Mark an option as selected - php

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>';
}

Related

how to get value from database and show in dropdown list

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 ^^

Retrieve value from DB and display in a drop down on edit form in php

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>";

Show all tables from database into select form in php

Hiho,
I have a little problem with showing all table names on page in form .
Code bellow:
<select name="users" onchange="showTables(this.value)">
<option value="">Select a table:</option>';
$result = mysql_list_tables($db_name);
for ($i = 0; $i < count(mysql_num_rows($result)); $i++){
echo '<option value="' . mysql_tablename($result, $i) . '">' . mysql_tablename($result, $i) . '</option>';
}
echo '</select>
</form>
<br>
<div id="tablesDb"><b>All content from selected table will be listed there.</b></div></div>';
I try with mysqli too ,and i get results but still without names of the tables and nothing can be select.
Maybe someone know how to get this work.
First, you definitely want to be using MySQLi, as the MySQL extension is deprecated. Second, you can probably get all of the data you need in a simple SELECT query like this:
select `TABLE_NAME` from information_schema.tables
Try this
$mydbname = 'database_name';
$con=mysqli_connect("localhost","my_user","my_password",$mydbname);
// Check connection
if (mysqli_connect_errno()){
echo "Failed to connect to MySQL: " . mysqli_connect_error();
exit();
}
$options = '';
// for the query you can use two of the following lines (line1+line2 OR line3+line4)
//Use line (line1+line2)
$result = mysqli_query($con,"SHOW TABLES");
$column_name ='Tables_in_'.$mydbname;
// OR (line3+line4
$result = mysqli_query($con,"SELECT TABLE_NAME AS tbl FROM information_schema.tables" WHERE TABLE_SCHEMA = \"'.$mydbname.'\"; ");
$column_name ='tbl';
while($row = mysql_fetch_array($result))
$options .= '<option value="' . $row[$column_name] . '">' . $row[$column_name] . '</option>';
echo '<select name="users" onchange="showTables(this.value)">';
echo '<option value="0">Select a table:</option>';
echo $options;
echo '</select>';
I modify code to this:
<select name="db_tablelist" onchange="showTables(this.value)">
<option value="">Select a table:</option>';
$result = mysqli_query($con,"SHOW TABLES");
while($row = mysqli_fetch_array($result)) { echo '<option value="' . $row[0] . '">'.$row[0].''; } echo '';
echo '</select>
</form>
<br>
<div id="tablesDb"><b>All content from selected table will be listed there.</b></div></div>';
And now everything works .

PHP: select current option

I have a DB table with all categories (id, category) and table with all of events (id, event, categoryID).
And on the event editing form I have the select field with all of the categories (getting from the DB). But sinse I'm developing an editing form, I need to select current category by default.
This is my select field (PHP method that gets all the categories from DB and puts them in the following order):
<option value="1">Cat1</option>
<option value="2">Cat2</option>
<option value="3">Cat3</option>
<option value="4">Cat4</option>
<option value="5">Cat5</option>
Let's say, current event is under category 3, so I need the following HTML to be generated:
<option value="1">Cat1</option>
<option value="2">Cat2</option>
<option value="3" selected>Cat3</option>
<option value="4">Cat4</option>
<option value="5">Cat5</option>
How do I achieve it with PHP, if I have the catID?
Hopefully, this question is clear enough. Sorry for my bad explanation
UPD: This is my PHP code that generates category list:
public function getCatList($conf) {
$mysqli = $this->dbConnect($conf);
// Quering...
$query = "SELECT * FROM categories";
$result = $mysqli->query($query);
while($row = mysqli_fetch_array($result)) {
echo '<option value="' . $row['id'] . '">' . $row['category'] . '</option>';
}
}
Appending to your while iteration a condition will solve this for you:
while($row = mysqli_fetch_array($result)) {
$isSelected = $row['id'] == $catID;
echo '<option '.($isSelected ? 'selected="selected"' : '').' value="' . $row['id'] . '">' . $row['category'] . '</option>';
}
You're making a comparison if the current value is the same as that stored in $catID - and store the boolean result in a variable. In the echo you're just doing a conditional forking and appending the selected attribute if the value was true, otherwise not appending any empty string.
You can do it like
while($row = mysqli_fetch_array($result)) {3
echo '<option value="' . $row['id'] . '"';
//if condition is met then make the option selected
if($row['categoryID'] == 3) {
echo " selected='selected' ";
}
echo '>' . $row['category'] . '</option>';
}
you should add the selected catID as a parameter to your getCatList-function. Then just change the creation of your HTML to include the selected-attribute:
public function getCatList($conf, $catID = 0) {
$mysqli = $this->dbConnect($conf);
// Quering...
$query = "SELECT * FROM categories";
$result = $mysqli->query($query);
while($row = mysqli_fetch_array($result)) {
echo '<option value="' . $row['id'] . '"' . ($row['id']==$catID?' selected':'') . '>' . $row['category'] . '</option>';
}
}
now you can just pass the catID to your function:
$myConf = "something";
$myCatID = 3;
getCatList($myConf, $myCatID);
//query for the categories and options
$current_value = 3;
$total_num_of_options = mysql_num_rows($your_query_result);
for($count = 1; $count <= $total_num_of_options){
echo "<option value='".$count."' ";
if($count == $current_value)
echo "selected";
echo ">cat".$count."</option>";
}

Find previously selected option in while loop

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.

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