I have issue with select menu on PHP. I tried to get mysql database to select menu. However, it displays none.
Here is my code:
default:
mysql_select_db($database_conn, $conn);
$query_Rsenroll = "SELECT * FROM `tbl_enroll` WHERE `tbl_enroll`.`courseid` ='".$_GET['courseid']."'";
$Rsenroll = mysql_query($query_Rsenroll, $conn) or die(mysql_error());
$row_Rsenroll = mysql_fetch_assoc($Rsenroll);
$totalRows_Rsenroll = mysql_num_rows($Rsenroll);
$courseid = $row_Rsenroll['courseid'];
$er_staffid = "";
break;
}
?>
<select name="courseid">
<option value="" SELECTED>Selected Course ID</option>
<?php
foreach( $Course as $course_id) {
if ( $course_id == $courseid) {
$selected = " SELECTED";
} else {
$selected = "";
}
?>
<option value="<?php echo $course_id; ?>"<?php echo $selected; ?>><?php echo $row_Rsenroll['courseid']; ?></option>
<?php
}
?>
</select>
Thank you for any help and advice.
Assuming courseid is being passed as a variable in the sending URL (file.php?courseid=COURSEID), I think this should do what you want:
This might clean up your script a little (though I switched it to mysql_fetch_array as I'm more familiar with that than mysql_fetch_assoc. Feel free to use assoc):
<?php
$cid = '6116';
?>
<select name="courseidMenu">
<option value="" SELECTED>Selected Course ID</option>
<?php
$query = mysql_query("SELECT * FROM tbl_enroll WHERE courseid = '$cid'", $conn)or die(mysql_error());
$total_rows = mysql_num_rows($query);
while($row = mysql_fetch_array($query)){
$courseId = $row['courseid'];
?>
<option value="<?=$courseId?>" ><?=$courseId?></option>
<?
}
?>
</select>
updated use this it is working on my portal <select>
<option value=''>Select Provider</option>
<?php
$server="server name";
$user="user name";
$password="password";
$database="database";
$conn=mysql_connect($server,$user,$password) or die("connection failed");
mysql_select_db($database,$conn);
$query_Rsenroll = "SELECT * FROM `tbl_enroll` WHERE `tbl_enroll`.`courseid` ='".$_GET['courseid']."'";
$result= mysql_query($query_Rsenroll, $conn) or die(mysql_error());
$n=mysql_num_rows($result);
if($n>0)
while($row=mysql_fetch_array($rs))
echo"<option value='$row['courseid']'>$row['courseid']</option>";
mysql_close($conn);
?>
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>
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 }?>
i have 2 tables
domains_info and tb2
i have got the form working well and entering data into the database
here is the top of my page
<?php
$action = isset($_POST['action']) ? $_POST['action'] : "";
if($action=='create'){
//include database connection
include 'db_connect.php';
//write query
$query = "insert into domains_info
set
domain = '".$mysqli->real_escape_string($_POST['domain'])."',
domain_account = '".$mysqli->real_escape_string($_POST['domain_account'])."',
renew_date = '".$mysqli->real_escape_string($_POST['renew_date'])."'";
if( $mysqli->query($query) ) {
//if saving success
header("Location:domains.php");
}else{
echo "Database Error: Unable to create record.";
}
$mysqli->close();
}
here is the form
<select id="domain_account" name="domain_account" class="txtBox">
<option value="">-select-</option>
<option value="a">a</option>
<option value="b">b</option>
<option value="c">c</option>
</select>
i tried and changed the top of my page like this
<?php
$action = isset($_POST['action']) ? $_POST['action'] : "";
if($action=='create'){
//include database connection
include 'db_connect.php';
//write query
$query = "insert into domains_info
set
domain = '".$mysqli->real_escape_string($_POST['domain'])."',
domain_account = '".$mysqli->real_escape_string($_POST['domain_account'])."',
renew_date = '".$mysqli->real_escape_string($_POST['renew_date'])."'";
if( $mysqli->query($query) ) {
//if saving success
header("Location:domains.php");
}else{
echo "Database Error: Unable to create record.";
}
$mysqli->close();
}
$query = "select id, data
from tb2
where id='".$mysqli->real_escape_string($_REQUEST['id'])."'
limit 0,1";
$result = $mysqli->query( $query );
$row = $result->fetch_assoc();
$id = $row['id'];
$data = $row['data'];
and updated my form as this
<select id="domain_account" name="domain_account" class="txtBox">
<option value="">-select-</option>
<option value="<?php echo$data; ?>"><?php echo$data; ?></option>
</select>
as you can tell i am very new to this and its not working.
sorry i didnt explain what i was trying to achieve, i am trying to display a drop down form with data from a database.
Try this for populating the dropdown using database:
<?php
$query = "select id, data from tb2";
$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['data']; ?>"><?php echo $row['data']; ?></option>
<?php
}
echo "</select>";
?>
<?php
mysql_query("SELECT * FROM page where page_id = '$page_id'") or die(mysql_error());
while($land = mysql_fetch_array($resultxx)){
$land =$land['land'];
?>
output $land is NL,DE,BE,AL..
How is it possible to splits the output
example
<?php
echo "
<select>
<option value='NL'>NL</option>
<option value='DE'>DE</option>
<option value='BE'>BE</option>
<option value='AL'>AL</option>
</select>";
}
?>
Try This
<?php
mysql_query("SELECT * FROM page where page_id = '$page_id'") or die(mysql_error());
while($land = mysql_fetch_array($resultxx)){
$land =$land['land'];
$landarray = explode(",",$land);
$count = count($landarray);
echo "<select>";
for($i=0;$i<$count;$i++){
echo '<option value='.$landarray[$i].'>'.$landarray[$i].'</option>';
}
echo "</select>";
}
?>
I have a database that I need to use to do select.
The select is repeat many times (days of the week).
I can't do a loop for because it can have some "gap" in the id so I use the while.
The problem is that for each select I need to call the database again and I think it's not really a good solution. I would like to find another solution which isn't so "heavy"
Here is my code:
<select name="monday">
<option value="none">Monday </option>
<?php
$Requete2 = "SELECT * FROM `record`";
$Result2 = mysql_query($Requete2) or die(mysql_error());
$rows2 = mysql_num_rows($Result2);
while($row2 = mysql_fetch_array($Result2)){ ?>
<option value="<?php echo ($row2['id'] - 1); ?>"><?php echo $row2['name']; ?></option>
<?php } ?>
</select>
<select name="tuesday">
<option value="none">Tuesday </option>
<?php
$Requete2 = "SELECT * FROM `record`";
$Result2 = mysql_query($Requete2) or die(mysql_error());
$rows2 = mysql_num_rows($Result2);
while($row2 = mysql_fetch_array($Result2)){ ?>
<option value="<?php echo ($row2['id'] - 1); ?>"><?php echo $row2['name']; ?></option>
<?php } ?>
</select>
Thanks
If you want to print out the same for every day you can use this(I havn't tested it for syntax errors):
<?php
$sql_query = "SELECT * FROM `record`";
$query = mysql_query($sql_query) or die(mysql_error());
$optionHtml = '';
while($row = mysql_fetch_array($query)){
$optionHtml .= '<option value="' . ($row["id"] - 1) . '">' . $row["name"] . '</option>';
}
?>
<select name="monday">
<option value="none">Monday</option>
<?php echo $optionHtml; ?>
</select>
<select name="tuesday">
<option value="none">Tuesday</option>
<?php echo $optionHtml; ?>
</select>