How to populate select options from DB

Question

What am I doing wrong here? When it populates the options it shows $name[0] and not the info from the DB. Though the correct number of options seem to be available.

<?php
//connect to database
$conn = mysqli_connect("example.com", "timemin", "Pass123", "timesheet");

//query database for items to populate
$sql = "SELECT DIS_NAME, NAME FROM INVITEM";
$query = mysqli_query($conn, $sql);

echo '<select>';
echo '<option value="">Choose your favorite fruit</option>';
while($name = mysqli_fetch_assoc($query)){
echo '<option value="' . '$name[1]' . '">' . '$name[0]' . '</option>';
}
echo '</select>';
echo $query;
?>

Just a note: If you use `$name['DIS_NAME']` and `$name['NAME']` rather than `$name[0]` etc, when you come to add another column into a query you dont have to worry about the positioning. — RiggsFolly, Aug 26 '15 at 22:18
That must be it @RiggsFolly, he indeed used `fetch associative`, so array uses named keys. — DeDee, Aug 26 '15 at 22:20
@DeDee You add this to your question afterall you did most of the work. — RiggsFolly, Aug 26 '15 at 22:24
@Elegance I explained why it wants names instead of numbers. — DeDee, Aug 26 '15 at 22:26
See [this manual page](http://php.net/manual/en/mysqli-result.fetch-assoc.php) for the reason or @DeDee answer. Fetch assoc does not return numbered array, only named associative array.Names being the column names from the table — RiggsFolly, Aug 26 '15 at 22:27

DeDee · Accepted Answer · 2015-08-26T22:23:20.007

2

Should be as follows, you quoted variables, that was the problem.

echo '<option value="' . $name['DIS_NAME'] . '">' . $name['NAME'] . '</option>';

Also as RiggsFolly mentioned, you used fetch_assoc so the array keys will be named accordingly DIS_NAME and NAME. Credit to RiggsFolly for spotting this.

edited Aug 26 '15 at 22:23

answered Aug 26 '15 at 22:01

DeDee

1,972
21
30

What is your database structure? Your query is probably incorrect – Iowa15 Aug 26 '15 at 22:07
@Iowa15 according to http://php.net/manual/en/mysqli.quickstart.connections.php its correct... is there reason to assume divergence? – Elegance Aug 26 '15 at 22:10
`example.com` should possibly be `localhost` in the connection `mysqli_connect("localhost", "timemin", "Pass123", "timesheet");` in the connection – RiggsFolly Aug 26 '15 at 22:11
@RiggsFolly I changed if from the actual url for examples sake. – Elegance Aug 26 '15 at 22:12
Hmm, try and just echo out the values in your variable: echo print_r($name); – Iowa15 Aug 26 '15 at 22:12
Just do `print_r($name);` the `echo` is superfluous – RiggsFolly Aug 26 '15 at 22:13
@Iowa15 It does not return anything – Elegance Aug 26 '15 at 22:13
Paste results in your queston, unless the problem is obvious of course – RiggsFolly Aug 26 '15 at 22:16
Your code appears to be proper. It must be database problem. – DeDee Aug 26 '15 at 22:16
Let's see if there's an sql error. echo mysqli_error($conn) after you do the query – Iowa15 Aug 26 '15 at 22:17
@Elegance See RiggsFolly's comment on your question. – DeDee Aug 26 '15 at 22:21

score 2 · Answer 2 · edited May 23 '17 at 10:26

2

In php, there are two types of quotes: single quotes and double quotes. Single quotes will not parse variables, double quotes will.

If you do want to use quotes, you could do something like this:

echo "<option value="."$name[1]".">$name[0]</option>";

So here, the double quotes will tell php to parse the variable names

However, I would recommend doing this:

echo '<option value="' . $name[1] . '">' . $name[0] . '</option>';

See this SO post for more.

edited May 23 '17 at 10:26

Community

1
1

answered Aug 26 '15 at 22:04

Iowa15

3,027
6
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35

use `print_r($name);` to check your array. If there's nothing you have error in your SQL. If there is records, use array key names like `$name['firstname']`. – Aleksandar Aug 26 '15 at 23:18

How to populate select options from DB

2 Answers2