从下拉列表中选择 - >比较 - >如果匹配 - >显示图片

Question

Good day all,

I've got a code that reads the users from a database and puts them in a dropdown menu:

<?php           
mysql_connect('', '', '');
mysql_select_db ("");

$sql = "SELECT id,name FROM jos_users";
$result = mysql_query($sql);


echo "<select name='deelnemers' onchange='copyId2textinput(this);'>";
while ($row = mysql_fetch_array($result)) {


echo "<option value='" . $row['id'] . "'>" . $row['name'] . "</option>";
}
echo "</select>";
?>

Now i've got another database called jos_comprofiler with also an ID and also a avatar (image).

I was wondering if somebody give me some advise to compare the ID's from the 2 tables and then show the picture.

So for example, if i click on a user in the dropdown, it must look if there's a ID match with the other table, and if there is, show the picture from 'avatar'.

Thank you for your help and excuse me for my bad english!

Answer 1

There would be some ways but I'll show fast one.

<?php           
mysql_connect('', '', '');
mysql_select_db ("");

$sql = "SELECT u.id, name, avatar FROM jos_users AS u LEFT JOIN jos_comprofiler USING(id)";
$result = mysql_query($sql);

echo "<div id='imgContainer'></div>";

echo "<select name='deelnemers' onchange='showAvatar(this.value);'>";
$avatars = array();
while ($row = mysql_fetch_array($result)) {
    if($row['avatar']){
        $avatars[$row['id']] = $row['avatar'];
}
echo "<option value='" . $row['id'] . "'>" . $row['name'] . "</option>";
}
echo "</select>";

?>
<script>
    var avatars = <?=json_encode($avatars)?>;
    //alert(avatars[5]);
    var avatarContainer = document.getElementById('imgContainer');
    function showAvatar(id) {
        if(avatars[id]===undefined) return false;
        avatarContainer.innerHTML = '<img src="/path/'+avatars[id]+'" />';
    }
</script>

This should work, with some modification for your code: img path, etc..