$sql = "select custName from customer where active = 1 order by custName";
sult=mysql_query($sql);
echo"<select name='custNameColo'>";
while($row = mysql_fetch_array($result)) {
if($row['custName']==$_GET['defaultCust']) {
echo "<option value=".$row['custName']."selected = 'selected'>".$row['custName']."</option>";
}
else {
echo "<option value=".$row['custName'].">".$row['custName']."</option>";
}
}
echo "</select>";
I want set a default value a drop down menu, but it doesn't work, please help me.
<?php
$sql = "select custName from customer where active = 1 order by custName";
$result=mysql_query($sql);
while($row = mysql_fetch_array($result)){
?>
<option value="<?php echo $row['custname'] ?>" <?php if($row['custname']==$_GET['defaultCust']) { echo "selected";}?>>
<?php echo $row['custname']; ?></option>
<?php
}
?>
<? echo "TRY this it will work";
$sql = "select custName from customer where active = 1 order by custName";
$result=mysql_query($sql);
?>
<select name="custNameColo">
<?
while($row = mysql_fetch_array($result)){?>
<option value="<?=$row['custname'];?>" <? if($row['custname']==$_GET['defaultCust']){?> selected="selected" <? }?>><?=$row['custname'];?></option>
<? } ?>
</select>
You didn't added single quotes in the value of option tag before and after. I done it for you. The default value was not selected because value attribute was not properly closed.
$sql = "select custName from customer where active = 1 order by custName";
$result=mysql_query($sql);
echo "<select name='custNameColo'>";
while($row = mysql_fetch_array($result)){
if($row['custName']==$_GET['defaultCust']){
echo "<option value='".$row['custName']."' selected = 'selected'>".$row['custName']."</option>";
}
else {
echo "<option value='".$row['custName']."'>".$row['custName']."</option>";
}
}
echo "</select>";
Related
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>";
I want a dynamic option list which reads from database table lets say table of : Students,
It must show to user the list of student_name and when it is selected by user it must send student_id of that student to the database. For example:
Students table :
student_id student_name
----------------------------
1 John
2 Edward
In users option list must be included only John, Edward.. but when user selects John, the option picker must send only student_id('1') to database.
My current code , but is not fetching list from db :S :
yes, this is my code but for some reasons it ain't work :
<select Name='student_id'>
<option value="">--- Select ---</option>
<?
mysql_connect ("localhost","root","");
mysql_select_db ("mydb");
$select="student_name";
if (isset ($select)&&$select!=""){
$select=$_POST ['student_name'];
}
?>
<?
$list=mysql_query("select * from students order by student_name asc");
while($row_list=mysql_fetch_assoc($list)){
?>
<option value="<? echo $row_list['student_id']; ?>"<? if($row_list['student_name']==$select){ echo "selected!"; } ?>>
<?echo $row_list['student_name'];?>
</option>
<?
}
?>
</select>
may this one help you
$result = mysqli_query($con,"SELECT * FROM Students");
echo "<select>";
while($row = mysqli_fetch_array($result)) {
echo "<option id=".$row['student_id'].">" . $row['student_name'] . "</option>";
}
echo "</select>";
mysqli_close($con);
?>
You should fetch your database from your PHP, then build your list out of it
//YOU MUST ADD YOUR DATABASE CONNECTION
mysql_connect('localhost','username','password');
mysql_select_db("dbname");
//HERE IS YOUR SQL REQUEST
$SQL_request = "SELECT * FROM `student_table`";
$req = mysql_query($SQL_request) or die(mysql_error().'<br/>'.$SQL_request);
echo '<select name = "students">';
while($result = mysql_fetch_assoc($req)){
echo '<option value="'.$result['student_id'].'">'.$result['student_name'].'</option>';
}
echo '</select>';
?>
Try below code.
$con = mysqli_connect('localhost','root','','mydb');
$sql="SELECT student_id ,student_name FROM students";
$result = mysqli_query($con,$sql);
?>
<select name = "student">
<?php
while($row = mysqli_fetch_array($result)) {
?>
<option value="<?php echo $row['student_id']; ?>"><?php echo $row['student_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 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!
How do I keep the selected item after I submit the page? I have the country dropdown list with the following code.
<?php
$SQL = "SELECT countr_id, country_name FROM countries";
$Result = mysql_query($SQL) or die(mysql_error());
?>
<select name="country" style="width:400px"> <option value='-1'></option>
<?php
while($row = mysql_fetch_array($Result))
{
echo "<option value=\"".$row["country_id"]."\">".$row["country_name"]."</option>";
}
?>
<?php
$SQL = "SELECT countr_id, country_name FROM countries";
$Result = mysql_query($SQL) or die(mysql_error());
?>
<select name="country" style="width:400px"> <option value='-1'></option>
<?php
while($row = mysql_fetch_array($Result))
{
echo "<option value=\"".$row["country_id"]."\"";
if($_POST['country'] == $row['country_id'])
echo 'selected';
echo ">".$row["country_name"]."</option>";
}
?>
<?php
while($row = mysql_fetch_array($Result))
{
if ($_POST['country'] == $row["country_id"]) {
echo "<option value=\"".$row["country_id"]."\" selected="selected">".$row["country_name"]."</option>";
} else {
echo "<option value=\"".$row["country_id"]."\">".$row["country_name"]."</option>";
}
}
?>
<?php
$SQL = "SELECT countr_id, country_name FROM countries";
$Result = mysql_query($SQL) or die(mysql_error());
?>
<select name="country" style="width:400px"> <option value='-1'></option>
<?php
while($row = mysql_fetch_array($Result))
{
echo "<option ";
if($_REQUEST['country'] == $row["country_id"]) echo 'selected="selected" ';
echo "value=\"".$row["country_id"]."\">".$row["country_name"]."</option>";
}
?>
This code depends if you are trying to keep the value after you submit back to the original page.
<?php
$SQL = "SELECT countr_id, country_name FROM countries";
$Result = mysql_query($SQL) or die(mysql_error());
?>
<select name="country" style="width:400px">
<option value='-1'></option>
<?php
while($row = mysql_fetch_array($Result))
{
echo "<option ";
if($_REQUEST["yourSelectName"] ==$row["country_id"])
echo ' selected = "selected" ';
echo " value=\"".$row["country_id"]."\">".$row["country_name"]."</option>";
}
?>
</select>
Or you can do it all inline... less code, doesn't require all the if/then/else stuff
Change
echo "<option value=\"".$row["country_id"]."\">".$row["country_name"]."</option>";
To
echo "<option ". (($_POST['country'] == $row["country_id"]) ? 'selected ' : '') ."value=\"".$row["country_id"]."\">".$row["country_name"]."</option>";
If it is get (passed in URL), the use GET
<?php
$selectedid = 5; //example of selected if before submitting
$SQL = "SELECT countr_id, country_name FROM countries";
$Result = mysql_query($SQL) or die(mysql_error());
?>
<select name="country" style="width:400px"> <option value='-1'></option>
<?php
while($row = mysql_fetch_array($Result))
{
echo "<option value=\"".$row["country_id"]." ".(($selected==$row["country_id"])?"SELECTED":"")."\">".$row["country_name"]."</option>";
}
?>
//assume you have $result = array(your result list);
<select name='question'>
<?php
foreach ($result as $question) {
if ($_POST['question'] == $question) {
$selected = "selected";
} else {
$selected = '';
}
echo "<option value='" . $question . "' $selected>$question</option>";
}
?>
</select>