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How to insert new id or return and insert a duplicate id

Thread began 9/23/2013 2:59 pm by jurinski334466 | Last modified 10/07/2013 11:03 am by Jason Byrnes | 1479 views | 17 replies |

jurinski334466

How to insert new id or return and insert a duplicate id

I have an insert form with two insert server record behaviors in it that uses two tables. Table 1 has a drop down menu showing existing selections and a text field to enter a new value. This field and form work when entering a new value and correctly shows an error when entering a duplicate value. But, how do I get the duplicate id to use to insert into table 2? How do I enter either an existing value or a new value?

I did not find anything in the WA forum or materials about how to do this, but I did find code on the internet using "if...else" around the insert statement; however, I was unsuccessful trying to adapt it to the WA code. I'm sorry, I don't know enough about what I'm doing to figure this out or whether there is a better way to accomplish what I'm trying to do so I'm hoping you can help me out and point me in the right direction. Thank you.

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Jason ByrnesWebAssist

send a copy of the page please.

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jurinski334466

Thank you for looking at this for me.

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jurinski334466

Please let me know if you need any additional files or information. Thanks!

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Jason ByrnesWebAssist

add the following code after the first insert and before the second one:

php:
<?php if((!isset($_POST["newTitle"]) || $_POST["newTitle"] == "") && (isset($_POST['CourseTitleID']) && $_POST['CourseTitleID'] != "")) $_SESSION['WADA_Insert_coursetitle'] = $_POST['CourseTitleID']; ?>
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jurinski334466

Thank you for getting back to me. Unfortunately, I couldn't get this code to work. I still get the error, " Duplicate entry 'A New Course' for key 'coursetitle' ". I tried placing the code between the two inserts (and several other places and hiding other bits of code as well), but no luck.

I was thinking that somehow code could be used to bypass the first insert if a duplicate were found and just return the id of the duplicate. (Similar to how the id of a new insert is returned?), and then use the id of the duplicate in the second insert? Does the error message mean the code is not getting past the first insert? I apologize for not understanding this better. Or maybe I just goofed something up...

I tried working with this code last weekend. I don't know if it could be helpful. I wasn't able to make it work. With my luck, it was probably meant to do something else entirely... :-)

Maybe I have the database or validation set up wrong?

I would appreciate if you could let me know what I am doing wrong and look at this again for me.
Thanks again!

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Jason ByrnesWebAssist

Edit the Course title insert behavior, for the trigger, select the New title form element so a new title will only be inserted if one is enterred

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jurinski334466

Thank you! Success! That worked! I couldn't be more appreciative. I'm impressed with how quick, helpful, and patient you've been. Thank you again!

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jurinski334466

I thought I was in the clear with the insert page working, but I'm having trouble with the update page now. I thought I could adapt the changes made to the insert page to the update page, but I have not had any success using two update server behaviors on one page to update two tables linked by the foreign key.

I tried making the page with a text field, a select list, both a text field and a select list, combining the code from the working coursetitle_update and the coursetitlesdescriptions_update page, but I'm not sure what approach I should take to set up the form, what needs renaming and where, and how to correctly pass the information between the two updates. The information in the forum hasn't contained enough detail for me to get a handle on what I'm doing wrong and how to fix it.

Could you please tell me what steps I need to take to update the parent table and then update the foreign key in the child table to update both tables? I attached a copy of my form. Please let me know if you need more information. Thank you!

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Jason ByrnesWebAssist

Use a select list for the existing course titles, use a recordset to populate the list. and a text box for new course title.

Use an insert record behavior to insert the new course title. For the trigger, click the lightning bolt icon, and select the New Course title text box. This will insert a new record if a new course title is entered, but skip the insert if new course title is left blank.

In the insert record behavior, set the save Inserted ID in Session option to a session named:
NewTitle

after the insert behavior, use the following code to set the CourseTitle form element to use the inserted ID if it exists:
<?php if((isset($_SESSION['NewTitle'])?$_SESSION['NewTitle']:"") != "") $_POST['CourseTitle'] = $_SESSION['NewTitle']; ?>

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