what trying this: when select color dropdown list, submited database , car table. have setup database.
this part of html code.
<form action="rinnad.php" method="post"> car color: <select name="rinnad"> <option>red</option> <option>blue</option> <option>green</option> </select> <input type="submit" value="lisa" /> </form>
here php code using enter car color database:
<?php $connect = @mysql_connect ("localhost", "root", "") or die("fail!!!! :d:d:d"); mysql_select_db("tibid") or die("there no such db"); if($connect){ echo "connected db"; }else{ echo "didnt connect db"; } $car = "red" $query = mysql_query("insert test values('','$car')"); ?>
what need do? thank you.
first need assign values options:
<option value="red">red</option>
http://www.w3schools.com/tags/tag_option.asp
after need assign value form $car variable:
$car = $_post['rinnad'];
or, more securely, protection against sql-injections:
$car = mysql_real_escape_string($_post['rinnad']);
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