Currently creating this project and having problems on how to create a search in select list. I would like to search from option values. how can i create this kind of search on select list?
<?php
error_reporting(0);
//connect to database
$conn = mysqli_connect("localhost", "root", "", "waterdis_stbwd");
if(mysqli_connect_errno($conn)) {
echo "Unable to connect to database server";
}
//query database for items to populate
$sql = "SELECT id FROM usrs";
$query = mysqli_query($conn, $sql);
echo '<select id="slct">';
echo '<option value="">Select ID</option>';
while($id = mysqli_fetch_assoc($query)){
echo "<option>{$id['id']}</option>";
}
echo '</select>';
?>
You will want to use a plugin, like select2
https://select2.github.io/examples.html
https://select2.github.io/
<?php
error_reporting(0);
//connect to database
$conn = mysqli_connect("localhost", "root", "", "waterdis_stbwd");
if(mysqli_connect_errno($conn)) {
echo "Unable to connect to database server";
}
//query database for items to populate
$sql = "SELECT id FROM usrs";
$query = mysqli_query($conn, $sql);
echo '<select id="slct">';
echo '<option value="">Select ID</option>';
while($id = mysqli_fetch_assoc($query)){
echo "<option value="{$id['id']}">{$id['id']}</option>";
}
echo '</select>';
?>
Related
I am little confused about this because everywhere I look for it, they have it same as me, but I have problem that I am listing various of options by <select> and <option> and I need to get back the value of what I have clicked, by it doesn't seem to be working, so if there is someone who knows?
$connect = mysqli_connect("localhost", "root", "", "uklidy");
$sql = "SELECT * FROM budovy ORDER BY id ASC";
$result = mysqli_query($connect, $sql);
echo '<select name="budova_name">
';
while($row = mysqli_fetch_array($result)) {
echo '<option value="'.$row["id"].'">'.$row["jmeno"].'</option>';
}
echo '</select>';
$budova = $_POST["budova_name"];
echo $budova;
You need to submit it with a form.
<?php
$connect = mysqli_connect("localhost", "root", "", "uklidy");
$sql = "SELECT * FROM budovy ORDER BY id ASC";
$result = mysqli_query($connect, $sql);
echo '<form method="post" action="mypage.php">';
echo '<select name="budova_name">';
while($row = mysqli_fetch_array($result)) {
echo '<option value="'.$row["id"].'">'.$row["jmeno"].'</option>';
}
echo '</select>';
echo '<input type="submit">';
echo '</form>';
and if you're talking about getting the selected value on the same page you're gonna have to use javascript for that. PHP is a server side language.
I'm trying to retrieve all table names from a MySQL database and link them to a dropdown list. Then I want to store those table names in a table and preview them.
here's my code...
<?php
$sql = "SHOW TABLES FROM $ip";
$result = mysql_query($sql);
?>
<form id="form1" name="form1" method="post" action="newloay.php">
<select name="select" id="select" required>
<?php
while($row = mysql_fetch_array($result)){
?>
<option> <?php print $row[0] ?> </option>
<?php
} ?>
}
</select>
MySQL :
SELECT
table_name
FROM
my_schema.tables
Where my_schema is your schema name , $ip, I guess.
Follow the following code to get the Name of the Tables in your Database :
<?php
$dbname = 'mysql_dbname';
if (!mysql_connect('mysql_host', 'mysql_user', 'mysql_password')) {
echo 'Could not connect to mysql';
exit;
}
$sql = "SHOW TABLES FROM $dbname";
$result = mysql_query($sql);
if (!$result) {
echo "DB Error, could not list tables\n";
echo 'MySQL Error: ' . mysql_error();
exit;
}
while ($row = mysql_fetch_row($result)) {
echo "Table: {$row[0]}\n";
}
mysql_free_result($result);
?>
Hope this helps, for more details on this, please check out THIS PAGE
I've created drop down list with value name from the database. When I select the value from the drop down list, other data will appear in other textfield based on the database. The submit process was doing fine except when I check on the list. The drop down list value didn't appear in the list but other data did.
This is my adding form:
<tr><td width="116">Medicine name</td><td width="221">
<center>:
<select name="name" id="name" >
<option>--- Choose Medicine ---</option>
<?php
mysql_connect("localhost", "root", "");
mysql_select_db("arie");
$sql = mysql_query("SELECT * FROM tabelmedicine ORDER BY name ASC ");
if(mysql_num_rows($sql) != 0){
while($row = mysql_fetch_assoc($sql)){
$option_value = $row['priceperunit'] . ',' . $row['stock'];
echo '<option value="'.$option_value.'">'.$row['name'].'</option>';
}
}
?>
</select ></center>
This is a script to display other database value in other textfield when the drop down list is selected:
<script>
var select = document.getElementById('name');
var priceperunit = document.getElementById('priceperunit');
var stock = document.getElementById('stock');
select.onchange = function()
{
var priceperunit_stock = select.value.split(',');
priceperunit.value = priceperunit_stock[0];
stock.value = priceperunit_stock[1];
}
</script>
This is my inserted data into database process:
<?php
$host = "localhost";
$user = "root";
$pass = "";
$db = "arie";
$connect = mysql_connect($host, $user, $pass) or die ('Failed to connect! ');
mysql_select_db($db);
$name=$_POST['name'];
if ($name === "")
{
echo "Please fill all the data";
}
else
{
$query="INSERT INTO `tabelout`(`name`)
VALUES ('$name');";
$result = mysql_query($query) OR die (mysql_error());
echo "You have successfully added new medicine to the database.";
}
?>
This is my list page, where the name didn't show up:
<?php
$con=mysqli_connect("localhost","root","","arie");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM tabelout");
echo "<table border='1'>
<th>name</th>";
while($row = mysqli_fetch_array($result))
{
echo "<td><center>" . $row['name'] . "</center></td>";
}
echo "</table>";
mysqli_close($con);
?>
Make sure your database table has records, If it has records, then change the table structure, Add tr tags where required.
echo "<table border='1'>
<tr><th>name</th></tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr><td><center>" . $row['name'] . "</center></td></tr>";
}
echo "</table>";
I am trying to retreive data from a database and populate a list from it, there doesnt seem to be errors in my code and still the list is not populating, my code
<?php
$hostname = "localhost";
$username = "username";
$password = "password";
$dbase = "db";
$link = # mysql_connect($hostname, $username, $password);
$db_selected = # mysql_select_db($dbase);
?>
<?php
include("scripts/dbconnect.php");
$query="select class from school";
$result=mysql_query($query);
$numrows=mysql_num_rows($result);
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){
echo '<option value="'.$row['class'].'">'.$row['class'].'</option>';
}
?>
any help much appreciated, thanks
Options must be wrapped with select tag:
echo "<select name='class'>";
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){
echo '<option value="'.$row['class'].'">'.$row['class'].'</option>';
}
echo "</select>";
Additionally do:
if (!$link) {
die('Could not connect: ' . mysql_error());
break;
}
To see if there are any errors of mysql connection.
You need to echo a select element.
<?php
include("scripts/dbconnect.php");
$query="select class from school";
$result=mysql_query($query);
$numrows=mysql_num_rows($result);
echo "<select>"
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){
echo '<option value="'.$row['class'].'">'.$row['class'].'</option>';
}
echo "</select>"
?>
I have a form where I am getting values from previous page in $GET. I want to fetch around 3 different values from database and display them in textbox. How will I do it? Following is my code for getting data from database.
<?php
require_once('config.php');
$id = $_GET['id'];
$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
if(!$link) {
die('Failed to connect to server: ' . mysql_error());
}
//Select database
$db = mysql_select_db(DB_DATABASE);
if(!$db) {
die("Unable to select database");
}
$query = "select question,price,sequence from questions where status = 1 and qid =".$id;
//echo $query;
$result=mysql_query($query);
if (!$result) {
die('Invalid query: ' . mysql_error());
}
$num_rows = mysql_num_rows($result);
if($num_rows==0) {
echo '<center><font color="red"><b>No record found!!</b></font></center>';
}
else {
$row = mysql_fetch_array($result);
echo $row['question'];
echo $row['sequence'];
echo $row['price'];
}
?>
Thanks
Pankaj
<textarea><?php echo htmlentites($row['question']);?></textarea>
or
<input type="text" value="<?php echo htmlentites($row['question']);?>" />
Depending on your fancy.