show selected value from dropdown - php

Hi i have been working on a form wherein there's a dropdown menu and it's values are from the database. My problem is it doesnt show the value selected after submitting the form. what maybe the problem?
<select name="professional" />
<option value="">Choose one</option>
<?php
$result2 = mysql_query("SELECT * FROM professional");
while($row2 = mysql_fetch_array($result2))
{
$prc = $row2['name'];
$prof = $row2['prcno'] ."\t"."|\t". $row2['name'] ."\t"."|\t".$row2['profession'];
echo "<option value ='$prc'>$prof</option>";
}
?>
</select>
<select name="professional" disabled/>
<option value="">Choose one</option>
<?php
$result2 = mysql_query("SELECT * FROM professional");
$i=0;
while($row2 = mysql_fetch_array($result2))
{
$prc = $row2['name'];
$p1[$i] = $prc;
$prof = $row2['prcno'] ."\t"."|\t". $row2['name'] ."\t"."|\t".$row2['profession'];
$p2[$i] = $prof;
if($_POST['professional'] == $p1[$i])
{
echo "<option selected value ='$p1[$i]'>$p2[$i]</option>";
}
else
{
echo "<option value ='$p1[$i]'>$p2[$i]</option>";
}
}
?>
</select>

It seems to me, you're not incrementing $i, so you keep overwriting $p1[0] and $p2[0] in each iteration of the while-loop.
So add $i++ at the beginning or the end of your loop - or drop the whole use of these to arrays ($p1 and $p2) and use $prc and $prof just as you do in the first code-block - or do you need them for something?
Another thing, try removing the space between value and ='$p1[$i]' - but I'm not sure if that's a problem.

Try
selected="selected"
in stead of
selected

change these lines to
echo "<option selected value ='<?php echo $p1[$i]; ?>'><?php echo $p2[$i]; ?></option>";
and do not forget to increment your $i too
Hope it will help :)

Related

Duple select from database in same table but diffent proposes

First option of select must be the name referring to the ID. The remaining select options are the remaining names
<select class="input" name="client_id">
<?php
$sel_client_detail="Select * from client WHERE client_id=".$id."";
$result_detail = mysqli_query($con,$sel_client_detail);
while($new_record_row = mysqli_fetch_assoc($result_detail)) { ?>
<option selected><?php echo $row['nome'];?></option>
<?php };?>
<?php
$sel_client="Select * from client";
$result = mysqli_query($con,$sel_client);
?>
<option>-----------</option>
<?php while($new_record_row = mysqli_fetch_assoc($result)) { ?>
<option><?php echo $new_record_row['nome'];?></option>
<?php };?>
</select>
Output:
<select>
<option selected> Izzi (current ID name)</option>
<option> ____________</option>
<option> Other existing clients</option>
<option> Other existing clients</option>
<option> Other existing clients</option>
<option> Other existing clients</option>
</select>
If you want the user to be first in your option list just run the query once and build the HTML parts in 2 seperate strings. Then once the loop is complete put them together and echo them
<?php
echo '<select class="input" name="client_id">';
$itsme = '';
$others = '<option>-----------</option>';
$sql = "Select * from client";
$result = $con->query($sql);
while($row = $result->fetch_assoc()){
if ( $id == $row['id'] ) {
$itsme = "<option selected='selected'>$new_record_row[nome]</option>";
} else {
$others += "<option>$new_record_row[nome]</option>";
}
}
// put the option tags together in the order you specified
echo $itsme . $others . '</select>';
Here's a different, but more conventional, approach to this common scenario:
Why not just make the chosen ID selected when you get to it in the list? Then it will still show to the user first. It's more efficient than having two separate queries.
Like this:
<select class="input" name="client_id">
<?php
$sel_client="Select * from client";
$result = mysqli_query($con,$sel_client);
?>
<option>-----------</option>
<?php while($new_record_row = mysqli_fetch_assoc($result)) { ?>
<option <?php echo ($new_record_row["client_id"] == $id ? "selected": ""); ?> ><?php echo $new_record_row['nome'];?></option>
<?php };?>
</select>

code for fetching value to select option

I have select option. this option has multiple value from database. I want to update something from database, this value i want to update is exist on the select option I have.
this is my option code
$id = $_GET['update'];
$query = mysql_query("SELECT * FROM transaction where id = '$id'") or die ("could not search");
$count = mysql_num_rows($query);
while ($rows = mysql_fetch_array($query)) {
$id = $rows['id'];
$tranid = $rows['tranid'];
$trandate = $rows['trandate'];
$patientid = $rows['patientid'];
$transactiontype = $rows['transactiontype'];
$trandescription = $rows['trandescription'];
$tranquantity = $rows['tranquantity'];
$tranunitprice = $rows['tranunitprice'];
$tranamount =$rows['tranamount'];
$gettrandescription = $rows['trandescription'];
}
}
if (isset($_POST['selectmedicine'])) {
$gettrandescription=$_POST['medicineid'];
}
if (isset($_POST['selectroomquantity'])) {
$tranquantity=$_POST['quantity'];
}
?>
<script type="text/javascript">
$('#collapseone').collapseone({
toggle: true
});
<option value="<?php echo $trandescription; ?>" <?php if($trandescription==$gettrandescription){ echo "selected";} ?> ><?php echo $gettrandescription; ?></option>
<option value="<?php echo $tranquantity; ?>" <?php if($tranquantity==$tranquantity){ echo "selected";} ?> ><?php echo $tranquantity; ?></option>
this has value results, but i cant fetch this value to my existing select option.
If you want to "make that variable an array" as aldrin27 said, append [] to the name attribute of the select tag. The selected value of the option with name selectroomquantity will be available in your script as $_POST["selectroomquantity"] (this is the varible).
<select multiple name="selectroomquantity[]">
<option value="...">...</option>
</select>
It should only be necessary if multiple options can be selected however.
Also, there seems to be a typo:
<?php if($tranquantity==$tranquantity)
That check will always return true. It should probably be:
<?php if($tranquantity==$gettranquantity)
hi all i just got the code on how to fecth the value to dropdown. actually i made a wrong word. pre-selected is the right one, sorry for that. here;s the working code.
<select name="selectmedicine" class="form-control col-sm-4" id="medicinename">
<option id="0" style="width:100px"></option>
<?php
$medicine = mysql_query("SELECT * FROM medicine");
while ($row = mysql_fetch_array($medicine)) {
echo '<option id="' . $row['medicinename'] . '"';
echo ' value="' . $row['medicineid'] . '"';
if($row['medicinename'] == $trandescription) {
echo ' selected="selected"';
}
echo '>';
echo $row['medicinename'];
echo '</option>';
}
?>
</select>
thanks everyone, whos trying to help me on this. actually this is my five revised question sorry for that. and finally i got the right one.

echo the selected value from the dropdown menu?

I have a form and here is code for the dropdown menu. Can you help me make a code to show the selected value after submitting the form? im using php
<select name="professional" />
<option value="">Choose one</option>
<?php
$result2 = mysql_query("SELECT * FROM professional");
while($row2 = mysql_fetch_array($result2))
{
$prc = $row2['name'];
$prof = $row2['prcno'] ."\t"."|\t". $row2['name'] ."\t"."|\t".$row2['profession'];
echo "<option value ='$prc'>$prof</option>";
}
?>
</select>
<?php
echo $_REQUEST['professional'];
?>
you might wish to use $_GET or $_POST instead of $_REQUEST, please check: http://php.net/manual/en/reserved.variables.request.php

PHP : Get the value of Dropdown and pass it to another dropdown

I have a problem right now [PHP]. I have a dropdown and its loading my database for the first page, when I proceed to the next page it also have a dropdown where its also loading the my database and also I can get the value of my dropdown in the first page using an echo only.
This is the scenario:
I choose in the dropdown first page the "Letter A" and when I click the button it will proceed to the next page. The dropdown in the second page loaded the items in the database but instead of "-select-" is the first index in the dropdown I want is "Letter A" will be the first index.
This is my code in first page for drowdown:
<select name="id">
<option value="" >- select -</option>
<?php
include 'connect.php';
$q = mysql_query("select fldNetname from tblnetwork");
while ($row1 = mysql_fetch_array($q))
{
echo "<option value='".$row1[fldNetname]."'>".$row1[fldNetname]."</option>";
}
?>
</select>
and this is my code in second page for dropdown:
if ($get_ID != "")
{
echo "<br/>";
echo $get_ID;
//echo "show()";
}
else
{
echo "No Network Selected";
echo "<br/>";
//echo "hide()";
}
?>
<option value="">- select -</option>
<?php
include 'connect.php';
$q = mysql_query("select fldNetname from tblnetwork");
while ($row1 = mysql_fetch_array($q))
{
echo "<option value='".$row1[fldNetname]."'>".$row1[fldNetname]."</option>";
}
?>
</select>
Thanks in advance!
On your second page, you need to check whether the value is the same as the one you received from the first page:
echo "<option value='".$row1[fldNetname]."' " . (($row1[fldNetname] == $get_id)?"selected":"") . ">".$row1[fldNetname]."</option>";
if $get_ID, is actually your $_GET['id'] value, then just do...
while ($row1 = mysql_fetch_array($q))
{
echo "<option value='".$row1['fldNetname']."'";
if($row1['fldNetname']==$get_ID){echo "selected='selected'";}
echo ">".$row1['fldNetname']."</option>";
}
Or as one line...
while ($row1 = mysql_fetch_array($q))
{
echo "<option value='".$row1['fldNetname']."' " . (($row1['fldNetname'] == $get_ID)?"selected='selected'":"") . ">".$row1'[fldNetname']."</option>"
}
Have you tried?
changing (on the second page)
<option value="">- select -</option>
to
<option value="<?php echo $yourVar; ?>"><?php echo $yourVar; ?></option>
Then in the while loop, skip the value selected to prevent a duplicate choice.
Something like:
while ($row1 = mysql_fetch_array($q))
{
if($yourVar != $row1[fldNetname]){
echo "<option value='".$row1[fldNetname]."'>".$row1[fldNetname]."</option>";
}
}

Show selected value in dropdown after submit

hi guys i added dropdown field to form however after i submit the form if there is any error dropdown resets itself how can keep to value after validation thanks a lot for your any helps and idea
here is my code
<td><select id="country" name="country" style="width:150px;">
<option value="-1">Select</option>
<?php
$query = "SELECT country_id, name FROM countries ";
$result = mysql_query($query);
while($row = mysql_fetch_array($result))
{
echo "<option value=\"".$row['country_id']."\" >".$row['name']."</option>\n ";
}
?>
</select></td>
<td><?php echo $form->error("country"); ?></td>
Typically you would set the default option with attribute selected that is tied to the current selected value. So in this case, the option that equals the value of $_POST['country']:
while ($row = mysql_fetch_array($result))
{
if ($row['country_id'] == $_POST['country'])
$selected = "selected=\"selected\"";
else
$selected = "";
echo "<option value=\"".$row['country_id']."\" $selected>".$row['name']."</option>\n ";
}
Which would render as the following on the appropriate option:
<option value="123" selected="selected">456</option>

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