<?php
$sql = "(SELECT dari FROM mapping WHERE author = '$_SESSION[user]')
UNION
(SELECT ke FROM mapping WHERE author = '$_SESSION[user]')";
$run = mysqli_query($conn, $sql);
while ($rows = mysqli_fetch_assoc($run)) {
if(!empty($rows['dari'])) {?>
<option value="<?php echo $rows['dari']; ?>"><?php echo $rows['dari']; ?></option> <?php
} else {?>
<option value="A1">A1</option><?php
}
}
?>
menu option with value A1 can not appear in the form
You need enter to the block at least once
do
{
$rows = mysqli_fetch_assoc($run);
if (null !== $rows && !empty($rows['dari']))
{ ?>
<option value="<?php echo $rows['dari']; ?>"><?php echo $rows['dari']; ?></option> <?php
}
else
{ ?>
<option value="A1">A1</option><?php
}
} while ($rows);
Related
I am trying to insert the selected value from a drop down that was populated from a reference table in my database. I followed a tutorial for a dynamic dropdown but now I would like to take the value and insert it. The problem is it keeps taking the echo the tutorial uses. Is there a way I can make that selected value a new variable? It currently inserts "< php echo $team_name"
<div>
<label>Home Team</label>
<select name="home_team" style="width:125px;>
<option value="">Select Team</option>
<?php
$query = "SELECT * FROM team";
$results = mysqli_query($db, $query);
mysqli_query($db, "SELECT * FROM team_name");
// loop
foreach ($results as $team_name) {
?>
<option value="<php echo $team_name["cid"]; ?><?php echo $team_name["team_name"]; ?></option>
<?php
}
?>
</select>
How I attempted to insert:
$db = mysqli_connect('localhost', 'root', 'root', 'register');
if(mysqli_connect_errno())
{
echo "failed" . mysqli_connect_error();
}
//var_dump($_POST);
$home_team = mysqli_real_escape_string($db, $_POST['home_team']);
$home_team = $home_team;
$query = "INSERT INTO game_table (home_team)
VALUES('$home_team')";
mysqli_query($db, $query);
//echo $query;
//echo $home_team;
//header('location: index.php');
please follow this.
<select name="home_team" style="width:125px;>
<option value="">Select Team</option>
<?php
$query = "SELECT * FROM team";
$results = mysqli_query($db, $query);
while($row = mysqli_fetch_assoc($results)) {
?>
<option value="<php echo $row['cid']; ?>"><?php echo $row["team_name"]; ?></option>
<?php } ?>
</select>
may be this should work
Try this. There was a couple of missing " and ? in your code.
<select name="home_team" style="width:125px;">
<option value="">Select Team</option>
<?php
$query = "SELECT * FROM team";
$results = mysqli_query($db, $query);
foreach ($row = mysqli_fetch_assoc($results)) {
?>
<option value="<?php echo $row["cid"]; ?>">
<?php echo $row["team_name"]; ?>
</option>
<?php
}
?>
</select>
I know only this method. this method is assume that you know the values in all <option>
<select name="agama" id="agama">
<option value="Islam"<?php if ($rows['agama'] === 'Islam') echo ' selected="selected"'>Islam</option>
<option value="Khatolik"<?php if ($rows['agama'] === 'Khatolik') echo ' selected="selected"'>Khatolik</option>
<option value="Protestan"<?php if ($rows['agama'] === 'Protestan') echo ' selected="selected"'>Protestan</option>
<option value="Hindu"<?php if ($rows['agama'] === 'Hindu') echo ' selected="selected"'>Hindu</option>
<option value="Buddha"<?php if ($rows['agama'] === 'Buddha') echo ' selected="selected"'>Buddha</option>
<option value="Lain-Lain"<?php if ($rows['agama'] === 'Lain-Lain') echo ' selected="selected"'>Lain-Lain</option>
</select>
.... the above code is example from other people not mine.
but My case is the <option> is select from database too.
I have 2 table, oav_event and oav_album
the oav_album has foreign key (event_id) from oav_event table
I want to check if row['event_id'] from oav_album table is equal to option value (from oav_event table) if true, then set selected="selected"
while($row = mysqli_fetch_assoc($result)) { ?>
<option value="<?php echo $row['event_id']; ?>" >Event: <?php echo $row['event_date']; ?> </option>
<?php } ?>
the option will change depend on change in database table, so I don't know the value in option. How should I do?
<select name="event_id">
<?php
$sql = "SELECT * FROM oav_event";
$result = mysqli_query($conn, $sql);
while($row = mysqli_fetch_assoc($result)) {
$selected = "";
if($row['event_id'] == $Yourmatchvalue)
{
$selected = "selected";
}
?>
<option value="<?php echo $row['event_id']; ?>" selected="<?php echo $selected; ?>" >Event: <?php echo $row['event_date']; ?> </option>
<?php } ?>
</select>
may this helps your. you need to replace $Yourmatchvalue variable with your variable.
You can use $_GET as the method on your form and pass the id of the record using it:
while($row = mysqli_fetch_assoc($result)) {
if (!empty($_GET['event_id']) && $row['event_id'] == $_GET['event_id']) {
$selected = 'selected = "selected"';
} else {
$selected = '';
}
echo '<option '.$selected.' value="'.$row["event_id"].'">'.$row["event_date"].'</option>';
}
Here is a solution,
$selected_value = 'Hindu'; // This will come from database
Change option tag with this
<option value="<?php echo $row['event_id']; ?>" <?php echo ($row['event_id'] == $selected_value) ? 'selected="selected"' : ''; ?> >Event: <?php echo $row['event_date']; ?> </option>
Create one function which will create options list like this:
function setDropdownValue($selectQueue_list,$selectedVal)
{
$queueVal = '';
$selectQueue_list_res=$db->query($selectQueue_list);
while($selectQueue_list_res_row=$db->fetchByAssoc($selectQueue_list_res))
{
$val = $selectQueue_list_res_row['id'];
$name = $selectQueue_list_res_row['name'];
if($val == $selectedVal)
{
$queueVal .= "<option value='$val' selected='selected' label='$name'>$name</option>";
}
else
{
$queueVal .= "<option value='$val' label='$name'>$name</option>";
}
}
return $queueVal;
}
Then create a query:
$get_value_query="SELECT id, name FROM table";
$dropdown_selected_value = !empty($dropdown_value) ? $dropdown_value: ''; // Pass value which you want to be selected in dropdown
Then call this function:
$dropdown_options = setDropdownValue($get_value_query, $dropdown_selected_value);
Later when you get dropdown options in $dropdown_options, use jquery to populate the dropdown, like this:
$('#dropdown_select_id').html("$dropdown_options");
Give it a try, and let me know.
I want if the user clicks submit to return the the selected value when the form reloads
that what I try and no luck
<?php
$getData = $db->prepare("SELECT * FROM first_university_degree_list");
if ($getData->execute()) {
$res = $getData->get_result();
while ($data = $res->fetch_array()) {
?>
<option value="<?php echo $data['univ_degree']; ?>" <?php if($FirstUniversityDegree == $data['univ_degree']) echo 'selected="selected"'; ?>><?php echo $data['univ_degree']; ?></option>
<?php
}
}
?>
clearing the question
I have a validation for every field on the form so if any of this fields return false the options on the drop menu will return to the first item
I need it to return on the selected item the users selected
As far i understood your problem,i think this is the solution you are looking for,
<?php
$getData = $db->prepare("SELECT * FROM first_university_degree_list");
if ($getData->execute()) {
$res = $getData->get_result();
while ($data = $res->fetch_array()) {
if($FirstUniversityDegree == $data['univ_degree'])
{
?>
<option value="<?php echo $data['univ_degree']; ?>" selected="selected"><?php echo $data['univ_degree']; ?></option>
<?php
}
else
{
?>
<option value="<?php echo $data['univ_degree']; ?>" ><?php echo $data['univ_degree']; ?></option>
<?php
}
?>
<?php
}
}
?>
<?php
$getData = $db->prepare("SELECT * FROM first_university_degree_list");
if ($getData->execute()) {
$res = $getData->get_result();
while ($data = $res->fetch_array()) {
?>
<option value="<?php echo $data['univ_degree']?>" <?php if($FirstUniversityDegree==$data['univ_degree']) print 'selected="selected"'; ?>>
<?php echo $data['univ_degree']; ?></option>
<?php
}
}
?>
hoe it helps
<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>
We managed to get the drop down list menu however, we are having difficulties getting the data from sql. So far, this is what we got.
<select>
<option id="">--Select jobscope--</option>
<?php
$con=mysqli_connect("host","user","password","database");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$getIT = mysqli_query("SELECT job_title FROM `job_details`");
while($viewIT = mysqli_fetch_array($getIT)) {
}
?>
<option id="<?php echo $viewIT['job_title']?>"<?php echo $viewIT['job_title']?></option>
</select>
Shouldn't be like this ? with tag inside WHILE LOOP
while($viewIT = mysql_fetch_array($getIT)) {
<option id="<?php echo $viewIT['job_title']?>"<?php echo $viewIT['job_title']?></option>
}
$query = "SELECT * FROM test_groups_tb WHERE user_id='$userid'";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_assoc($result))
{
$dd .= "<option value='{$row['group_id']}'>{$row['group_name']}</option>";
}
Try this,
<option value="">--select--</option>
<?php
while($rec = mysql_fetch_assoc($result)) {
?>
<option value="<?=$rec['job_title']?>"><?=$rec['job_title']?></option>
<?php }
}?>
</select>
I am not from php background. Try this.
<?php
$query = "SELECT job_title FROM job_details";
$result = $mysqli->query( $query );
echo '<select id="domain_account" name="domain_account" class="txtBox">';
echo '<option value="">-select-</option>';
while ($row = $result->fetch_assoc()){
?>
<option value="<?php echo $row['job_title']; ?>"><?php echo $row['job_title']; ?></option>
<?php
}
echo "</select>";
?>
Better use PDO or MYSQLi . MYSQL* is depriciated
U need to use that <option id="<?php echo $viewIT['job_title']?>"<?php echo $viewIT['job_title']?></option> line b/w while loop
$getIT = mysql_query("SELECT job_title FROM `job_details`");
while($viewIT = mysql_fetch_array($getIT)) {?>
<option id="<?php echo $viewIT['job_title']?>"<?php echo $viewIT['job_title']?></option>
<?hp }?>