How to populate Dropdown values from database in php [duplicate] - php

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This question already has answers here:
Using PHP to populate a <select></select> dropdown? [duplicate]
(7 answers)
Closed 7 months ago.
Hello Friends how to get data from database in Drop-down list?
This is my code but i can't get data in Drop-down list,
This is my php code.
public function System_parameter_dropdown_value() {
$sql= " SELECT * FROM SYSTEM_PARAMETERS WHERE PARAMETER_KEY='ROLE_TYPE'";
$rs= mysql_query($sql) or die(mysql_error());
$option_list="<option value=0>Select Role Type</option>";
while($data= mysql_fetch_assoc($rs)) {
$option_list.="<option value='$data[PARAMETER_KEY]'>
$data[PARAMETER_VALUE]</option>";
}
return $option_list;
}
This is my html option list, When i run the code it's show nothing in Dropdown list
<label class="control-label" for="roleid" style="padding-right: 80px;">ROLE TYPE</label>
<select id="" name="roleid">
<?php echo $option_list['PARAMETER_KEY']['PARAMETER_VALUE'];?>
</select>
I am Stuck on this code , please Guide me How to Resolve it.

Answer 1

Try some thing like this:
$arr = array(1 => 'MP', 2 => 'UP');
echo '<select>';
foreach($arr as $key => $val)
{
echo '<option value="'.$key.'">'.$val.'</option>';
}
echo '</select>';
It will generate html like:
<select>
<option value="1">MP</option>
<option value="2">UP</option>
</select>
In your case it is like:
<select id="" name="roleid">
<?php echo System_parameter_dropdown_value();?>
</select>

Answer 2

that's because $option_list in not array but string:
<?php $option_list = $your_object->System_parameter_dropdown_value(); ?>
<label class="control-label" for="roleid" style="padding-right: 80px;">ROLE TYPE</label>
<select id="" name="roleid">
<?php echo $option_list;?>
</select>
Method should be changed too
public function System_parameter_dropdown_value()
{
$sql= " SELECT * FROM SYSTEM_PARAMETERS WHERE PARAMETER_KEY='ROLE_TYPE'";
$rs= mysql_query($sql) or die(mysql_error());
$option_list="<option value=0>Select Role Type</option>";
while($data= mysql_fetch_assoc($rs))
{
$option_list.=sprintf("<option value='%s'>%s</option>", $data['PARAMETER_KEY'], $data['PARAMETER_VALUE']);
}
return $option_list;
}

Answer 3

If you are writing the function System_parameter_dropdown_value inside a class, you have to create an object of the class and then call the function.
Like this
$classObj = new className();
echo $classObj->System_parameter_dropdown_value();
class className {
public function System_parameter_dropdown_value() {
$sql= " SELECT * FROM SYSTEM_PARAMETERS WHERE PARAMETER_KEY='ROLE_TYPE'";
$rs= mysql_query($sql) or die(mysql_error());
$option_list="<option value=0>Select Role Type</option>";
while($data= mysql_fetch_assoc($rs)) {
$option_list.="<option value='$data[PARAMETER_KEY]'>
$data[PARAMETER_VALUE]</option>";
}
return $option_list;
}
}
I think this might help you.

Answer 4

I hope this one helps, it's complete example with the db connection.
mysql_connect('hostname', 'username', 'password');
mysql_select_db('database-name');
$sql = "SELECT record FROM table";
$result = mysql_query($sql);
echo "<select name='tagName'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['tagName'] ."'>" . $row['tagName'] .</option>";
}
echo "</select>";