using php code within html to dynamically load a dropdown list

Question

I am trying to use php code within html to grab content from a database to populate the values for a drop down menu (see below). When I run the php script on its own (test.php) I get all the expected values. When nested within html I get nothing but a single blank value under Select a Species. I would expect this to run through the while loop more than once as there are about 7 values returned and I would also expect the contents to include data derived from the table.

What I am attempting is possible correct?? Is it just an error with code (no errors popping up in the logs)

<label for="select">Species Observed : </label>
<select name="select" class="textfields" id="species">
    <option id = "0">-- Select a Species -- </option>
    <?php
        $con = mysqli_connect("localhost","xxx","xxxx","databases");
        $result = mysqli_query($con, "SELECT * FROM bc_species WHERE 1"); 
        while ($row = $result->fetch_assoc()){
     ?>
     <option><?php echo $row['species'];?></option> 
     <?php } ?>
</select>

whats that json_encode for? just echo it combining the markup option — Kevin
– Kevin, Commented Mar 2, 2016 at 5:09

Peeje · Accepted Answer · 2016-03-02 05:09:49Z

1

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

answered Mar 2, 2016 at 5:09

Peeje

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Comments

Manik Thakur · Accepted Answer · 2016-03-02 05:48:52Z

0

<?php 
        $optionData = '<option id = "0">-- Select a Species -- </option>';

        $con = mysqli_connect("localhost","xxx","xxxx","databases");
        $result = mysqli_query($con, "SELECT * FROM bc_species WHERE 1"); 
        while ($row = $result->fetch_assoc()){
          $optionData .= "<option>".$row['species']."</option>" ;
        } 
?>
<label for="select">Species Observed : </label>
<select name="select" class="textfields" id="species">
    <?php echo $optionData;?>
</select>

answered Mar 2, 2016 at 5:48

Manik Thakur

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Comments

jeff macalino · Accepted Answer · 2016-03-02 05:54:56Z

0

//Put your fetched data on an array then loop them like this

<?php 
                        foreach ($list as $data  => $item) {
                    ?>
                        <option value="<?php echo $data?>">
                            <?php echo $data?>
                        </option>
                   <?php
                        }
                    ?>

answered Mar 2, 2016 at 5:54

jeff macalino

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Comments

diggabledork · Accepted Answer · 2016-03-07 14:27:59Z

0

This should work:

<label for="select">Species Observed : </label>
<select name="select" class="textfields" id="species">
    <option id = "0">-- Select a Species -- </option>
    <?php
        $con = mysqli_connect("localhost","xxx","xxxx","databases");
        $result = mysqli_query($con, "SELECT * FROM bc_species"); 
        while ($row = $result->fetch_assoc()){
           echo '<option>'. $row["species"].'</option> 
     } ?>
</select>

You were not iterating through the values within the HTML <option> tags, so I moved those into the while loop. (notice the PHP section includes echoing those tags).

I also removed the unnecessary "SQL injection-looking" where 1 clause from your query.

edited Mar 7, 2016 at 14:27

answered Mar 2, 2016 at 21:07

diggabledork

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