本文介绍了过滤datagridview时如何将数字转换为字符串的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
当我使用员工编号而不是其他内容搜索时,我发现这是错误,即名称或电子邮件地址'Employee_ID'运算符后缺少操作数?请协助提供过滤gridview的正确指南
Hi I get this is Error when I search with employee number instead of something else i.e name or email address Missing operand after 'Employee_ID ' operator? Please assist in providing guideline for the rightway of filtering gridview
private void txtSearch_KeyUp(object sender, KeyEventArgs e)
{
string outputInfo = "";
string[] keyWords = txtSearch.Text.Split(' ');
try {
foreach (string word in keyWords)
{
if (outputInfo.Length == 0)
{
/*Convert Employee number to string if search text is numbers*/
outputInfo = "(Firstname LIKE '%" + word + "%' OR LastName LIKE '%" +
word + "%' OR Emailaddress LIKE '%" + word + "%' OR GID LIKE '%" + word + "%' OR '" + string.Format("'Employee_ID LIKE '%" + word + "%')");
}
else
{
/*Convert Employee number to string if search text is numbers*/
outputInfo = "(Firstname LIKE '%" + word + "%' OR LastName LIKE '%" +
word + "%' OR Emailaddress LIKE '%" + word + "%' OR GID LIKE '%" + word + "%' OR '" + string.Format("'Employee_ID LIKE '%" + word.ToString() + "%')");
}
}
//Applies the filter to the DataView
myView.RowFilter = outputInfo.ToString();
}catch(Exception ex){
MessageBox.Show("Result " + ex.Message.ToString() + " Not Found");
}
}
我是什么尝试过:
我试图将Employee_ID转换为字符串,不确定我是否正确使用
What I have tried:
I have tried to convert Employee_ID to string not sure if i am doing it right
outputInfo = "(Firstname LIKE '%" + word + "%' OR LastName LIKE '%" +
word + "%' OR Emailaddress LIKE '%" + word + "%' OR GID LIKE '%" + word + "%' OR '" + string.Format("'Employee_ID LIKE '%" + word + "%')");
推荐答案
OR '" + string.Format("'Employee_ID
在<$ c之后看起来像额外的单引号$ c> OR 之前 Employee_ID
。
/*Convert Employee_ID column to string before using like */
outputInfo = "(Firstname LIKE '%" + word + "%' OR LastName LIKE '%" + word + "%' OR Emailaddress LIKE '%" + word + "%' OR GID LIKE '%" + word + "%' OR "+ string.Format("CONVERT(Employee_ID,'System.String')")+" LIKE '%" + word + "%')";
这很好用。我希望有人发现它有用。感谢所有人提前获得帮助
That works just fine. I hope someone finds it useful. Thanks all in advance for assistance
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