[英]Format a label in a gridview to hide first 5 digits of a social security number
I have a data-bound label inside a template field of a griview that displays Social Security Numbers. 我在显示社会保险号的griview的模板字段内有一个数据绑定标签。 It currently shows the SSN's as such, 123-45-6789. 当前显示的是SSN,即123-45-6789。 I need to hide the first 5 digits so it show's like ###-##-6789. 我需要隐藏前5位数字,使其显示为###-##-6789。 Does anyone have an example of how I might accomplish this. 有没有人举我如何做到这一点的例子。
Here is the mark-up for the template fied: 这是模板字段的标记:
<asp:TemplateField HeaderText="Social Security Number" SortExpression="SSN">
<ItemTemplate>
<asp:Label ID="lblSSN" runat="server" Text='<%# Bind("SSN") %>'></asp:Label>
</ItemTemplate>
</asp:TemplateField>
Code block that adds decrypted SSN to grid: 将解密的SSN添加到网格的代码块:
table.Columns.Add("SSN");
foreach (DataRow row in table.Rows)
{
row["SSN"] = DecryptSSN(row["EncryptedSSN"].ToString());
row.AcceptChanges();
}
I'm new to programming so any help would be greatly appreciated. 我是编程新手,所以将不胜感激。
Here is a helper method. 这是一个辅助方法。
public static string GetMaskedNumber(string number)
{
if (String.IsNullOrEmpty(number))
return string.Empty;
if (number.Length <= 5)
return number;
string last5 = number.Substring(number.Length - 5, 5);
var maskedChars = new StringBuilder();
for (int i = 0; i < number.Length - 5; i++)
{
maskedChars.Append(number[i] == '-' ? "-" : "#");
}
return maskedChars + last5;
}
123-123-1234 to ###-##-1234 123-123-1234至###-##-1234
Use this and replace row["SSN"]
the code below you can use to test it's functionality so something like this would work if the value were "123-45-6789"
使用此代码并替换row["SSN"]
下面的代码即可用来测试其功能,因此,如果值是"123-45-6789"
则类似的代码就可以工作
foreach (DataRow row in table.Rows)
{
var newMaskedSSN = DecryptSSN(row["EncryptedSSN"].ToString().Replace("-", string.Empty));
if (newMaskedSSN.Length > 4)
{
Console.WriteLine(
string.Concat(
"".PadLeft(9, '*'),
newMaskedSSN.Substring(newMaskedSSN.Length - 4)));
row["SSN"] = newMaskedSSN;
row.AcceptChanges();
}
}
results: *****6789
这个怎么样?
select '###-##-'+right(ssn_coumn,4) as ssn from your_table
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