Question

我正在编写一个程序来访问表单2中的表单1的数组列表。在以下程序中我可以访问它但在表单2中我正在访问的数组列表（表单1）显示为空白。可能是这个原因？ form1的程序如下：

public partial class Form1 : Form
    {
        public  ArrayList hop = new ArrayList();

        public Form1()
        {
            InitializeComponent();
        }      

        private void button1_Click(object sender, EventArgs e)
        {
            hop.Add("2016");
            hop.Add("2015");
            Form2 f = new Form2();
            f.checkedListBox2.Text = this.textBox1.Text;
            f.Show();
        }
    }

表格2的

如下：

 public partial class Form2 : Form
    {
        ArrayList hop2 = new ArrayList();
        public Form2()
        {
            InitializeComponent();        
        }
        private void Form2_Load(object sender, EventArgs e)
        {
            hop2.Add("2016");
            Form1 fp = new Form1();
           // fp.hop.Add("kite");
            if (hop2[1] == fp.hop[1])
                MessageBox.Show("equal");
            else
                MessageBox.Show("not equal");
        }
    }

Answer 1

将Z=[{A{1,1} B{1,1} C{1,1}};... {A{2,1} B{2,1} C{2,1}};... {A{3,1} B{3,1} C{3,1}};... {A1{1,1} B1{1,1} C1{1,1}};... {A1{2,1} B1{2,1} C1{2,1}};... {A1{3,1} B1{3,1} C1{3,1}};... {A2{1,1} B2{1,1} C2{1,1}};... {A2{2,1} B2{2,1} C2{2,1}};... {A2{3,1} B2{3,1} C2{3,1}}]; Tstat = cell2table(VV); Tstat.Properties.VariableNames = {'ok1' 'ok2' 'ok3'}; Tstat.Properties.RowNames = {'one' 'two' 'three' 'four' 'five' ... 'six' 'seven' 'eight' 'nine'}; writetable(Tstat, 'TstatOverview.xlsx')传递给Form1构造函数：

Form2

在public partial class Form1 : Form { public ArrayList hop = new ArrayList(); public Form1() { InitializeComponent(); } private void button1_Click(object sender, EventArgs e) { hop.Add("2016"); hop.Add("2015"); Form2 f = new Form2(this); f.checkedListBox2.Text = this.textBox1.Text; f.Show(); } }中获取Form1的实例：

Form2

注意：请注意，没有空值检查或其他任何内容。

Answer 2

使用constructer参数将值传递给form2。

public partial class Form1 : Form
    {
        public  ArrayList hop = new ArrayList();

        public Form1()
        {
            InitializeComponent();
        }      

        private void button1_Click(object sender, EventArgs e)
        {
            hop.Add("2016");
            hop.Add("2015");
            Form2 f = new Form2(hop);
            f.checkedListBox2.Text = this.textBox1.Text;
            f.Show();
        }
    }

Form2代码

public partial class Form2 : Form
    {
        private ArrayList _hopForm1;
        ArrayList hop2 = new ArrayList();
        public Form2(ArrayList hopForm1)
        {
            InitializeComponent();  
            _hopForm1 = hopForm1
        }
        private void Form2_Load(object sender, EventArgs e)
        {
            hop2.Add("2016");
            Form1 fp = new Form1();
           // fp.hop.Add("kite");
            if (hop2[1] == _hopForm1[1])
                MessageBox.Show("equal");
            else
                MessageBox.Show("not equal");
        }
    }

Answer 3

您应该在Form1加载时创建新的Form2。使用Application.OpenForms作为：

private void Form2_Load(object sender, EventArgs e)
 {
        hop2.Add("2016");   
         var f1 = (Form1)Application.OpenForms[0];//<--this references Form1  current instance   
        if (hop2[1] == f1.hop[1])
            MessageBox.Show("equal");
        else
            MessageBox.Show("not equal");
}

Answer 4

ArrayList hop2 = new ArrayList();

应该是

public ArrayList hop2 = new ArrayList();

C＃：通过form2访问form1的arraylist时出错

4 个答案: