jamesnkk Posted March 27, 2006 Share Posted March 27, 2006 Hi,I have an entry form that contain some text entry and one drop down menu which I retrieve the data from a table.The problem is when I click the submit button, It go to the 2nd form to validate for error, if error it will return to the entry form, I could manage to retain all text entry, so that user donlt have to re-type all over again.But I cannot retain the selected item from a drop down menu, is there a way ?, please help me.Thanks Link to comment https://forums.phpfreaks.com/topic/5901-drop-down-menu/ Share on other sites More sharing options...
micah1701 Posted March 27, 2006 Share Posted March 27, 2006 what method are you using to re-populate the text field? if your are just echoing the posted values (value="<? echo $_POST['value'] ?>") then for the drop down just do:<select name="dropdown"> <option value="option1" <? if($_POST['dropdown'] == 'option1'){ echo "selected"; } ?>> Option 1 </option> <option value="option2" <? if($_POST['dropdown'] == 'option2'){ echo "selected"; } ?>> Option 2 </option> <option value="option3" <? if($_POST['dropdown'] == 'option3'){ echo "selected"; } ?>> Option 3 </option></select>get the idea? Link to comment https://forums.phpfreaks.com/topic/5901-drop-down-menu/#findComment-21064 Share on other sites More sharing options...
jamesnkk Posted March 27, 2006 Author Share Posted March 27, 2006 [!--quoteo(post=358756:date=Mar 27 2006, 11:25 AM:name=micah1701)--][div class=\'quotetop\']QUOTE(micah1701 @ Mar 27 2006, 11:25 AM) [snapback]358756[/snapback][/div][div class=\'quotemain\'][!--quotec--]what method are you using to re-populate the text field? if your are just echoing the posted values (value="<? echo $_POST['value'] ?>") then for the drop down just do:<select name="dropdown"> <option value="option1" <? if($_POST['dropdown'] == 'option1'){ echo "selected"; } ?>> Option 1 </option> <option value="option2" <? if($_POST['dropdown'] == 'option2'){ echo "selected"; } ?>> Option 2 </option> <option value="option3" <? if($_POST['dropdown'] == 'option3'){ echo "selected"; } ?>> Option 3 </option></select>get the idea?[/quote]Thanks for the prompt reply, here the code to populate from a table on form-1(create_partno.php)<select name="aid" id="aid" size="1"> <option selected value="">Please select</option> <? while ($company = mysql_fetch_array($companys) ) { $aid=$company['company']; $aname=htmlspecialchars($company['company']); echo "<option value='$aid'>$aname</option>\n"; } ?> </select>Then on the 2nd form (create_part.php) I use header to return, - header("Location: create_partno.php?part_no=".$_POST['part_no']."&supp_partno=".$_POST['supp_partno']."&part_desp=".$_POST['part_desp'] ."&cost_price=".$_POST['cost_price']."&sell_price=".$_POST['sell_price'] ."&brand=".$_POST['brand']."&package=".$_POST['package']."&qty_bal=".$_POST['qty_bal'] ."$aid=".$_POST[aid] ."&qty_order=".$_POST['qty_order']."&re_order=".$_POST['re_order']."&error=".$_POST['error'].$error1 ); When return to form-1, Just could not retain the selected value from the drop down menu, Link to comment https://forums.phpfreaks.com/topic/5901-drop-down-menu/#findComment-21070 Share on other sites More sharing options...
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