Using PHP to Populate a <Select></Select> Dropdown

Using PHP to populate a select /select dropdown?

What about something like this :

echo '<select name="select">';
while($row=mysql_fetch_array($result))
{
    echo '<option value="' . htmlspecialchars($row['column_for_value']) . '">' 
        . htmlspecialchars($row['column_for_label']) 
        . '</option>';
}
echo '</select>';

Of course, up to you to decide which items from $row should be used for the value and the text of each <option>

Just make sure you are escaping the data that comes from your DB -- especially if it can contain HTML ; as you are outputting HTML, this can be done with htmlspecialchars or htmlentities.

Note that those might take a couple of additionnal parameters that I didn't use in my example -- setting those can be useful, depending on the charset you're using.

How to populate old select option values?

you can do like this

  @if(count($category_list)>0)
   <option value="null">Pick a Category</option>
   @foreach($category_list as $cl)
       <option value="{{$cl->id}}" @if(old('category_id') == $cl->id) selected="selected" @endif>{{$cl->name}}</option>
   @endforeach
 @else
   <option>No categories found</option>
@endif

How to use PHP to populate a select Dropdown with json data

you can use ' and " together:

  <select id="printer" name="printer">
        <option value="none">Select Printer</option>
        <?php 
          foreach($test as $t) {
            echo '<option value="'. $t['Name'].  '">'. $t['Name'].'</option>';      
          }
        ?>
    </select>

Populate select drop down from a database table

$query = "SELECT volunteers_2009.id, volunteers_2009.comments, volunteers_2009.choice1, volunteers_2009.choice2, volunteers_2009.choice3, volunteers_2009.lname, volunteers_2009.fname, volunteers_2009.venue_id, venues.venue_name FROM volunteers_2009 AS volunteers_2009 LEFT OUTER JOIN venues ON (volunteers_2009.venue_id = venues.id) ORDER by $order $sort";

$res = mysql_query($query);
echo "<select name = 'venue'>";
while (($row = mysql_fetch_row($res)) != null)
{
    echo "<option value = '{$row['venue_id']}'";
    if ($selected_venue_id == $row['venue_id'])
        echo "selected = 'selected'";
    echo ">{$row['venue_name']}</option>";
}
echo "</select>";

Populating dropdown - PHP Ajax MySQL

If you want a more dynamic solution (that will accommodate changes to the background DB) you can do something like this on your page:

<script>
        function changeSecond(first){
        var xmlhttp;
        if (window.XMLHttpRequest)
          {// code for IE7+, Firefox, Chrome, Opera, Safari
          xmlhttp=new XMLHttpRequest();
          }
        else
          {// code for IE6, IE5
          xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
          }
        xmlhttp.onreadystatechange=function()
        {
          if (xmlhttp.readyState==4 && xmlhttp.status==200)
            {
            var res=xmlhttp.responseText;
            document.getElementById("second").innerHTML=res;
            }
          }
        xmlhttp.open("GET","second_script.php?first="+first,true);
        xmlhttp.send();
        }
        </script>
...
<select onChange="changeSecond(this.value)">
<option value="Degree">Degree</option>
<option value="City">City</option>
</select>
<div id="second"><select><option value=""></option></select></div>

and then a script similar to:

<?php
//database connection
$first=mysql_real_escape_string($_REQUEST["first"]);
$query="SELECT ".$first." FROM tablename GROUP BY ".$first;
$data=mysql_query($query);
echo "<select>";
while($row=mysql_fetch_row($data)){
echo "<option value=\"".$row[0]."\">".$row[0]."</option>";
}
echo "</select>";
?>

I guess for real flexibility you'd also want to dynamically populate that first one using mysql_field_name in another script similar to above

Laravel-5 how to populate select box from database with id value and name value

Laravel provides a Query Builder with lists() function

In your case, you can replace your code

$items = Items::all(['id', 'name']);

with

$items = Items::lists('name', 'id');

Also, you can chain it with other Query Builder as well.

$items = Items::where('active', true)->orderBy('name')->lists('name', 'id');

source: http://laravel.com/docs/5.0/queries#selects

---

Update for Laravel 5.2

Thank you very much @jarry. As you mentioned, the function for Laravel 5.2 should be

$items = Items::pluck('name', 'id');

or

$items = Items::where('active', true)->orderBy('name')->pluck('name', 'id');

ref: https://laravel.com/docs/5.2/upgrade#upgrade-5.2.0 -- look at Deprecations lists

Populate data from database in Select tag

This should do the trick:

<?php
    $q1 = "select * from bloodbank_facility";
    $options=[];
    $q = $conn->query($q1);
    while($row1 = $q->fetch_array($q1)){
        $options[] = "<option>{$row1[1]}</option>";
    }
?>
<select name="facility_name" required>
  <?php echo implode($options) ?>
</select>

The main thing you didn't do was have the curly braces around $row1[1] which meant PHP thought you just wanted $row1 rather than the array value.

Notice: I already fixed it and thank you for your help. This is the solution that I came up with.

<select name="facility_name" required style="width:260px;">
          <?php
              $q1 = $conn->query("SELECT bloodbank_name FROM bloodbank_facility");
            
          ?>
          <?php
            while($rows = $q1->fetch_assoc()){
              $facility_name= $rows['bloodbank_name'];
              echo "<option value='$facility_name'>$facility_name</option>";
            }
          ?>
          </select>

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