Hi I'm trying to implement a dropdown menu in php with mysql and having lots of trouble

Question

The database name is sloganstreet and the table is slogan, and the column i'm trying to pull from is called text. Any help would be appreciated. I have established the connection to the database but I can't seem to get the dropdown to populate with the info from mysql. thanks.

<label>Select a slogan:</label>
<select name="select_slogan">

    <?php
    $query = "SELECT text from slogansstreet";
    $result = mysqli_query($conn, $query);
    echo "<select name='Slogan'>";
    echo "<option value='0'>-Select-</option>";
    while($row = mysqli_fetch_array($result)) 
    {
        echo "<option value='".$row['text']."'></option>";
    }
echo "</select>";
?>

Is your problem that the dropdown is empty, or that there are no elements in it at all? Because you have no text to appear in the dropdown, as there is nothing between <option></option>, you only assign a value (which is hidden to the user). You also likely want mysqli_fetch_assoc instead of mysqli_fetch_array (based on how you use the index). — Qirel
– Qirel, Commented Jun 5, 2016 at 21:38
the drop down is empty and it spits out a huge error on my index: "Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in..." — orangutat
– orangutat, Commented Jun 5, 2016 at 21:40
I tried what you said, but when i remove label, and <select> i lose the dropdown box and get a new error :"arse error: syntax error, unexpected '' (T_ENCAPSED_AND_WHITESPACE), expecting identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING) in ..." — orangutat
– orangutat, Commented Jun 5, 2016 at 21:43
@orangutat Update your question with the code you are actually using, and which line the error is generated on. — Qirel
– Qirel, Commented Jun 5, 2016 at 21:44

RiggsFolly · Accepted Answer · 2016-06-05 22:59:13Z

You seem to have done everything twice, but missed putting something in the dropdown that the user can see.

You dont need this line, as you do this in the PHP code as well, so remove it

<select name="select_slogan">

All you need is this

I added some error processing as well, as you seem to be getting a error in that query.

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

    $query = "SELECT text from slogansstreet";
    $result = mysqli_query($conn, $query);

    if ( $result === FALSE ) {
        echo mysqli_error($conn);
        exit;
    }


    echo "<select name='Slogan'>";
    echo "<option value='0'>-Select-</option>";

    // add MYSQLI_ASSOC as the second param, so you dont get 
    // a numeric and an associative array returned
    // and waste memory
    while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
        echo "<option value='{$row['text']}'>{$row['text']}</option>";
    }
    echo "</select>";
?>

Collectives™ on Stack Overflow

Hi I'm trying to implement a dropdown menu in php with mysql and having lots of trouble

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related