Editing Drop Down Menu In Php

[apple 0]

suppose i make a dropdown menu having value 1, 2, 3, 4, 5, one option can be selected, and selected option is stored in database.
now i create an edit page, how do i display the selected value in the menu and other values in dropdown,
for example,
Menu is like
Select One
1
2
3
4
5

and i select 3, this 3 is stored in database, now on the edit page i want to show 3 as already selected,
something like that..

[quote]<select name="select_thing">	<option value="1">1</option>	<option value="2">2</option>	<option value="3" selected>3</option>	<option value="4">4</option>	<option value="5">5</option></select>[/quote]

I will be very thankful, if someone could help by explaining how do i make option 3 as selected.. eg. how i know option 3 is selected and stored in db.
Edited April 21, 2012 by moderator (see edit history)

[Saint_Michael 3]

Here are some links to help you out.

This is a wizard that designs a php drop down menu

https://www.thesitewizard.com/wizards/navigationmenu.shtml

This is a script that creates a dhtml drop down menu

http://www.scriptdungeon.com/script.php?ScriptID=1010

This site talks about what goes into a php drop down menu and provides files as well

http://www.phpclasses.org/package/1562-PHP-Generate-a-drop-down-menu-with-HTML-tables.html

Hopefully that helps give you an idea on what to look for in designing a php drop down menu

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[galexcd 0]

Couldn't you just make an if statement? If you know how to get and set the values in the database already, all you'd have to do is somthing like:

<select name="select_thing"><option value="1"<?phpif(mysql_result($result,"","value")==1) echo" selected";?>>1</option><option value="2"<?phpif(mysql_result($result,"","value")==2) echo" selected";?>>2</option><option value="3"<?phpif(mysql_result($result,"","value")==3) echo" selected";?>>3</option><option value="4"<?phpif(mysql_result($result,"","value")==4) echo" selected";?>>4</option><option value="5"<?phpif(mysql_result($result,"","value")==5) echo" selected";?>>5</option></select>

If that's what you meant, here's a more code-conserving way of writing it:

<?php ${mysql_result($result,"","value)}=" selected";?><select name="select_thing"><option value="1"<? echo $1 ?>>1</option><option value="2"<? echo $2 ?>>2</option><option value="3"<? echo $3 ?>>3</option><option value="4"<? echo $4 ?>>4</option><option value="5"<? echo $5 ?>>5</option></select>

[hitmanblood 0]

OK this solution will work also I encourage you to try it it looks more simple to me then others. I cannot tell is it because I use it and I am familiar with it or just because it is simpler then other.

Good luck with whatever you are doing

For the script I think that it doesn't need any more explanation also I would like to point out that I took in count that when you have loaded this value from your database that then you have saved it in the initial_var that is initial variable.

<select name="testlist">  <option <? if(initial_var == "1"){ echo 'selected'; } >?  >1</option>  <option <? else if(initial_var == "2"){ echo 'selected'; } >? >2</option>  <option <? else if(initial_var == "3"){ echo 'selected'; } >? >3</option>  <option <? else if(initial_var == "4"){ echo 'selected'; } >? >4</option>  <option <? else{ echo 'selected'; } >? >5</option>  </select>

[iGuest 3]

i have not fired a query from which i can get $result....do u have any other solution?plss help me out