select id from the select gender dropdown list - php

I have a dropdown list of gender. I am getting the values from my table 'candidate' and in this table i have a field which is actually a foreign key to another table.
The field is gender_code_cde, and the table name is gender_code. Now gender_code contains 2 rows. and id and a description. Its structure is like:
1->Male
2->Female
Its being displayed in dropdown list. but I want to get the id 1 if male is selected and id 2 if female is selected. My code is below:
<p>
<label for ="gender">Gender:</label>
<?php
$query = mysql_query("select * from gender_code"); // Run your query
echo '<select name="GENDER">'; // Open your drop down box
//Loop through the query results, outputing the options one by one
while ($row = mysql_fetch_array($query))
{
echo '<option value="'.$row['gender_cde'].'">'.$row['gender_dsc'].'</option>';
}
echo '</select>';
?>
</p>
I am working in codeIgniter.

// this will alert value of dropdown whenever it will change
<script>
function getvalue(val)
{
alert(val);
}
</script>
<p>
<label for ="gender">Gender:</label>
<?php
$query = mysql_query("select * from gender_code"); // Run your query
echo '<select name="GENDER" id="Gender" onchange="getvalue(this.value)">'; // Open your drop down box
//Loop through the query results, outputing the options one by one
while ($row = mysql_fetch_array($query))
{
echo '<option value="'.$row['gender_cde'].'">'.$row['gender_dsc'].'</option>';
}
echo '</select>';
?>
</p>

In Javascript, you can get the value with native JS.
var e = document.getElementById("give-the-select-element-an-id");
var gender_cde = e.options[e.selectedIndex].value;
If, you want it back at the server, it will come in to PHP as $_GET['GENDER'] or $_POST['GENDER'] depending on if the form does a POST or a GET request.

If you are submitting a (POST) form and this is inside that form you can access the the value (id) from CodeIgniter's $_POST wrapper: $this->input->post('GENDER');. I don't use CodeIgniter so I could be wrong, but it will be located in your $_POST array from PHP.
If you just want to replicate the variable somewhere on the page you can just echo it. Or set via js/jQuery with document.getElementById('#something').value = value; or jQuery('#something').html(value);.
You should also change the following:
use a foreach in place of your while:
foreach ($query as $row) {
echo $row['gender_x'] . $row['gender_y'];
}
use the CodeIgniter SQL wrapper instead of depreciated PHP functions. The query part is $this->db->query('SELECT statement goes here');. You should look at your CodeIgniters' version documentation for a more in depth explanation.
EDIT: To make it a bit more clear, an example:
$query = $this->db->query('SELECT * FROM gender_code');
foreach ($query->result() as $row) {
echo '<option value="' . $row->gender_cde . '">' . $row->geneder_dsc . '</option>';
}
This is assuming you have setup the previous calls in CodeIgniter. Please see http://ellislab.com/codeigniter/user-guide/database/examples.html and you may wish to read http://ellislab.com/codeigniter/user-guide/

Related

PHP/MySQL Auto select option based on previous page

I have a music database with a PHP front end where you can add/edit/delete artists, albums, tracks using the web client. The last real problem I have is getting a select box to automatically select an option I pass to the page.
Example:
On a page called 'createCD.php' I have this code:
echo "<td><a href=\"editCD.php?cdTitle=".rawurlencode($row['cdTitle'])."&cdID=$row[cdID]&artID=$row[artID]&cdGenre=".rawurlencode($row['cdGenre'])."&cdPrice=$row[cdPrice]\" </a>Edit</td>";`
This is used as a link to the next page, and collects all the information about an album in the database and sends in to a page called 'editCD.php'.
Now on this page, all the information is used to fill out the webpage as shown here (there is more but for the purposes of this post, only the first select box matters):
Artist Name:
<!-- dropdown with artist name -->
<?php
echo '<select name= "artID" id="artID">';
while ($row = mysqli_fetch_assoc($result)){
echo '<option value="'.$row['artID'].'">'.$row['artName'].'</option>';
}
echo '</select>';
?>
<p>
Album Title:
<input id="cdTitle" type="text" name="cdTitle" value ="<?php echo htmlspecialchars($cdTitle); ?>" />
</p>
What I would like is for the "selected" option for 'artID' to be the value that is passed to the page. Using the associative array, I was able to display the 'artName' associated with the 'artID'. Currently, all the information about the album appears correctly apart from the 'artName' and it defaults to the first value. This is a problem as if a user simply clicks "Update" it will update the name to the default name, therefore changing the database entry by accident.
I know I need to be using
<option selected ...>
but I'm not sure on the syntax to use.
<?php
$artID = $_GET['artID']; // get the artID from the URL, you should do data validation
echo '<select name= "artID" id="artID">';
while ($row = mysqli_fetch_assoc($result)){
echo '<option value="'.$row['artID'].'"';
if ($artID == $row['artID']) echo ' selected'; // pre-select if $artID is the current artID
echo '>'.$row['artName'].'</option>';
}
echo '</select>';
?>
$artId = $_GET['artID'];
while ($row = mysqli_fetch_assoc($result)) {
$selected = $artId == $row['artID'] ? 'selected' : '';
echo '<option value="'.$row['artID'].'" '.$selected.'>'.$row['artName'].'</option>';
}
First you get the id via $_GET['artID']. (In a real scenario use intval or something to prevent sql injection)
Then check in the loop if the id from database is the same as the id from GET and when it is print "selected, else nothing.

Assign and return select option dynamically?

I have a form select which is generated based on results returned from a mysql query.
Q1) How do I assign a name identity (integer value) based on the teamID pull out of the database?
Q2) How could I then get the option selected and add it to a php variable which could then be used to update a table based on the users selection?
Below is the code I have so far for creating the dynamic drop down list, which get the results out of the database.
<?php
$data= mysql_query("SELECT * FROM teams
WHERE teamID NOT IN (
SELECT TeamID
FROM leagueInformation
WHERE leagueID = 1
)
") or die(mysql_query());
echo "<select name=\"team\" class=\"col-lg-12\" style=\"padding:10px; background:#e1e1e1;\">\n";
while($info = mysql_fetch_array( $data ))
{
$teamID = $info['teamID'];
echo "<option name=" . $team . " value=" . $teamID . ">" .$info['teamName'] . "</option>";
}
echo "</select>\n";
?>
Q1) How do I assign a name identity (integer value) based on the teamID pull out of the database?
This question is not so clear, but I think this is what you are trying to do?
<?php $data= mysql_query(
"SELECT * FROM teams WHERE teamID NOT IN (
SELECT TeamID
FROM leagueInformation
WHERE leagueID = 1)
") or die(mysql_query());
?>
<form name="teams_form" method="POST">
<select name="team" class="col-lg-12" style="...">
<?php while($info = mysql_fetch_array( $data )): ?>
<option value="<?php echo $info['teamID'] ?>">
<?php echo $info['teamName'] ?>
</option>
<?php endwhile; ?>
</select>
</form>
Q2) How could I then get the option selected and add it to a php variable which could then be used to update a table based on the users selection?
First you will need to wrap your select inside a form (like I did), then you can check the selected value like this:
if(isset($_POST['team']) && !empty['team']){
$selected_team = $_POST['team'];
mysql_query("Do what you want with the selected team");
}
Some notes about your code:
Avoid to use mysql_*, use mysqli_* or PDO, and always prevent SQL injections
<option> tags don't have name properties, and in this case $team is not defined
Embed PHP inside HTML, not HTML in PHP.
Your first question doesn't make sense to me yet, I'll update the answer when it does.
When the user submits the form, you can get the option selected using the $_GET or $_POST (depending your your form's method) variables.
Like this:
$teamId = $_POST['team'];
put everything in a <form/> with an action pointing to the page, and a method (get or post), then in the script get the variable by the <select/> name value reading $_GET or $_POST variables, and I quote the comment about the "name" in option as invalid attribute.

Dropdown menu not displaying change when I change product

I want to access the value selected by user for further processing.
Hence I am using post method to pass the values of whole form.
But GET to access cust_id so that I can reflect change in
further parts of my form. Hence I had to post the following line:
<select id='fullname' onChange="window.location='sp_menu.php?product='+this.value" name='fullname'>
outside php code. But now, once I select some option from dropdown menu, URL changes accordingly, but dropdown menu does not reflects the change
<?php
$query = "SELECT Cust_id, Cust_Name, Cust_City FROM Customers";
$result = mysql_query($query);
?>
<select id='fullname' onChange="window.location='sp_menu.php?product='+this.value" name='fullname'>
<?php
while($row = mysql_fetch_assoc($result)) {
echo '<option value="'.$row['Cust_id'].'">'.$row['Cust_Name'].','.$row['Cust_City'].'</option>';
}
echo '</select>';
?>
How can I, in the same form, access the address of the particular customer id from database when user selects customer name from this dropdown menu?
I think you mean when you change dropdown, the value is not retained, it obviously won't be because your page is being refresh, you need to GET the value from url and put a selected attribute to have that value selected.
Do it this way:
<?php
$query = "SELECT Cust_id,Cust_Name,Cust_City FROM Customers" ;
$result = mysql_query($query);
//checking if GET variable is set, if yes, get the value
$selected_option = (isset($_GET['product'])) ? $_GET['product'] : '';
//we will store all the dropdown html code in a variable and display it later
$select = "<select id='fullname' onChange=\"window.location='sp_menu.php?product='+this.value\" name='fullname'>";
while($row = mysql_fetch_assoc( $result )) {
//checking if the cust_id matches the GET value,
//if yes, add a selected attribute
$selected = ($selected_option==$row['Cust_id'])?'selected':'';
echo '<option value="'.$row['Cust_id'].'"'. $selected. '>' . $row['Cust_Name'] .' , '.$row['Cust_City']. '</option>';
}
$select .= '</select>';
//display the dropdown
echo $select;
?>

Setting combo box to value dynamically retrieved from database in PHP

I have a task that I need to retrieve data from the database and set it in the Combo Box. Fortunately, I have done it.
Now, I have a Search Button which retrieves the data relevant in these text and combo boxes. My Issue is, After I click Search Button all my combo box and text box selected values become empty. How can I set those same data after clicking Search button ?
My Code Effort is,
<?php
$sql="select cat_id,cat_name from disease_category group by cat_id ";
foreach ($dbo->query($sql) as $row){
if(isset($_REQUEST['cat_name'])&&$_REQUEST['cat_name']==$row[cat_name])
{
echo "<option value=$row[cat_id] selected='selected' >$row[cat_name]</option>";
}
Else
{
echo "<option value=$row[cat_id]>$row[cat_name]</option>";
}
}
?>
My SEARCH button code,
<?php
include 'config.php';
if(isset($_REQUEST['SUBMIT']))
{
$cat=$_REQUEST['cat'];
$subcat=$REQUEST['subcat']
$sel=mysql_query("SELECT * from table_name where cat_id like '$cat%' AND sub_id like '$sub_cat%'AND survier like '$survier%' ")
}
It should be pretty simple. I still don't fully understand what you're trying to do. But if all you want is to dynamically populate an options list based on the results of a SQL query.
<?php
$sql = '
SELECT
*
FROM
`my_table`
';
$query = mysql_query($sql) OR die(mysql_error());
print '<select name="dropdown">';
while ($row = mysql_fetch_assoc($query)) {
print '<option value="'. $row['cat_id'] .'"';
if (
isset($_REQUEST['cat_name']) &&
$_REQUEST['cat_name'] == $row['cat_name']
) { print ' selected="selected"'; }
print '>'. $row['cat_name'] .'</option>';
}
print '</select>';
?>
You should be able to modify the SELECT query to fit your needs, and modify content within the while() loop, as well. That should get you going if I understand what you're trying to do.

How to check a check box if it's value is in DATABASE. PHP

I have inserted some check box values in mysql database using PHP
And in the below image i have fetch the values:
Now i need the o/p like the below image: The values which i inserted in the database should be checked
Hope now its clear.
Thanks in advance..
You should have a table of available options (in this case, something like a cities table), and then a user-to-cities look-up table. Then you can loop over the cities, but also fetch which cities the user has checked.
A sample, without knowing your database structure, would be as follows:
$uid = $_SESSION['user']['id']; // your logged in user's ID
$cities = array();
// get an array of cities
$sql = "SELECT id, name FROM cities";
$res = mysql_query($sql);
while ($row = mysql_fetch_object($res)) {
$cities[$row->id] = $row->name;
}
// get an array of cities user has checked
$sql = "SELECT DISTINCT city_id FROM users_cities WHERE user_id = '$uid'";
$res = mysql_query($sql);
while ($row = mysql_fetch_object($res)) {
$checked[] = $row->city_id;
}
// this would be templated in a real world situation
foreach ($cities as $id => $name) {
$checked = "";
// check box if user has selected this city
if (in_array($checked, $id)) {
$checked = 'checked="checked" ';
}
echo '<input type="checkbox" name="city[]" value="'.$id.'" '.$checked.'/>';
}
If I understand you question properly, the obvious and simplest approach is that you need to fetch records from database and when producing HTML [in a loop ot something similar] check if that value exists in array to results. You haven't given us any examples of your DB structure or code, so you must figure it our yourself how to do it.
Usually, you insert the values into the database. After inserting, you should have access to the same values again. It's not clear how you set up your script, so let's assume you redirect to another script.
What you need to do is retrieve the values for the checkboxes from your database again. Then you know which are selected. This can be used to determine if your checkbox need to be checked or not.
Note:
I assume here that the result of your query is an array with
the selected Ids as a value.
I assume here that your fields are stored in the result of
some query and is basically an array
with Field Id as key and Field Name
as Value.
E.g., something like this:
<?php
// Retrieve values from database.
$resultArray = ... some code ... ;
?>
<?php foreach ($field_types as $field_name => $field_value): ?>
<input type="checkbox" name="<?php echo $field_name; ?>" value="<?php echo $field_value ?>" <?php if (in_array($field_name, $resultArray)) { echo 'checked'; }/>
<?php endforeach; ?>
This results in a checkbox which is checked, if the field_name is inside the result array (with your already checked results). Otherwise they're just rendered as unchecked checkboxes. Hope this is clear enough.

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