I can't see the result in the select area
<?php
require 'config.php';
$query = "SELECT cat_id, category FROM categories LIMIT 1";
$result = mysqli_query($con,$query);
if(!$result){
echo 'Query failed : '.mysqli_error();
exit(0);
}
while($row = mysqli_fetch_assoc($result)) {
echo '<select name="cat_id">
' . $row['cat_id'] . '
</select>';
}
mysqli_close($con);
?>
I try to add the HTML select tag but it still isn't working.
This is not a good format for select
echo '<select name="cat_id">
' . $row['cat_id'] . '
</select>';
It should be
echo '<select name="cat_id">';
while($row = mysqli_fetch_assoc($result)) {
echo '<option value="'.$row['cat_id'] .'">'.$row['cat_id'].'</option>';
}
echo '</select>';
Related
I already try many ways but the value didn't show in dropdown list
Here, this is my code. can you suggest me anything that i was wrong
<?php
$result = mysqli_query($con,"SELECT * FROM project");
if( mysqli_num_rows( $result )==0){
echo "<tr><td>No Rows Returned</td></tr>";
}else{
$row = mysqli_fetch_assoc( $result );
$pos = 0;
echo "<select name=Pname >";
while($pos <= count ($row)){
echo "<option value="$row["project_no"]">"$row["project_name"]"</option>";
$pos++;
}
echo "</select>";?>
And i write as .php file. Thanks for your help.
Try this out:
$output = '';
if(mysqli_num_rows($result) == 0){
// echo error;
} else {
while($row = mysqli_fetch_assoc($result)){
$project_no = $row['project_no'];
$project_name = $row['project_name'];
$output .= '<option value="' . $project_no . '">' . $project_name . '</option>";
}
}
Then inside of your HTML, print your $output variable inside of your <select> element:
<select>
<?php
print("$output");
?>
</select>
It should print all options for every row that you have requested from the database.
Hope this helps :)
Try this:
$result = mysqli_query($con,"SELECT * FROM project");
if( mysqli_num_rows( $result )==0){
echo "<tr><td>No Rows Returned</td></tr>";
}else{
echo "<select name=Pname >";
while ($row = mysqli_fetch_assoc($result)) {
echo "<option value="$row["project_no"]">"$row["project_name"]"</option>";
}
echo "</select>";
}
This is the result code that i can run it. I put this code in a form code of html
$result = mysqli_query($con,"SELECT * FROM project"); ?>
<?php
$output = '';
if(mysqli_num_rows($result) == 0){
// echo error;
} else {
echo " <select name = Pname>";
while($row = mysqli_fetch_assoc($result)){
$project_no = $row['project_no'];
$project_name = $row['project_name'];
$output = "<option value=" . $project_no . "> ". $project_name ." </option>";
print("$output");
}
echo " </select>";
}
?>
Thank you every one for helping me ^^
This script does not display the DB value in a drop down on the edit form.
<?php
echo "<select name='assign' value=''><option>Select name</option>";
while ($r = mysql_fetch_array($result)) {
$value = $r['name'];
echo "<option value=" . $r['emp_id'] . ">" . $r['name'] . " if ($name=='$value') echo 'selected = 'selected''></option>";
}
echo "</select>";
It does not show any error. How it can write in a correct way.
You can try this :
$echoSting = '<select name="assign"><option value="">Select name</option>'.PHP_EOL;
while($r = mysql_fetch_array($result)) {
$value=$r['name'];
$echoSting .= '<option value="'.$r['emp_id'].'" '.($name==$value ? 'selected' : '').'>'.$r['name'].'</option>'.PHP_EOL;
}
$echoSting .= '</select>'.PHP_EOL;
echo $echoSting;
a side note, try looking into PDO for your database stuff : http://php.net/manual/en/book.pdo.php
Try this:
echo "<select name='assign' value=''><option>Select name</option>";
while($r = mysql_fetch_array($result)) {
$value=$r['name'];
echo "<option value='.$r['emp_id'].'>'.$r['name'].' "; if ($name=='$value') echo "selected = 'selected'";echo">$value</option>";
}
echo "</select>";
I am populating a dropdown menu with data in a database, using this code:
$query = "SELECT * FROM my_gallery";
$execute = mysqli_query($link, $query);
$results = mysqli_num_rows($execute);
if ($results!=0) {
echo '<label>The galleries are: ';
echo '<select id="galleries" name="galleries">';
echo '<option value=""></option>';
for ($i=0; $i<$results; $i++) {
$row = mysqli_fetch_array($execute);
$name = htmlspecialchars($row['galleryName']);
echo '<option value="' .$name. '">' .$name. '</option>';
}
echo '</select>';
echo '</label>';
}
I want to add the selected=selected to the option selected and then use that option in another query, but I have problem adding the selected tag in the actual selected entry.
More INFO:
I am using dropzone.js to upload images, and I want to select the category on the fly. The category has to be selected from the dropdown menu and used in the INSERT query.
if you have a form with POST then the code is
$selected = "";
if($_POST['galleries']==$name) $selected=" selected='selected'";
echo '<option value="' .$name. '"'.$selected.'>' .$name. '</option>';
Just as an example try this,
$query = "SELECT * FROM my_gallery";
$execute = mysqli_query($link, $query);
$results = mysqli_num_rows($execute);
if ($results!=0) {
echo '<label>The galleries are: ';
echo '<select id="galleries" name="galleries">';
echo '<option value=""></option>';
for ($i=0; $i<$results; $i++) {
$row = mysqli_fetch_array($execute);
$name = htmlspecialchars($row['galleryName']);
if( $i == 1 )
{
$sel = 'selected="selected"';
}
else
{
$sel = '';
}
echo '<option value="' .$name. '"'. $sel .' >' .$name. '</option>';
}
echo '</select>';
echo '</label>';
}
If you want to select a specific option then you are going to have a variable containing that specific option example:
$conditionalName = "Mark";
$query = "SELECT * FROM my_gallery";
$execute = mysqli_query($link, $query);
$results = mysqli_num_rows($execute);
if ($results!=0) {
echo '<label>The galleries are: ';
echo '<select id="galleries" name="galleries">';
echo '<option value=""></option>';
for ($i=0; $i<$results; $i++) {
$row = mysqli_fetch_array($execute);
$name = htmlspecialchars($row['galleryName']);
echo '<option value="' .$name. '"'; if($name=$conditionalName){ echo ' selected=selected';} echo '>'. $name. '</option>';
}
echo '</select>';
echo '</label>';
}
I'm trying to get a drop down menu to keep its selected value when the user hits submit, but it fails due to errors on the form.
I have a while loop returning values from a database to build the options for the drop down, but how do I echo "selected" on the right option?
I have tried if($district == $row["name"]) { echo "selected";} as you see below, but it doesn't work.
<?php
$result = mysql_query("SELECT dist.name FROM districts AS dist JOIN int_bd AS ibd ON dist.id = ibd.districts_id WHERE banners_id = 6 GROUP BY dist.id ORDER BY dist.id ASC", $connection);
if (!result) {
die("Database query failed: " . mysql_error());
}
while ($row = mysql_fetch_array($result)) {
echo '<option value="{$row["name"]}"'; if($district == $row["name"]) { echo "selected";} ; echo '>' . $row["name"] . "</option>";
}
?>
Sorry for the delay. None of the suggested answers worked for me. Any other ideas?
Can you try this,
<?php
$result = mysql_query("SELECT dist.name FROM districts AS dist JOIN int_bd AS ibd ON dist.id = ibd.districts_id WHERE banners_id = 6 GROUP BY dist.id ORDER BY dist.id ASC", $connection);
if (!result) {
die("Database query failed: " . mysql_error());
}
$district = $_REQUEST['name']; // You need pass the value you have been submitted
while ($row = mysql_fetch_array($result)) {
$selected ="";
if(trim($district) == trim($row["name"])) { $selected = "selected";}
echo '<option value="{$row["name"]}" '.$selected.' >' . $row["name"] . "</option>";
}
?>
Try this..
if($district == $row["name"])
{
echo "<option value='$district' selected>$district</option>";
}
I just found the answer. This is what I did:
while ($row = mysql_fetch_array($result)) {
echo '<option value="' . $row["name"] . '"';
if($row["name"] == $district) { echo 'selected';} ;
echo '>' . $row["name"] . '</option>';
}
It seems to have been this line
echo '<option value="{$row["name"]}"';
that was causing the problem.
I am new to php, i created drop down which calling data from mysql data base, user selects option and its save to data base.
Problem Arises in edit form in which its do not showing selected value.
Drop Down code is below:
$query = 'SELECT name FROM owner';
$result = mysql_query($query) or die ('Error in query: $query. ' . mysql_error());
//create selection list
echo "<select name='owner'>\name";
while($row = mysql_fetch_row($result))
{
$heading = $row[0];
echo "<option value='$heading'>$heading\n";
}
echo "</select>"
Please advise solution for the edit form.
Thanks in Advance
you must close <option> tag:
echo "<option value='$heading'>$heading</option>";
$query = 'SELECT name FROM owner';
$result = mysql_query($query) or die ('Error in query: $query. ' . mysql_error());
//create selection list
echo "<select name='owner'>\name";
while($row = mysql_fetch_row($result))
{
$heading = $row[0];
?>
<option <?php if($heading=="SOMETHING") { echo "selected='selected'"; } ?> value="SOMETHING">SOMETHING</option>
<option <?php if($heading=="SOMETHING2") { echo "selected='selected'"; } ?> value="SOMETHING2">SOMETHING2</option>
<option <?php if($heading=="SOMETHING3") { echo "selected='selected'"; } ?> value="SOMETHING3">SOMETHING3</option>
<?php
}
echo "</select>"
I'd do it this way.
$numrows = mysql_num_rows($result);
if ($numrows != 0){
echo "<select name='owner'>\name";
while ($x = mysql_fetch_assoc($result)){
echo "<option value='".$x['heading']."'>".$x['heading']."</option>";
}
echo "</select>";
}
$x['heading'] is using the value of the row 'heading' in the database
It's much more efficient and simply looks more sophisticated.