Php script question

[sudan4ever]

I have a php script that uploads multiple images and assign it a value image 1 image 2 on the form drop down menu

so I have an upload for an image and a drop down that assign image1 to that upload

my php script will insert the image to a database I wanted a script that basically inserts to different table on that database based on the image value

so if I have table1 table 2 table 3 ........

I want if image 1 selected from the drop down menu during the upload to be inserted to table 1 and so on

any help will be appreciated

[freshnet]

Maybe start by posting the code you have for assigning 'image1' to these images for a start, but it's definitely possible. If you have a variable you could just use all or part of that variable as the table name, e.g.

Code: Select all

INSERT INTO '$varforimage1'...

or
//look at the last character of the image variable

Code: Select all

$table = 'Table'.substr($image,-1,1);

or
//look at the last character of the image variable

Code: Select all

if (substr($image,-1,1)==1)
{}
else if (substr($image,-1,1)==2)
{}
etc.


That last one is your best option if the correspondence between $image and $table is not regular (e.g. if you have $image3 -> $table5, $image4 -> $table7 etc.)