this is my code which populates a drop down menu, all working perfectly, but when editing a database record, i want the first value in the drop down to be what is currently in the database, how would i do this?
<li class="odd"><label class="field-title">Background <em>*</em>:</label> <label><select class="txtbox-middle" name="background" />
<?php
$bgResult = mysql_query("SELECT * FROM `backgrounds`");
while($bgRow = mysql_fetch_array($bgResult)){
echo '<option value="'.$bgRow['name'].'">'.$bgRow['name'].'</option>';
}
?>
</select></li>
You would set the selected="selected" attribute on the relevant <option>. Presumably there would be some sort of check in your while loop, checking against the variable that contains the current value.
You can do like:
$counter = 1;
while($bgRow = mysql_fetch_array($bgResult)){
if ($counter === 1)
{
echo '<option value="'.$bgRow['name'].'" selected="selected">'.$bgRow['name'].'</option>';
}
else
{
echo '<option value="'.$bgRow['name'].'">'.$bgRow['name'].'</option>';
}
$counter++;
}
As can be seen I have added selected="selected" for you so it will work automatically for you :)
Correct me if I'm wrong, I believe that you have another database table which holds the selected background (e.g. users table with background field), you need to do a query to get the background from the other table and add selected="selected" attribute to the option tag of the background, please check the code below (hope it helps):
<?php
$result = mysql_query("SELECT `background` FROM `users` LIMIT 1");
$myBg = mysql_fetch_array($result, MYSQL_ASSOC);
$bgResult = mysql_query("SELECT * FROM `backgrounds`");
while($bgRow = mysql_fetch_array($bgResult)){
if($myBg['background'] == $bgRow['name'])
echo '<option value="'.$bgRow['name'].'" selected="selected">'.$bgRow['name'].'</option>';
else
echo '<option value="'.$bgRow['name'].'">'.$bgRow['name'].'</option>';
}
?>
Related
I can't seem to get this right. I am trying to make a select box that gets it values from the database sticky please see my code below :
<select name="cob_bank">
<?php while($row = mysqli_fetch_array($query,MYSQLI_ASSOC)){
if($row['Id']==$myId) {
echo '<option value="'.$row['Id'].'"selected="selected">'.$row['Name'].'</option>';
} else {
echo '<option value="'.$row['Id'].'">'.$row['Name'].'</option>';
}
}?>
</select>
after the select box is filled in and the submit button is pressed where I save my Post variable
$myId = $_POST['cob_name'];
The above code does not work I please assist
the no space could be causing the problem as a space before selected
if($row['Id']==$myId) {
echo '<option value="'.$row['Id'].'" selected="selected">'.$row['Name'].'</option>';
}
The select name should be set as:
<select name='cob_name'>
Is there a way to find out what is the value of the <option> in a <select> tag?
I have a form that contains a tag, with several options. Then, I want to insert the selected value on the database.
Is this doable with PHP? Or do i need JS/Jquery?
EDIT: this is what i have tried:
This creates the variable and stores the value of the option.
$query=mysql_query("Select cargo from utilizadores where nome<>'admin'");
echo '<select name="icargo[]">';
while ($row = mysql_fetch_array($query, MYSQL_ASSOC))
{
foreach ($row as $cargo){
echo '<option value="$cargo" name="icargo[]">'.$cargo.'</option>';
}
}
echo '</select>';
I send the variable this way, and then i try to receive it with a foreach:
$countcargo = 0;
foreach ($_POST["icargo"] as $cargoo){
$cargoarray[$countcargo] = $cargo;
$countcargo++;
}
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;
?>
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.
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/