在选择组合框值时，必须在从数据库中选择查询后动态显示标签
问题描述：
In a file following code is written which populate the value of combobox :
<select id="company" required="required" class="form-control" value = "select" name="subject_code" placeholder="Select" >
<?php
//Iterating list of subjects available to be filled .
echo "<option> Select </option>";
foreach ($subjects_to_fill as $subject_id => $subject_name) {
# code...
echo "<option value=".$subject_id."> ".$subject_name." </option>";
}
?>
</select>
On selecting a particular item from the combobox, I want to display the faculty_name dynamically from faculty_table on the basis of $subject_id.
table structure:
faculty_table(faculty_id,faculty_name,subject_id)
subject_table(subject_id,subject_name,faculty_id)
答
You will need to use ajax for this task.
Add a <div> and below script in your current file.
<div id="faculty"></div>
<script type="text/javascript">
function get_faculty()
{
var id = document.getElementById("company").value;
var dataString = 'id='+ id;
$.ajax
({
type: "POST",
url: "get_faculty.php",
data: dataString,
success: function(html)
{
$("#faculty").html(html);
}
});
}
</script>
Now create another get_faculty.php file in same folder which will be called on onchange event of company dropdown.
Write below code in that file
if($_POST['id'])
{
$id=$_POST['id'];
$con=mysqli_connect("localhost","username","pass","dbname");
$sql=mysqli_query($con,"SELECT faculty_name from faculty_table where subject_id = ".$id);
while($row=mysqli_fetch_array($sql))
{
echo $row['faculty_name'];
}
}
Dont forget to write mysqli_connect credentials and query accordingly.