<div>
Hi! Do it something like this<br />

<pre><code id="code-369578">            using (System.Diagnostics.Process process = new System.Diagnostics.Process())
            {
                string fileName = @&quot;C:\Program Files\OpenOffice.org 3\program\scalc.exe&quot;; //the path where you're scalc.exe located
                string argument =  @&quot;C:\yourCSVFile.csv&quot;; //you may change the path 
                process.StartInfo.UseShellExecute = false;
                process.StartInfo.WindowStyle = System.Diagnostics.ProcessWindowStyle.Normal;
                process.StartInfo.FileName = fileName;
                process.StartInfo.Arguments = argument;
            
</code></pre>
</div>


Hi! Do it something like this


            using (System.Diagnostics.Process process = new System.Diagnostics.Process())
            {
                string fileName = @"C:\Program Files\OpenOffice.org 3\program\scalc.exe"; //the path where you're scalc.exe located
                string argument =  @"C:\yourCSVFile.csv"; //you may change the path 
                process.StartInfo.UseShellExecute = false;
                process.StartInfo.WindowStyle = System.Diagnostics.ProcessWindowStyle.Normal;
                process.StartInfo.FileName = fileName;
                process.StartInfo.Arguments = argument;
            


<div>
<div class="content wysiwyg-content">
Hello,<br />
<br />
I want to call the scalc.exe and pass the path+filename to it. That does not work.<br />
<br />
ProcessStartInfo proc = new ProcessStartInfo();<br />
proc.Filename = &quot;scalc.exe 'myPathToCSVFile'; OR &quot;soffice -calc 'myPathToCSVFile<br />
Process.Start(proc);<br />
<br />
does not work... :/<br />
<br />
Can someone tell me please what is the proper commandline parameter to open a .csv File with scalc.exe ? The OO help is not very good...<br />

</div>
</div>


Hello,

I want to call the scalc.exe and pass the path+filename to it. That does not work.

ProcessStartInfo proc = new ProcessStartInfo();
proc.Filename = "scalc.exe 'myPathToCSVFile'; OR "soffice -calc 'myPathToCSVFile
Process.Start(proc);

does not work... :/

Can someone tell me please what is the proper commandline parameter to open a .csv File with scalc.exe ? The OO help is not very good...


Open a .CSV file with OpenOffice` scalc.exe

C# is an object-oriented programming language created in conjunction with Microsoft’s .NET framework. Compilation is usually done into the Microsoft Intermediate Language (MSIL), which is then JIT-compiled to native code (and cached) during execution in the Common Language Runtime (CLR).