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inserting multiple record using drop down

Thread began 1/27/2016 1:38 pm by Kumel | Last modified 3/15/2016 1:02 pm by Ray Borduin | 380 views | 8 replies |

Kumel

inserting multiple record using drop down

Hi,
Lets say a product needs to be added & product got multiple categories. so from dropdown he chooses multiple categories.
But this multiple categories i want to be inserted in different table ei: CatProduct
It contains ID, ProductCatID, ProductID

ProductCatID will be category id choosen while adding product. & ProductID is current ProductID

Now when i choose multiple categories i want multiple records to be added in CatProduct table. But it isnt happening!
How can it be achieved?

Current when i test only one record is getting added in CatProduct.

If i choose 4 categories then 4 records must be added in CatProduct.

I choosed "insert multiple record" but isnt accepting it. (dw) even tho repeat region is added in multiple dropdown option!

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Ray BorduinWebAssist

If you set the name of the select list using brackets like: name="ProdCatID[]"

Then multi-selected values will automatically be submitted as an array. That way you can simply apply the insert for the CatProduct table and then manually add a FOR loop around it like:

php:
<?php

if (isset($_POST['ProdCatID'])) {
  for (
$x=0$x<sizeof($_POST['ProdCatID']); $x++)  {
?>
[YOUR INSERT CODE HERE]
<?php
  
}
}
?>


When you bind the column in the insert server behavior, set it to:

php:
<?php echo($_POST['ProdCatID'][$x]); ?>
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Kumel

After making select name as "ProdCatID[]" .

then manually add a FOR loop around it like...?

I didnt understanding this. Does this require single record insert or multiple record inserts? Cos even after changing name i m not able to use multiple record inserts.

& what does [YOUR INSERT CODE HERE] means? which codes?

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Ray BorduinWebAssist

The for loop is in the code I showed you. There is no multiple record insert in mysqli currently. So it uses the single record insert with the manual for loop I've given an example of in the post above.

[Your insert code here] is where you would have the chunk of code you want to be repeated (insert server behavior)

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Kumel

i added this. doesnt work. file attached.

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Kumel

its working.. there was a small change required.. So how about it edit page? will it work the same way?

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Ray BorduinWebAssist

The best bet on the edit page is to delete all of the rows for that record and then re-insert the ones selected with the same code... So you use the same insert code, just add the delete record code before it so that the old selections are effectively cleared out.

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Kumel

delete all rows? reinsert the ones selected?
So delete all rows using hidden elements & have different form or within same new form?
How will we display already selected on select?
Place delete record trigger on submit button & before insert record?

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Ray BorduinWebAssist

You shouldn't need hidden elements to delete all rows... presumably they all share a master id that you can use to delete in one application of the delete server behavior.

Re-insert using the same logic as the insert page.

Re-selecting the previously selected would require:
1) create a recordset to get the selected values
2) loop through it and create an array of selected values
3) Use in_array() to check if the value was selected in the <option> tags

Yes, delete should use the same trigger and be added before the looped insert.

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