小编Pho*_*ian的帖子

class CoOrds
{
    public int x, y;

    public CoOrds() {
        x = 0;
        y = 0;
    }

    public CoOrds(int x, int y) {
        this.x = x;
        this.y = y;
    }
}

public CoOrds toto() {
    CoOrds B = new CoOrds(3, 2);
    return B;
}

private void result_Click(object sender, EventArgs e) {
    l6.Text = "";
    CoOrds D = new CoOrds();
    D = toto();
    l6.Text = "(" + D.x + "," + D.y + ")";
}

我收到以下错误:

Error 2 Inconsistent accessibility: return …

我有这个错误:

错误1"johny.Form1"不包含关于"Form1_Load的"和没有扩展方法的定义"Form1_Load的"接受类型"johny.Form1"的第一个参数可以找到(是否缺少using指令或程序集引用？)

这是表单设计者的代码:

    // 
    // Form1
    // 
    this.AutoScaleDimensions = new System.Drawing.SizeF(6F, 13F);
    this.AutoScaleMode = System.Windows.Forms.AutoScaleMode.Font;
    this.ClientSize = new System.Drawing.Size(456, 411);
    this.Controls.Add(this.l6);
    this.Controls.Add(this.label1);
    this.Name = "Form1";
    this.Text = "Form1";
    this.Load += new System.EventHandler(this.Form1_Load);
    this.ResumeLayout(false);
    this.PerformLayout();

错误来自这一行:

 this.Load += new System.EventHandler(this.Form1_Load);

可能重复:
如何在c#中搜索数据库中的字符串

我想详细搜索一个名称..这就是搜索按钮的代码

private void button1_Click(object sender, EventArgs e)
{
    string connectionString = Tyres.Properties.Settings.Default.Database1ConnectionString;
    SqlConnection conn = new SqlConnection(connectionString);
    DataTable dt = new DataTable();
    SqlDataAdapter SDA = new SqlDataAdapter("SELECT * FROM table1 where Details like" + textBox1.Text, conn);
    SDA.Fill(dt);
    dataGridView1.DataSource = dt;
}

以及我的应用程序示例 单击此处查看图像http://i45.tinypic.com/35aqa6t.jpg

我收到错误:

An unhandled exception of type 'System.Data.SqlClient.SqlException' occurred in System.Data.dll

附加信息:在"likeelie"附近的预期条件的上下文中指定的非布尔类型的表达式.

我试过这个代码来搜索一个整数(比如Quantity),它对我很有用:

private void button1_Click(object sender, EventArgs e)
{
    string connectionString = Tyres.Properties.Settings.Default.Database1ConnectionString;
    SqlConnection conn = new SqlConnection(connectionString);
    DataTable dt = new DataTable();
    SqlDataAdapter SDA …