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> Editing Drop Down Menu In Php
apple
post Apr 25 2007, 11:35 PM
Post #1


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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>


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.
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Saint_Michael
post Apr 26 2007, 12:54 AM
Post #2


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Here are some links to help you out.

This is a wizard that designs a php drop down menu

http://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/browse/package/1562.html

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

GOOGLE KEYWORDS

"drop down menu in php"
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galexcd
post May 12 2007, 07:02 PM
Post #3


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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:

CODE
<select name="select_thing">
<option value="1"<?php
if(mysql_result($result,"","value")==1) echo" selected";
?>>1</option>
<option value="2"<?php
if(mysql_result($result,"","value")==2) echo" selected";
?>>2</option>
<option value="3"<?php
if(mysql_result($result,"","value")==3) echo" selected";
?>>3</option>
<option value="4"<?php
if(mysql_result($result,"","value")==4) echo" selected";
?>>4</option>
<option value="5"<?php
if(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:
CODE
<?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>
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hitmanblood
post May 18 2007, 08:00 PM
Post #4


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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.

CODE
<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>
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