I need to return the ID corresponding to a description on a drop down list.
I am displaying recipeDS and when I chose it I want to return the corresponding recipeID.
I do now know how to do this in my asp form.
I am using VBcript.
Thank you.
mgrichen
Last Activity: Aug 19 '14, 10:08 PM
Joined: Feb 14 '13
Location:
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Thanks anyway.
I found the error:
I removed
set myConnection = Nothing
and it is working without error. -
The following is returned by the web browser:
------------------------------------------
New York
Insertion Test: MySQL56 Instance
INSERT INTO TESTE (testeDS) VALUES ('New York')
An error occurred on the server when processing the URL. Please contact the system administrator.
If you are the system administrator please click here to find out more about this error.
--------------------------------------------------...Leave a comment:
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Error on SQL Insert Statement
I am using IIS on Windows 7 Enterprise.
The following lines of code are an asp page with vbscript calling a MySQL database.
...Code:<!--#INCLUDE file="header.asp"--> <html> <% iRegion=request.querystring("fRegion") response.write iRegion & "<br/>" Set myConnection = Server.CreateObject("ADODB.Connection") myConnection.ConnectionStringLast edited by Rabbit; Aug 14 '14, 10:37 PM. Reason: Please use [code] and [/code] tags when posting code or formatted data. -
I am a complete newbie on PHP
Nevertheless i wrote
echo "<td><a href=getPerson. php?identity=$s tr_id>".$row['nameNM']."</a></td>";
and it works.
Thank you for your didactical patience.Leave a comment:
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Here it goes
<a href="getPerson .php?identity=$ str_id">Link to Page</a>
How do I place this inside
echo "<td>" . $row['nameNM'] . "</td>";
The variable $str_id is the personID of the nameNM I want.Leave a comment:
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Exactly. I know how to do it in asp but not in PHP. I have tried several options but none works.Leave a comment:
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Create a dynamic hyperlink inside a cell and send variable to another page
Hi
I have the following piece aof code
...Code:while($row = mysql_fetch_array($event)) { echo "<tr>"; echo "<td>" . $row['surnameNM'] . "</td>"; $str_id = $row['personID']; echo "<td>" . $row['nameNM'] . "</td>"; echo "<td>" . $row['personID']
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