CODEWITHSUNDEEP
phphtmlmysqlhtml-select
anggimery
anggimery

Reputation: 41

how to set the selected value tag <select> html from database in php?

I'm trying to create a drop down menu that will select a value that is stored in the database. here's the code :

require 'koneksi.php';
    $sql_select = "SELECT * FROM supplier"; 
    $hasil = mysql_query($sql_select);
    if(!$hasil) {
        echo "data supplier not found ".mysql_error();
    }

    $id = $_GET["id"];
    $hasil2 = mysql_query("SELECT * FROM brg_supplier WHERE id_brg=".$id);
    $data = mysql_fetch_array($hasil2);
    if($hasil2) {
    $supplier = $data['nama_supplier'];
    }


<select name="supplier">
     <option value="">---pilih supplier---</option>
     <?php
        while($baris = mysql_fetch_array($hasil)){
     ?>
     <option value="<?php $baris['nama_supplier'] ?>" <?php if     ($supplier==$baris['nama_supplier']) echo 'selected="selected"'; ?> > <?php echo  $baris['nama_supplier']; ?> </option>; 
<?php   }?>
</select>

the problem is my code creates a dropdown with nothing selected. here's the screenshot : link

i've tried all the solutions in the stackoverflow. but the dropdown value still nothing selected. i know that it has to be something simple that i am missing but seriously i cannot figure it out. please anyone help, thanks!

Upvotes: 1

Views: 16773

Answers (5)

Felipe Rugai
Felipe Rugai

Reputation: 1388

I spent some time trying to find the best solution for this, and came up with a tiny little jQuery code.

First of all you should be using PDO or mysqli instead of mysql, since it's deprecated. Let's assume you've fixed that.

Inside the <form> tag, add an <input type="hidden"/> so that it can storage your database value, for example:

HTML

<form>
    <input id="valueFromDatabase" type="hidden" value="<?php echo $stringFromDB ?>"/>
</form>

Note: in this case, $stringFromDB is a variable that holds your query's return from DB.

So now we have the value of our database inside our HTML code. Now we just need to check if any of the options inside the <select> tag is equal to this value. We'll be doing it with jQuery:

jQuery

$( document ).ready(function() {
    $('option').each(function(){
        if (this.value == $('#valueFromDatabase').val()){
            this.setAttribute('selected', 'selected');
        }
    });
});

What's happening here? We are telling to jQuery to analyze all the <option> tags in the HTML and compare its value with our value from database; if its equal, we add the selected attribute to the equivalent <option>.

You can it working here (used a calendar example).

Hope that helps!

Upvotes: 0

ksealey
ksealey

Reputation: 1738

I think the problem lies in this line:

<option value="<?php $baris['nama_supplier'] ?>" <?php if     ($supplier==$baris['nama_supplier']) echo 'selected="selected"'; ?> > <?php echo  $baris['nama_supplier']; ?> </option>;

You're missing an echo and it looks funny :/ Try instead:

<option <?php $val=$baris['nama_supplier']; echo "value='$val'";  if($supplier==$val) echo "selected='selected'>";echo $val;?> </option>;

Upvotes: 2

nicolaenichifor
nicolaenichifor

Reputation: 111

<option value="<?php $baris['nama_supplier'] ?>

should be

<option value="<?php echo $baris['nama_supplier'] ?>

Upvotes: 0

Vishal Sharma
Vishal Sharma

Reputation: 1372

Try this out

require 'koneksi.php';
$sql_select = "SELECT * FROM supplier"; 
$hasil = mysql_query($sql_select);
if(!$hasil) {
    echo "data supplier not found ".mysql_error();
}

$id = $_GET["id"];
$hasil2 = mysql_query("SELECT * FROM brg_supplier WHERE id_brg=".$id);
$data = mysql_fetch_array($hasil2);
if(mysql_num_rows($hasil2) > 0) {
$supplier = $data['nama_supplier'];
}


<select name="supplier">
 <option value="">---pilih supplier---</option>
 <?php
    while($baris = mysql_fetch_array($hasil)){
 ?>
 <option value="<?php echo $baris['nama_supplier'] ?>" <?php if      ($supplier==$baris['nama_supplier']) {echo 'selected="selected"';} ?> > <?php echo  $baris['nama_supplier']; ?> </option>; 
<?php   }?>
</select>

Upvotes: 0

dev4092
dev4092

Reputation: 2891

    Try this way..

$sel="select f.*,c.category from final f, category c where f.category_id=c.id and f.id='$id'";
$data=mysql_query($sel);
$res=mysql_fetch_assoc($data);    

     <select name="cat">
                       <?php

                        $sql = mysql_query("SELECT * FROM category");
                        while ($row = mysql_fetch_array($sql)){
                           if($row['id'] ==  $res['category_id']){
                        echo '<option value="'.$row['id'].'" selected="selected">'.$row['category'].'</option>';
                           }else{
                       echo '<option value="'.$row['id'].'">'.$row['category'].'</option>';
                           }
                        }
                       ?>
                  </select>

Upvotes: 0

Related Questions