如何增加计数值 [英] How do I increment count value
问题描述
protected void LinkButton1_Click(object sender, EventArgs e)
{
count = 0;
string s = Session["r"].ToString();
string s2 = Session["r2"].ToString();
string s3 = Session["r3"].ToString();
string s4 = Session["r4"].ToString();
string t="Select * from Test_paper Where Paper_id='8'";
DataTable dt3 = new DataTable();
dt3 = c2.bindgrid(t);
if (dt3.Rows[0][9].ToString() == s)
{
count = count + 1;
}
else
{
count = count + 0;
}
if (dt3.Rows[0][15].ToString() == s2)
{
count = count + 1;
}
else
{
count = count + 0;
}
if (dt3.Rows[0][20].ToString() == s3)
{
count = count + 1;
}
else
{
count = count + 0;
}
if (dt3.Rows[0][25].ToString() == s4)
{
count = count + 1;
}
else
{
count = count + 0;
}
Label2.Text =count.ToString();
}
推荐答案
我不确定我的建议是否以任何方式解决了您的问题,但检查是否这些有帮助。首先,如果你想让一个计数器在任何一个匹配上递增,那么为什么不呢?
I am not sure whether my suggestions any way solves your problem or not but check if these helps. First of all, if you want a single counter to be incremented on any of the match then why not this-
if ((dt3.Rows[0][9].ToString()) == s || (dt3.Rows[0][15].ToString() == s2) || (dt3.Rows[0][20].ToString() == s3) || (dt3.Rows[0][25].ToString() == s4))
{
count ++; //equivalent to count=count+1;
}
这段代码与你的4组if..else语句的结果相同。
其次,通过索引从数据表列访问值不是一个好习惯。如果您通过列名来访问它们会更具可读性且不易出错,例如
This piece of code yields same result as that of your 4 sets of if..else statements.
Secondly, it is not a good practice to access value from data table columns by index. It would be more readable and less error-prone if you access them by their column names like
//dt3.Rows[0][20].ToString() --bad idea
dt3.Rows[0]["columnName"].ToString();
此外,您还没有检查数据表是否包含任何行,这可能会导致异常。确保在所有这些操作之前进行了检查。
Also, you haven't checked if the data table contains any rows or not, which may lead to exception. Make sure that you have put check before all these operations.
if(dt3.Rows.Count>0)
{
// write rest of the logic here
}
这些只是建议更好的代码的建议,可能无法解决您的问题,但可以帮助您比以前更轻松地识别您的问题。
These are just suggestion for building better code and may not solve your problem but can help you identify your problem easier than earlier.
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