Set the value of a <select> as it is popualted - php

I am populating a select list with a PHP MySQL query. At the same time, I also want to set the select value to the current value in the db. It goes as such
1) Query the associated table and populate the <select> with each entry as an <option>
2) Set the <select> value to the current entry in the database
Right now I am doing this, but when I view the page the <select> is simply at the first option value.
echo '
<select name = "arena" type = "text" value = "'.$row['arena'].'"> ';
try{
$retrieveArena = $db->prepare("SELECT arena_id AS key1, arena_name AS val1 FROM arenas");
$retrieveArena->execute();
$retrieveArenaResult = $retrieveArena->fetchAll(PDO::FETCH_ASSOC);
if ($retrieveArena->rowCount() > 0){
foreach ($retrieveArenaResult as $arena){
echo '<option value = "'.$arena['key1'].'">'. $arena['val1'].'</option>
';
}
}else{
returnMsg("No arenas found");
}
}
catch(PDOException $e){
echo $e->getMessage();
}
echo '
</select>';

You need to give the correct <option> tag the selected attribute. So you will need to know what the selected option should be. Then you can use code like this to build the options.
foreach ($retrieveArenaResult as $arena){
echo '<option value = "' . $arena['key1'] .'"';
if (this option is selected) {
echo ' selected';
}
echo '>'. $arena['val1'].'</option>
';
}

Related

select dropdown value according to previous valuesubmit new one

I use the following code to select all value from a DB table and put it in dropdown. Here is the code :
<select name="dis_name">
<?php
$qu="Select DISTINCT name from root";
$res=mysqli_query($con,$qu);
while($r=mysqli_fetch_row($res))
{
echo "<option value='$r[0]'> $r[0] </option>";
}
?>
Based on previous code a value is selected and submitted. I want to show previously selected value in dropdown on another page so i use following code:
<select name="dis_name">
<option value="RAM"<?php echo $r20 == "RAM" ? " selected" : ""; ?>>RAM</option>
<option value="ROHAN"<?php echo $r20 == "ROHAN" ? " selected" : ""; ?>>ROHAN</option>
</select>
This code work fine till no new value added in DB table. But whenever i add new value i have to update edit code manually. Is there any way, so that dropdown previous selection updated automatically in edit page.
Add the condition in your loop, eg:
<?php
while($r=mysqli_fetch_row($res))
{
echo "<option value='$r[0]'";
if ($r[0] == $r20) echo " selected";
echo "> $r[0] </option>";
}
?>
If I get what you want (assuming that $r20 is the previously selected value:
<?php
$qu="Select DISTINCT name from root";
$res=mysqli_query($con,$qu);
while($r=mysqli_fetch_row($res))
{
echo "<option value='$r[0]'";
if($r[0]==$r20){echo 'selected="selected"';}
echo "> $r[0] </option>";
}
?>
this way the select will display as selected the option that match the $r20 value dinamically.
BTW, you should use id in your code. This way you could also have more than one user with the same name (but different ID). With DISTINCT, as you are doing, if you have more than one user with the same name you will get only one.

How to Select Multiple Values from Database in php

How to solve
if i select one Value form dropdown. Biside I have another drop down and its values should be auto Updated According to Option I have selected from first dropdown ?
you can try to generate a lookup table whenever you create a drop-down list:
function createDropdown(&$ddlLookup) {
echo '<select multiple name="items[]">';
try {
$items = mysql_query("SELECT item_id,item_type FROM items");
while ($row = mysql_fetch_assoc($items)) {
echo '<option value="'.$row['item_type'].'"';
echo '>'. $row['item_type'] . '</option>'."\n";
$ddlLookup[$item_type] = $item_id;
}
}
catch(PDOException $e) {
echo 'No results';
}
echo '</select>';
}
Then whenever you need the id for a given description you use that table(array) to get it:
$mainDropdownLUT = array();
createDropdown($mainDropdownLUT);
var_dump($mainDropdownLUT['testCow']);
-> 734
Also, if you need to pass it to another page it can be serialized and added to a hidden field.
$mainDropdownLUT = serialize($mainDropdownLUT);
"<input type="hidden" value =\"$mainDropdownLUT\">"
-------------------------**OTHER PAGE **--------------
$mainDropdownLUT = unserialize($mainDropdownLUT);

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.

Showing current value in DB in a drop down box

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>';
}
?>

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