I am using the following PHP code to retrieve the team name from the mysql db table team_details which contains 2 columns,team_id and team_name
<?php
mysql_connect("", "", "");
mysql_select_db("db_name");
$data = mysql_query("SELECT team_name FROM team_details");
print "Team A:";
Print "<select name="dropdown">";
while($info = mysql_fetch_array( $data ))
{
Print "<option value='".$info['team_name']."'>".$info['team_name'] . "</option> ";
}
Print '</select>';
?>
Full code
Team and Players selection for the match
Choose the teams for the match
";
while($info = mysql_fetch_array( $data ))
{
echo "".$info['team_name'] . " ";
}
echo '';
?>
VS
Team B:
<select id="team2" disabled="true">
<option value="volvo">Volvo</option>
<option value="saab">Saab</option>
<option value="opel">Opel</option>
<option value="audi">Audi</option>
</select>
<button id="sub" disabled="true" name ="choose" onclick="load_players();"> Choose Players</button>
<button type="reset" value="Clear" onclick="reset_load();"> Clear</button>
</form>
</div>
</body>
you error may be here:
Print "<select name="dropdown">";
you need to escpae the double quotes:
Print "<select name=\"dropdown\">";
First of all, you have syntax error here:
Print "<select name="dropdown">";
Escape quotes like this:
Print "<select name='dropdown'>";
Second, use mysql_fetch_assoc instead of mysql_fetch_array, so your code looks like this:
while($info = mysql_fetch_assoc( $data ))
{
Print "<option value='".$info['team_name']."'>".$info['team_name'] . "</option> ";
}
And check your mysql_connect in order to have right values (Host name, user, password)
try following code, replace host, db_username, db_password and db_name
mysql_connect("host", "db_username", "db_password");
mysql_select_db("db_name");
$data = mysql_query("SELECT team_name FROM team_details");
echo "Team A:";
echo "<select name='dropdown'>";
while($info = mysql_fetch_array( $data ))
{
echo "<option value='".$info['team_name']."'>".$info['team_name'] . "</option> ";
}
echo '</select>';
Related
I have made a dropdown search form that is auto populated by my database content. The voices in the table would be for example types of woods with varing dimensions.
So there are repeatable wood names with diverse data.
To avoid repetition the dropdown is populated with wood types combined to be selected then displayed with all their variants.
The problem is, upon selecting an input, the results are of the item listed above and not the one selected.
<form action="search2.php" method="POST">
<select name="finit" onchange='this.form.submit()'>
<?php
include("connect.php");
$query = "SELECT finit FROM prime";
$info = mysqli_query($conn, $query);
$finit = '';
echo "<option value=\"\">Selezione Materiale</option>";
while($row = $info->fetch_assoc()){
if($row['finit'] != $finit) {
echo "<option value=\"$finit\">" . $row['finit'] . "</option>";
$finit = $row['finit'];
}
}
?>
</select>
<noscript><input type="submit" value="Submit"></noscript>
</form>
Since there are many variants(dimensions) associated with a single wood type, you have to first take the wood type as input from user(via dropdown list), and probably then you may want to display all possible variants(dimensions) of that particular wood type.
So, change the SQL query in the following way,
$query = "SELECT DISTINCT finit FROM prime";
and the while loop in the following way,
while($row = $info->fetch_assoc()){
$output = "<option value='" . $row['finit'] . "'";
if($row['finit'] == $_POST['finit']){
$output .= " selected='selected'";
}
$output .= ">" . $row['finit'] . "</option>";
echo $output;
}
Try this but change if condition according your default value and sql value should be match.
<form action="search2.php" method="POST">
<select name="finit" onchange='this.form.submit()'>
<?php
include("connect.php");
$query = "SELECT finit FROM prime";
$info = mysqli_query($conn, $query);
$finit = '';
?>
<option value="">Selezione Materiale</option>;
<?php
while($row = $info->fetch_assoc()){
if($row['finit'] == $finit) {
$selected = 'selected';
}else{
$selected = '';
$finit = $row['finit'];
}
?>
<option value="<?php echo $finit ?>" <?php echo $selected ?>><?php echo $row['finit']?></option>
<?php } ?>
</select>
<noscript><input type="submit" value="Submit"></noscript>
</form>
I am trying to create some code to do a list of components in mysql.
Something like this
decod - 1,2,3,4,5
and to make it in drop-down listlike this
decod (dorplist) 1
2
3
4
5
I tried this one:
<?php
$mysqli = mysqli_connect("host", "username", "password", "name");
$result = mysqli_query($mysqli, 'SELECT * FROM componnets') or die(mysqli_error(mysqli));
while ($row = mysqli_fetch_array($result)) {
$org = str_replace("," , "'>", $row['fill']."");
$replace = str_replace("," , "
".$org."</option><option value='", $row['fill']."'>");
echo "<input type='hidden' name='compo' value='".$row['compon']."'>".$row['compon']."";
echo "<select name='fill'>";
echo "<option value='".$replace."'>".$replace."</option>";
echo "</select>";
}
// Free result set
mysqli_free_result($result);
/* close connection */
$mysqli->close();
?>
and do make it output to an XML
but I don't know how! I output like this
<input type='hidden' name='compo' value='DECODERS'>DECODERS
<select name='fill'>
<option value='1010'>20'>30'>40</option>
<option value='2010'>20'>30'>40</option>
<option value='3010'>20'>30'>40</option>
<option value='40'>'>1010'>20'>30'>40</option>
<option value='2010'>20'>30'>40</option>
<option value='3010'>20'>30'>40</option><option value='40'></option>
</select>
<input type='hidden' name='compo' value='DECODERS'>DECODERS
<select name='fill'>
<option value='10'>10</option>
</select>
First create the <select> outside the while loop. Then iterate over the rows to add the elements.
I'm not sure what you're trying to accomplish with your str_replace calls, so I removed them.
Something like this might get you started in the right direction:
<?php
$result = mysqli_query($mysqli, 'SELECT * FROM componnets') or die(mysqli_error(mysqli));
echo "<select name='fill'>";
echo "<input type='hidden' name='compo' value='".$row['compon']."'>".$row['compon']."";
while ($row = mysqli_fetch_array($result)) {
echo "<option value='".$row['fill']."'>".$row['fill']."</option>";
}
echo "</select>";
mysqli_free_result($result);
$mysqli->close();
?>
I am trying to edit my form. I want to get selected value selected in selection list.
I have created function to store values in database, and it works. Below is html code I use and function below to insert values in database.
// insert values in database
<label>Dobavljač</label>
<select class="form-control" name="dobavljac" required>
<?php dobavljac() ?>
</select>
function dobavljac(){
$sql=mysqli_query($link, "SELECT * FROM `partneri` WHERE `Dobavljac`='1'
order by `PartnerId` asc ");
echo '<option value="">Izaberi dobavljača</option>';
while($record = mysqli_fetch_array($sql)) {
echo '<option value= "' .$record['PartnerId']. '">' . $record['PartnerNaziv'] . ' </option>';
}
}
// edit values
First I retrieve information from database
$id=$_GET['id'];
$sql = "SELECT * FROM materijali where Id=$id ";
$q = $conn->query($sql);
$r = $q ->fetch();
if ($r) {
$dobavljac=$r['Dobavljac'];
I want to get selected value in box
<label>Dobavljač</label>
<select class="form-control" name="dobavljac" value="<?php echo $dobavljac; ?>">
<?php dobavljac() ?>
</select>
Probably I am not doing it the right way, any advice would be appreciated
Try this
<label>Dobavljač</label>
<select class="form-control" name="dobavljac" value="<?php echo $dobavljac; selected?>">
<option value=<?php echo $dobavljac?> selected>
<?php dobavljac() ?>
</option>
</select>
Try this...
$id=$_GET['id'];
$sql = "SELECT * FROM materijali where Id=$id ";
$q = mysql_query($query);
echo "<select name="dobavljac" class="form-control">";
while (($row = mysql_fetch_row($q)) != null)
{
echo "<option value = '{$row['Dobavljac']}'>";
echo $row['Dobavljac'];
echo "</option>";
}
echo "</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 wanted to make a dropdown area by getting the values from a mysql table. This code does not seem to work; all it prints out is the box for the dropdown without the content, how can I get this to work? or is there an alternative procedure in doing this?
<?
$connection = mysql_connect("localhost", "root", "");
mysql_select_db("test", $connection);
$query = "SELECT full_name FROM test";
$names = mysql_query($query);
function dropDown($content, $result)
{
while($row = mysql_fetch_row($result))
{
$name = $row[0];
$content .= "<option value='$name'>$name</option>";
}
}
$content=<<<EOT
<html>
<body>
<select name="name">
EOT;
dropDown($content, $names)
$content.=<<<EOT
</select>
</body>
</html>
EOT;
echo $content;
?>
return the string. PHP is not C where you use out parameters just because they are sometimes handy.
function dropDown($result, $fieldName)
{
$content = '';
while($row = mysql_fetch_row($result)) {
$name = $row[0];
$content .= "<option value='$name'>$name</option>";
}
return '<select name="'.$fieldName.'">'.$content.'</select>';
}
$content = '<html><body>';
$content .= dropDown($names, 'name');
here database details
mysql_connect('hostname', 'username', 'password');
mysql_select_db('database-name');
$sql = "SELECT username FROM userregistraton";
$result = mysql_query($sql);
echo "<select name='username'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['username'] ."'>" . $row['username'] ."</option>";}
echo "</select>";
here username is the column of my table(userregistration)
it works perfectly
or simply the code is
<?
$sql = "SELECT * FROM userregistraton ";
$result = mysql_query($sql); ?>
<select name="username" id="username">
<option value="">Select user</option>
<?php
while ($row = mysql_fetch_array($result))
{ ?>
<option value="<?php echo $row['uid']?>" <?php if($row['uid']==$_POST["username"]) echo "selected"; ?> >
<?php echo $row['username'];?></option>
<?php
} ?>
<?php echo $form->dropDownList($model,'courseprefer', array(
'(*value in database table*'=>'displaying value',
)); ?>
you can manually add them just make sure that the first value in array is in the database table.. :)
to avoid error!