head>
<title>Untitled Page</title>
</head>
<body>
<form action="http://www.pay-pos.com/test.php" method="post">
<textarea name="datafromuser">
<CC5Request><Name>Mert</Name><Password>123</Password><ClientId>1058</ClientId><IPAddress>213.456.654.456</IPAddress><Adress>asdsa</Adress><OrderId>123</OrderId><Type>Auth</Type>
<Number>4522115422659661</Number><ExpiresAy>01</ExpiresAy><ExpiresYil>13</ExpiresYil><Cvv2Val>123</Cvv2Val><Total>10</Total><Taksit></Taksit><Kdv>18</Kdv><BankaID>1</BankaID>
<TcKimlik>12345678912</TcKimlik></CC5Request>
</textarea>
<input type="submit" value="aaa" />
</form>
</body>
</html>
这不起作用>
String ali = "<CC5Request><Name>Mert</Name><Password>123</Password><ClientId>xxxx</ClientId><IPAddress>213</IPAddress><Adress>asdsa</Adress>" +
"<OrderId>123</OrderId><Type>Auth</Type><Number>1234567891234567</Number><ExpiresAy>01</ExpiresAy><ExpiresYil>13</ExpiresYil><Cvv2Val>123</Cvv2Val>" +
"<Total>10</Total><Taksit></Taksit><Kdv>xx</Kdv><BankaID>1</BankaID><TcKimlik>12345678912</TcKimlik></CC5Request>";
WebRequest req = WebRequest.Create("http://www.pay-pos.com/test.php?datafromuser=" + ali);
req.Credentials = CredentialCache.DefaultCredentials;
req.Method = "POST";
HttpWebResponse loWebResponse = (HttpWebResponse)req.GetResponse();
Encoding enc = Encoding.GetEncoding(1254);
StreamReader loResponseStream =
new StreamReader(loWebResponse.GetResponseStream(), enc);
string lcHtml = loResponseStream.ReadToEnd();
loWebResponse.Close();
loResponseStream.Close();
编辑:实验1
String postData = …我正在尝试创建这个lambda,(x => x.MenuItemId)但我不知道该怎么做.
var item = Expression.Parameter(typeof(MenuItem), "x");
var prop = Expression.Property(item, "MenuItemId");
var lambda = Expression.Lambda<Func<MenuItem, object>>(x => x.MenuItemId);
我有3个具有相同来源的下拉列表2,第3个具有所选列表.如果选中的列表不为null,我应该从第一个中删除选定项,并在第三个中绑定.第二个是静态的,显示原始列表.
但我无法弄清楚如何在一个物体中前进.
错误:foreach语句不能对'object'类型的变量进行操作,因为'object'不包含'GetEnumerator'的公共定义
public void DataSource(object source, object select, string Value = "UId", string Text = "Text")
{
ddlThis.DataSource = source;
ddlThis.DataTextField = Text;
ddlThis.DataValueField = Value;
ddlThis.DataBind();
ddlThisHidden.DataSource = source;
ddlThisHidden.DataTextField = Text;
ddlThisHidden.DataValueField = Value;
ddlThisHidden.DataBind();
if (select != null)
{
ddlOther.DataSource = select;
ddlOther.DataTextField = Text;
ddlOther.DataValueField = Value;
ddlOther.DataBind();
foreach (var item in ddlOther.DataSource)
ddlThis.Items.Remove(item);
}
}
版本2
public void DataSource(IList source, IList select, string Value = "UId", string Text = "Text")
{
ddlThis.DataSource = source; …Caused by: java.lang.NoClassDefFoundError: org.ksoap2.serialization.SoapObject
我需要ksoap 2上的新东西,所以我更新到2.6.5,但我不工作.给了我这个错误.但它的构建也毫无例外.我怀疑这个不确定; 我添加jar后我在库中有ksoap2它也出现在root(src - gen - android 4.0.3 - ksoap2 - assets bla bla)
谢谢你提前的时间.
我正在尝试填充会话,如果它的null,在null检查时得到错误
public static int ReportGroupId()
{
if (System.Web.HttpContext.Current.Session["ReportGroupId"] == null) <ERROR HERE>
System.Web.HttpContext.Current.Session["ReportGroupId"] = BL.Reporting.GET.ReportGroupId();
return Convert.ToInt32(System.Web.HttpContext.Current.Session["ReportGroupId"]);
}
我有一个无法使用的项目System.Linq，所以我尝试创建一个与 linq 完全相同的 cutsomWhere 方法，我该怎么做？
var x = y.ToList().Where(t => t.Title != "Foo");
解决方案1
protected void Page_Load(object sender, EventArgs e)
{
var x = y.ToList().Where<Microsoft.SharePoint.Navigation.SPNavigationNode>(x => x.Title != "Foo");
}
public static class Extensions
{
public static IEnumerable<T> Where<T>(this IEnumerable<T> source, Func<T, bool> filter)
{
foreach (var item in source)
{
if (filter(item))
yield return item;
}
}
}
注意：.NET v 2.0