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>
Related
<select id="section" name="section">
<?php
include("Nethost.php");
$section = "";
$yr = "";
$sql = mysql_query("SELECT DISTINCT * FROM section ORDER BY yrlvl, section");
while ($row = mysql_fetch_array($sql)){
$section = $row['section'];
$yr = $row['yrlvl'];
?>
<option value="">Select</option>
<option <?php $result2 = mysql_query("SELECT section FROM student WHERE idnumber = '$idnumber'");
if(mysql_num_rows($result2) > 0) { ?>
selected="selected" <?php } ?> value="<?php print $section; ?>"><?php print $yr; ?> - <?php print $section; ?></option>
<?php } ?>
</select>
Above is the php code of the select option. Populated with data from database table, how can I set that the first value if empty. I tried adding a Select but this is the result.
The option select keeps repeating. What to do with this?
Try this
<select id="section" name="section">
<?php
include("Nethost.php");
$section = "";
$yr = "";
$sql = mysql_query("SELECT DISTINCT * FROM section ORDER BY yrlvl, section");
?>
<option value="">Select</option>
<?php
while ($row = mysql_fetch_array($sql)){
$section = $row['section'];
$yr = $row['yrlvl'];
?>
<option <?php $result2 = mysql_query("SELECT section FROM student WHERE idnumber = '$idnumber'");
if(mysql_num_rows($result2) > 0) { ?>
selected="selected" <?php } ?> value="<?php print $section; ?>"><?php print $yr; ?> - <?php print $section; ?></option>
<?php } ?>
</select>
Just move that option line "Select" out of your php code:
<select id="section" name="section">
<option value="">Select</option>
<?php
...your php code
?>
</select>
My SELECT looks like the following:
<?php
$query = "SELECT * FROM Rec_SW2_Rel AS a JOIN SW2 b ON a.Sbj_ID = b.IDsbj GROUP BY a.Sbj_ID ORDER BY b.Descriptor";
$result = mysql_query($query);
?>
<select name="country" onchange="getState(this.value)">
<?php
while ($line = mysql_fetch_array($result, MYSQL_ASSOC))
{
?>
<option value="<?php echo $line['Sbj_ID']; ?>">
<?php echo $line['Descriptor']; ?>
</option>
<?php
}
mysql_close();
?>
</select>
Querying the DB and setting up the drop-down works. The problem is that the value listed first isn't automatically selected. If a user wants to use it, for further navigation, they must first select a different one and then select the first once again.
I couldn't alter the values in the DB. If I insert selected='selected' it returns the last value of the result set, but always without being selected.
You maybe want this? First selected option when the form is loaded is blank.
<select name="country" onchange="getState(this.value)">
<option value=""></option>
<?php
while ($line = mysql_fetch_array($result, MYSQL_ASSOC))
{
or select the selected data from the database? selected column with selected value.
<select name="country" onchange="getState(this.value)">
<?php
$first = true;
while ($line = mysql_fetch_array($result, MYSQL_ASSOC))
{
?>
<option value="<?php echo $line['Sbj_ID']; ?>" <?php echo ($line['selected']=='selected') ? 'selected="selected"' : '' ; ?>>
you can test with respect to $line['Sbj_ID'] if this is = to the value you want by default
<?php
$query = "SELECT * FROM Rec_SW2_Rel AS a JOIN SW2 b ON a.Sbj_ID = b.IDsbj GROUP BY a.Sbj_ID ORDER BY b.Descriptor";
$result = mysql_query($query);
?>
<select name="country" onchange="getState(this.value)">
<?php
while ($line = mysql_fetch_array($result, MYSQL_ASSOC))
{
?>
<option value="<?php echo $line['Sbj_ID']; ?>" <?php if($line['Sbj_ID']==value_you_want_selected){?>selected<?php } ?>>
<?php echo $line['Descriptor']; ?>
</option>
<?php
$i++; }
mysql_close();
?>
I can see the query returning results, but I can't seem to be able to put them into a html dropdown box. Also, the dropdown box has just as many entries as the query returns, but THEY ARE ALL WHITE SPACES. HOWEVER, the page source shows correct option values such as
<option value="3 John"></option>
<option value="Jude"></option>
<option value="Revelation"></option>
Can somebody help me out? Why dont they actually show in the dropdown box?
<html>
<?php
//Connect to the database
$mysqli = new mysqli("localhost", "root", "", "bible");
//Return an error if we have a connection issue
if ($mysqli->connect_error) {
die('Connect Error (' . $mysqli->connect_errno . ') '
. $mysqli->connect_error);
}
//Query the database for the results we want
$query = $mysqli->query("select distinct bname as Name from kjv limit 1");
//Create an array of objects for each returned row
while($array[] = $query->fetch_object());
array_pop($array);
//Print out the array results
print_r($array);
?>
<h3>Dropdown Demo Starts Here</h3>
<select name="the_name">
<?php foreach($array as $option) : ?>
<option value="<?php echo $option->Name; ?>"></option>
</select>
<?php endforeach; ?>
Try This
<select name="the_name">
<?php foreach($array as $option) : ?>
<option value="<?php echo $option['Name']; ?>"><?php echo $option['Name']; ?></option>
<?php endforeach; ?>
</select>
After the query is executed use the while loop to add the options to select
$query = $mysqli->query("select distinct bname as Name from kjv limit 1"); ?>
<select>
<?php while($option = $query->fetch_object()){ ?>
<option><?php echo $option->Name; ?></option>
<?php } ?>
</select>
Not sure what the array_pop is doing in the code
AS TIM WAX SAID THIS IS THE SOLUTION
$query = $mysqli->query("select distinct bname as Name from kjv limit 1"); ?>
<select>
<?php while($option = $query->fetch_object()){ ?>
<option><?php echo $option->Name; ?></option>
<?php } ?>
</select>
<select name="the_name">
<?php foreach($array as $option) : ?>
<option value="<?php echo $option->Name; ?>"></option>
<?php endforeach; ?>
</select>
You ended your loop in a way that it also create <select> tag again and again. Change it and try again. I don't know much about .php but it could be a problem in showing your dropdown box.
here is mine .. im a beginner but it works for me,
$query = $mysqli->query("SELECT * FROM `student_type_db`"); //table of student type
echo "<select>";
while($row = $query->fetch_array()){
echo "<option>";
echo $row['student_type'] . " - " . $row['student_description'];
echo "</option>";
}
echo "</select>";
// student type = 1 | student description = regular
// output : 1 - regular
I have a list of checkboxes displayed below. This shows all the contactors and allows them to be selected via check box.
<?php
$query = "SELECT * FROM form_4 GROUP BY contractors ASC";
$result = mysql_query($query);
?>
<li><select multiple="multiple" size="10" name="contractors[]">
<option value="None Yet" selected="selected">None Yet
</option>
<?php
while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
?>
<option value="<?php echo $line['contractors'];?>"> <?php echo $line['contractors'];?> </option>
<?php
}
?>
</select></li>
I have an array saved in another place that I would like to generate the list above but with the items in the array below already checked/selected.
<?php
$options = unserialize('contractors');
$result = mysql_query("SELECT * FROM form_2 WHERE jobname = 'testjob' GROUP BY jobname ORDER BY biddate ASC LIMIT 0, 1");
while($row = mysql_fetch_array($result))
{
$contractors = unserialize($row['contractors']);
foreach ($contractors as $contractor)
echo "" . htmlspecialchars ($contractor).' - ';
?>
Any help would be greatly appreciated.
Try this :
<option value="<?php echo $line['contractors'];?>" <?php if(in_array($line['contractors'],$contractors)){?>checked="checked" <?php }?>> <?php echo $line['contractors'];?> </option>
How do I show the selected value of a dropdown list from my mysql database. The dropdown list is dependent to my Category dropdown list. These are the codes:
<?php $id = $_GET["id"];
if(!$pupcon){
die(mysql_error());
}
mysql_select_db($database_pupcon, $pupcon);
$getDropdown2 = mysql_query("select * from tblitemname where CategoryID = $id");
while($row = mysql_fetch_array($getDropdown2)){
echo "<option value=\"".$row["ItemID"]."\">".$row["Item_Name"]."</option>";
} ?>
Here are the codes for the first dropdown list (Category) which populates the Item Name dropdown.
<select name="Category" id="Category" class="field select large" onChange="loadXMLDoc(this.value);">
<?php do { ?>
<option value="<?php echo $row_Recordset1['CategoryID']?>"<?php if (!(strcmp($row_Recordset1['CategoryID'], $row_editRecordset['CategoryID']))) {echo "selected=\"selected\"";} ?>><?php echo $row_Recordset1['CategoryName']?></option>
<?php } while ($row_Recordset1 = mysql_fetch_assoc($Recordset1)); $rows = mysql_num_rows($Recordset1); if($rows > 0) {
mysql_data_seek($Recordset1, 0);
$row_Recordset1 = mysql_fetch_assoc($Recordset1);}?>
</select>
While you're listing out the drop down options, you can check to see if the current ItemID matches the passed id value. If it matches, throw a selected="selected" in there. Try:
$selected = ($row['ItemID'] == $id);
echo "<option value=\"".$row["ItemID"]."\" ".($selected ? " selected=\"selected\"":"").">".$row["Item_Name"]."</option>";
EDIT
I tried to clean up the code some...not sure why you're using a do...while because $row_Recordset1 wouldn't be available on the first iteration.
<select name="Category" id="Category" class="field select large" onChange="loadXMLDoc(this.value);">
<?php
while ($row_Recordset1 = mysql_fetch_assoc($Recordset1)) {
?>
<option value="<?php echo $row_Recordset1['CategoryID']; ?>"<?php if (!(strcmp($row_Recordset1['CategoryID'], $row_editRecordset['CategoryID']))) { echo " selected=\"selected\""; } ?>>
<?php echo $row_Recordset1['CategoryName']; ?>
</option>
<?php
}
?>
</select>
you can use this code inside while
$selected=$row["ItemID"]==$id ?'Selected':'';
echo "<option value=\"".$row["ItemID"]."\" {$selected} >".$row["Item_Name"]."</option>";;