Using pre-populated drop down list in PHP to send output to INSERT

Question

I am creating a form which when submitted sends the output to an INSERT statement. I am learning how to do this using:

<form action="somepage.php" method="post">
Example: <input type="text" name="something" />
<input type="submit" />
</form>

I am fairly comfortable doing this and then writing an INSERT statement containing $_POST.

I also have a field that I would like to present in a drop down which is populated with the results of a mysql_query, as below:

$sql = "SELECT name FROM customer";
$result4 = mysql_query($sql);

echo "<select name='name'>";
while ($row = mysql_fetch_array($result4)) {
echo "<option value='" . $row['name'] . "'>" . $row['name'] . "</option>";
}
echo "</select>";

However I am not sure how to use the value that the user selects. I.e. I would like to use the value that they select in my INSERT statement along with the other form fields.

How do I do this?

Please don't use mysql_* functions in new code. They were removed from PHP 7.0.0 in 2015. Instead, use prepared statements via PDO or MySQLi. See Why shouldn't I use mysql_* functions in PHP? for more information. — Madara's Ghost
– Madara's Ghost, Commented Sep 2, 2012 at 19:25

Peter · Accepted Answer · 2012-09-02 19:24:37Z

1

You specify in the <select> tag the name, like:

<select name="car">
    <option>Audi</option>
    <option>BMW</option>
</select>

And get the submitted value with $_POST['car']

In your case, after submitting you can use $_POST['name']

answered Sep 2, 2012 at 19:24

Peter

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2 Comments

user1620419 Over a year ago

Thanks for that. I'm still a little confused though (beginner...). Where do I put the drop down code in relation to the rest of the form/submit button? How does the code send the value to somepage.php without it being specified as with action="somepage.php"?

user1620419 Over a year ago

Would this work?: <form action="somepage.php" method="post"> Example: <input type="text" name="something" /> <?php echo "<select name='name'>"; while ($row = mysql_fetch_array($result4)) { echo "<option value='" . $row['name'] . "'>" . $row['name'] . "</option>"; } echo "</select>"; ?> <input type="submit" /> </form>

Makoto · Accepted Answer · 2012-09-02 19:39:56Z

0

Use $_POST['name'] to get the submitted value.

edited Sep 2, 2012 at 19:39

Makoto

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answered Sep 2, 2012 at 19:34

Alex Grossman

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1 Comment

user1620419 Over a year ago

The problem is I don't know how to put the bits of the form together. I.e. where is the 'Submit' button? Where is the action="..."?

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