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No values shown in <select> aft using php code

genzedu777

By genzedu777
in PHP Coding Help

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genzedu777 Regular Member

genzedu777

Posted
Posted

Hi all,

 

Currently my dob_day is not showing any value, even though there is a value in the database, it seems like it is unable to retrieve the value.

 

No value was shown in the <select> box. Do you guys have any idea? Thanks

 

 

<?php
$query = "SELECT name, nric, gender, picture, dob_day FROM tutor_profile WHERE tutor_id = '" . $_GET['tutor_id'] . "'";
    
$data = mysqli_query($dbc, $query)
or die(mysqli_error($dbc));

// The user row was found so display the user data
if (mysqli_num_rows($data) == 1) {
	$row = mysqli_fetch_array($data);

    if ($row != NULL) {
      $name = $row['name'];
      $nric = $row['nric'];
      $gender = $row['gender'];
  $old_picture = $row['picture'];
  $dob_day = $row['dob_day'];
    }
    else {
      echo '<p class="error">There was a problem accessing your profile.</p>';
    }

//HTML CODING
<label for="dob" class="label">Date Of Birth</label>
<select id="dob_day" name="dob_day" value="<?php if (!empty($dob_day)) echo $dob_day; ?>">
  <option>
			<option>01</option>
			<option>02</option>
			<option>03</option>
			<option>04</option>
			<option>05</option>
			<option>06</option>
			<option>07</option>
			<option>08</option>
			<option>09</option>
			<option>10</option>
			<option>11</option>
			<option>12</option>
			<option>13</option>
			<option>14</option>
			<option>15</option>
			<option>16</option>
			<option>17</option>
			<option>18</option>
			<option>19</option>
			<option>20</option>
			<option>21</option>
			<option>22</option>
			<option>23</option>
			<option>24</option>
			<option>25</option>
			<option>26</option>
			<option>27</option>
			<option>28</option>
			<option>29</option>
			<option>30</option>
			<option>31</option>
		</select>
?>

 

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atrum Regular Member

atrum

Posted
Posted

Add this above the line starting with "if ($row != NULL)"

 

print_r($row);

 

That should print the array of data you pulled down for tutor_id.

Show me what that gives you.

 

EDIT: Doh!, I missed <option> error that pikachu2000 pointed out :(.

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genzedu777 Regular Member

genzedu777

Posted
  • Author
Posted

Hi Atrum,

 

Sorry for the late response.

 

Hey I got values in the date of birth, but it is still not showing in the drop down box. What could possibly be happening? Thanks

 

Array ( [0] => MUHAMMAD  [name] => MUHAMMAD  [1] => S89362A [nric] => S89362A [2] => M [gender] => M [3] => belitapic.jpg [picture] => belitapic.jpg [4] => 16 [dob_day] => 16 )

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genzedu777 Regular Member

genzedu777

Posted
  • Author
Posted

Hi Pikachu2000,

 

Thanks for replying. However I still dont really get what you have meant

 

Currently the value for dob_day = '16'

 

However '16' is not reflected in my drop down box, apparently it is 'blank'. Under my option, I do have <option>16</option>, what do you mean by

Currently the <option>s do not have value= attributes.

 

Could you elaborate further? Thank you so much

 

<?php
<?php
$query = "SELECT name, nric, gender, picture, dob_day FROM tutor_profile WHERE tutor_id = '" . $_GET['tutor_id'] . "'";
    
   $data = mysqli_query($dbc, $query)
   or die(mysqli_error($dbc));
   
   // The user row was found so display the user data
   if (mysqli_num_rows($data) == 1) {
      $row = mysqli_fetch_array($data);

    if ($row != NULL) {
      $name = $row['name'];
      $nric = $row['nric'];
      $gender = $row['gender'];
     $old_picture = $row['picture'];
     $dob_day = $row['dob_day'];
    }
    else {
      echo '<p class="error">There was a problem accessing your profile.</p>';
    }

//HTML CODING
<label for="dob" class="label">Date Of Birth</label>
<select id="dob_day" name="dob_day" value="<?php if (!empty($dob_day)) echo $dob_day; ?>">
            <option value=''></option>
            <option>01</option>
            <option>02</option>
            <option>03</option>
            <option>04</option>
            <option>05</option>
            <option>06</option>
            <option>07</option>
            <option>08</option>
            <option>09</option>
            <option>10</option>
            <option>11</option>
            <option>12</option>
            <option>13</option>
            <option>14</option>
            <option>15</option>
            <option>16</option>
            <option>17</option>
            <option>18</option>
            <option>19</option>
            <option>20</option>
            <option>21</option>
            <option>22</option>
            <option>23</option>
            <option>24</option>
            <option>25</option>
            <option>26</option>
            <option>27</option>
            <option>28</option>
            <option>29</option>
            <option>30</option>
            <option>31</option>
         </select>
?>
?>

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genzedu777 Regular Member

genzedu777

Posted
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Posted

Hi Pikachu2000,

 

I have followed your advice, but it still didnt work, I have done up to option value = '16', but there is no value in my drop down box.

 

Did I miss out anything? Thanks

 

<td><select id="dob_day" name="dob_day" value="<?php if (!empty($dob_day)) echo $dob_day; ?>">
			<option value=''></option>
			<option value='1'>01</option>
			<option value='2'>02</option>
			<option value='3'>03</option>
			<option value='4'>04</option>
			<option value='5'>05</option>
			<option value='6'>06</option>
			<option value='7'>07</option>
			<option value='8'>08</option>
			<option value='9'>09</option>
			<option value='10'>10</option>
			<option value='11'>11</option>
			<option value='12'>12</option>
			<option value='13'>13</option>
			<option value='14'>14</option>
			<option value='15'>15</option>
			<option value='16'>16</option>

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PaulRyan Member

PaulRyan

Posted
Posted

Did no-one else spot the closing tag at the end of the script, that should infact be above the above the Label and Select?

 

<?php
$query = "SELECT name, nric, gender, picture, dob_day FROM tutor_profile WHERE tutor_id = '" . $_GET['tutor_id'] . "'";
    
   $data = mysqli_query($dbc, $query)
   or die(mysqli_error($dbc));
   
   // The user row was found so display the user data
   if (mysqli_num_rows($data) == 1) {
      $row = mysqli_fetch_array($data);

    if ($row != NULL) {
      $name = $row['name'];
      $nric = $row['nric'];
      $gender = $row['gender'];
     $old_picture = $row['picture'];
     $dob_day = $row['dob_day'];
    }
    else {
      echo '<p class="error">There was a problem accessing your profile.</p>';
    }
//HTML CODING BELOW
?>

<label for="dob" class="label">Date Of Birth</label>
<select id="dob_day" name="dob_day" value="<?php if (!empty($dob_day)) { echo $dob_day; } ?>">
     <option>
            <option>01</option>
            <option>02</option>
            <option>03</option>
            <option>04</option>
            <option>05</option>
            <option>06</option>
            <option>07</option>
            <option>08</option>
            <option>09</option>
            <option>10</option>
            <option>11</option>
            <option>12</option>
            <option>13</option>
            <option>14</option>
            <option>15</option>
            <option>16</option>
            <option>17</option>
            <option>18</option>
            <option>19</option>
            <option>20</option>
            <option>21</option>
            <option>22</option>
            <option>23</option>
            <option>24</option>
            <option>25</option>
            <option>26</option>
            <option>27</option>
            <option>28</option>
            <option>29</option>
            <option>30</option>
            <option>31</option>
         </select>

 

Try that? Unless you noticed this before the just disregard this post.

 

Regards, PaulRyan.

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