Why is my php script not dynamically populating an input field on my html form?

Question

I am building an html form that will need to dynamically fill in the first input field with the "facility" column values in my "doctors" table. The facility column contains the names of our 7 offices. However, when I run the code below, my input field is blank and I have verified there is data in my "doctors" table. After this is working, I need to be able to dynamically fill in the second input field (which I haven't coded for in the code below because I'm stuck with the issue of first input) with the "provider" column values, also from my "doctors" table. The "provider" column contains all the provider names in our practice. However, the providers should be filtered, so that only the providers at the facility from the first input field is showing.

<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
  <?php
     $link = mysqli_connect("localhost","USERNAME","PASSWORD");
     mysqli_select_db($link,"DB");
  ?>
  <head>
    <title> Untitled Doc</title>
    <meta http-equiv="content-type" content="text/html; charset=iso-8859-1">
  </head>
  <body>
    <form name "form1" action="" method="post">
      <select>
        <?php
        $res=mysqli_query($link,"select facility from doctors");
        while($row=mysqli_fetch_array($res)){?>
          <option> <?php echo $row ["facility"]; ?></option>
        <?php }?>
      </select>
    </form>
  </body>
</html>

This question needs more details. Are you seeing any errors? are you consoling out the results of $row? What are the results of $row? — Sari Rahal, Apr 06 '17 at 14:05
try removing the space `$row ["facility"];` should be `$row["facility"];` Also, Try to do `print_r($row);` — hungrykoala, Apr 06 '17 at 14:06
@RameshS no this is procedural, the db con must be first parameter — Masivuye Cokile, Apr 06 '17 at 14:10
This is working the problem might be your connection, juts tested the exact code — Masivuye Cokile, Apr 06 '17 at 14:13
Duplicated query, please follow below link for solution. http://stackoverflow.com/questions/14456529/mysqli-fetch-array-while-loop-columns — Shanmuk Kumar, Apr 06 '17 at 14:20
And your next depended field will never work as your option does not have value — Masivuye Cokile, Apr 06 '17 at 14:27
@MasivuyeCokile you are right...it was a connection issue. Now that the first part is working, may I get some help with the second part? I want to be able to filter for only the doctors at that facility. This data would be in the second input field. Ideally "Choose your Provider" should be the default. Thank you in advance! — George Benavente, Apr 06 '17 at 15:47
@GeorgeBenavente you must ask one question at a time... currently u hv not tried anything on the second one, I have given u an answer on the first problem now u need to do research on ajax — Masivuye Cokile, Apr 06 '17 at 15:50
Accept one of the answers below that helped u so ppl can be able to help u again — Masivuye Cokile, Apr 06 '17 at 15:54
Just accepted the answer..thank you to everyone for all the help — George Benavente, Apr 06 '17 at 15:59

score 1 · Accepted Answer · answered Apr 06 '17 at 14:49

It is important that you check if the connection have been established, as I have copied your code as it is and use it on my side and was working fine, therefore made me suspect that the problem might be connection related. also check how sensitive your server is maybe your server see this as an error : $row ["facility"] that space might be the problem as well, but it didn't on my side.

Check your server error log and also enable error reporting at the top of your page add

<?php 
ini_set('display_errors', 1); 
error_reporting(E_ALL);?>

That will enable error reporting, but use that on local server only

Then on live site send them to error log

error_reporting(E_ALL);
ini_set('display_errors',0);
ini_set('log_errors',1);

also to get the mysqli errors, before your connection

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

Important to check if your query does indeed return results before trying to display them.

<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
    <html>
    <?php
     $link = mysqli_connect("localhost", "root", "", "DB");
        if (mysqli_connect_errno()) {
            printf("Connect failed: %s\n", mysqli_connect_error());
            exit();
        }
    ?>
    <head>
    <title> Untitled Doc</title>
    <meta http-equiv="content-type" content="text/html; charset=iso-8859-1">
    </head>
    <body>
    <form name ="form1" action="" method="post">
    <?php
       $query = "SELECT facility FROM doctors";

        if ($res = mysqli_query($link, $query)) {
            echo "<select name=\"myselect\">";
            while ($row = mysqli_fetch_assoc($res)) {
        ?>

        <option value="<?php echo $row['facility'];?>"><?php echo $row['facility'];?></option>
           <?php
            }

            echo "</select>";
            mysqli_free_result($res);
        } else {

            printf("Error : %s\n", mysqli_error($link));
        }

        /* close connection */
        mysqli_close($link);
        ?>



    </form>
    </body>
    </html>

NB: For your own benefit, if you haven't used prepared statements, would suggest that you learn them as well, though they are not needed in this case

Chris Townsend · Answer 2 · 2017-04-06T14:20:53.973

0

When doing this kinda of query its important to catch errors so that you can debug easier. I have added a different way of connecting using mysqli object. Try this, this should determine if you have a connection error or if your query is not returning any results

<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<?php
    $link = new mysqli("localhost","USERNAME","PASSWORD", "DBNAME");
    // Check connection
    if ($link->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    } 
    ?>
    <head>
        <title> Untitled Doc</title>
        <meta http-equiv="content-type" content="text/html; charset=iso-8859-1">
    </head>
    <body>
        <form name "form1" action="" method="post">
            <select>
            <?php

                $res = $link->query("select facility from doctors");

                // Check to see if query returned results
                if($res->num_rows > 0 {
                    while($row = $res->fetch_assoc())
                    {
                        echo '<option>' .  $row["facility"] . '</option>';
                    }
                } else {
                    echo 'No results';
                }
            ?>
            </select>

        </form>
    </body>
</html>

edited Apr 06 '17 at 14:20

answered Apr 06 '17 at 14:11

Chris Townsend

2,586
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also suspect that he's connection is the problem – Masivuye Cokile Apr 06 '17 at 14:14
1

@Chris Townsend- you have a space between `$row` and `["facility"]` which shouldn't be there :). Probably you copied from the question like that.. – Marin N. Apr 06 '17 at 14:14
Thanks @MarinNedea – Chris Townsend Apr 06 '17 at 14:15
I also copied like that and changed the connection only, I still have the space and my dropdown is working well @MarinNedea – Masivuye Cokile Apr 06 '17 at 14:16
If the mysql connection doesn't connect, I do not believe it throws any exceptions. The only errors you'll get is from syntax errors or undefined variables – Chris Townsend Apr 06 '17 at 14:24
This is not my question, but its important to check the style that the OP is using – Masivuye Cokile Apr 06 '17 at 14:30

Marprin · Answer 3 · 2017-04-06T14:55:46.373

0

Based on your code, when you want to access the data in a array you should put no space between the variable and the square brackets [] or your application will not show anything. In your code when you want to echo use

<option><?php echo $row["facility"]; ?></option>

instead of

<option><?php echo $row ["facility"]; ?></option>

For another case of the second field depends on the first field, i suggest to use JavaScript to get the data by sending the first data because it will reduce the processed of getting the data.

UPDATED:

To access your data, please add another parameter in your mysqli_fetch_array to be like this

while($row = mysqli_fetch_array($res, MYSQLI_ASSOC))

this will make your data can be access using name on your table

edited Apr 06 '17 at 14:55

answered Apr 06 '17 at 14:16

Marprin

186
1
6

I have just copied his code as it is and ran on my localhost with that space and it work – Masivuye Cokile Apr 06 '17 at 14:20
@MasivuyeCokile, I just updated my answer. You can try again. Thanks. – Marprin Apr 06 '17 at 14:56
this is not my question – Masivuye Cokile Apr 06 '17 at 14:57

score -1 · Answer 4 · answered Apr 06 '17 at 14:12

-1

   <select>
    <?php
        $res=mysqli_query($link,"select facility from doctors");
        while($row = $result->fetch_assoc($res))
            {
    ?>
    <option> <?php echo $row ["facility"]; ?></option>
    <?php
        }   
    ?>
</select>

answered Apr 06 '17 at 14:12

Ramesh S

841
3
15
35

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any explanation? – Masivuye Cokile Apr 06 '17 at 14:13
`mysqli_fetch_assoc()` function is used to return an associative array representing the next row in the res – Ramesh S Apr 06 '17 at 14:16
You asking the wrong person.. you still have not explained why he should use `fetch_assoc()` instead of `mysqli_fetch_array()` The downvote is not mine – Masivuye Cokile Apr 06 '17 at 14:19

Why is my php script not dynamically populating an input field on my html form?

4 Answers4