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在文本框中显示Word文件的内容

textbox ms-word .net c#

Uniquedesign · 技术社区 · 7 年前

我有一个启用了muiltilne选项的文本框。我想在那个文本框中显示特定word文件的内容。我该怎么做?我使用这个代码,但它只是显示文件名。

private void btnOpen_LinkClicked(object sender, LinkLabelLinkClickedEventArgs e)
{
    OpenFileDialog openFileDialog1 = new OpenFileDialog();
    openFileDialog1.Title = "Open Word File";
    openFileDialog1.Filter = "Word Files (*doc)|*docx";   

    if (openFileDialog1.ShowDialog() == DialogResult.OK)
    {
        Microsoft.Office.Interop.Word.Application word = new Microsoft.Office.Interop.Word.Application();
        Microsoft.Office.Interop.Word.Document doc = new Microsoft.Office.Interop.Word.Document();

        object fileName = openFileDialog1.FileName;
        // Define an object to pass to the API for missing parameters
        object missing = System.Type.Missing;
        doc = word.Documents.Open(ref fileName,ref missing, ref missing);

        String read = string.Empty;
        List<string> data = new List<string>();
        for (int i = 0; i < doc.Paragraphs.Count; i++)
        {
            string temp = doc.Paragraphs[i + 1].Range.Text.Trim();
            if (temp != string.Empty)
            data.Add(temp);
        }
        doc.Close();
        word.Quit();   
        txtTxt.Text = data.ToString();

    } 
}

这是C#中的windows窗体应用程序。

请帮忙!

2 回复 | 直到 7 年前

Mohamed Najiullah 7 年前

您正在使用该行将数据添加到文本框中

txtTxt.Text = data.ToString();

data 是字符串列表。这样分配它是行不通的,除非您使。ToString()方法。

遍历列表并将其添加到文本框。像这样的

foreach(var item in data)
{
    txtTxt.Text += item;
}

Uniquedesign 7 年前

找到解决方案。只需要在末尾添加一个循环。感谢“Mohamed Najuillah”。

private void btnOpen_LinkClicked(object sender, LinkLabelLinkClickedEventArgs e)
{
    OpenFileDialog openFileDialog1 = new OpenFileDialog();
    openFileDialog1.Title = "Open Word File";
    openFileDialog1.Filter = "Word Files (*doc)|*docx";   

    if (openFileDialog1.ShowDialog() == DialogResult.OK)
    {
        Microsoft.Office.Interop.Word.Application word = new Microsoft.Office.Interop.Word.Application();
        Microsoft.Office.Interop.Word.Document doc = new Microsoft.Office.Interop.Word.Document();

        object fileName = openFileDialog1.FileName;
        // Define an object to pass to the API for missing parameters
        object missing = System.Type.Missing;
        doc = word.Documents.Open(ref fileName,ref missing, ref missing);

        String read = string.Empty;
        List<string> data = new List<string>();
        for (int i = 0; i < doc.Paragraphs.Count; i++)
        {
            string temp = doc.Paragraphs[i + 1].Range.Text.Trim();
            if (temp != string.Empty)
            data.Add(temp);
        }
        doc.Close();
        word.Quit();   
       foreach(var item in data)
{
    txtTxt.Text += item;
}

    } 
}