select box help

[techker]

hey guys i got to the part were it populates my select box.but  i need to echo what i select?

 

<?
// Connect database
mysql_connect("localhost","techker_","o");
mysql_select_db("o_test");
?>
<select name="categoryID">
<?
$sql = "SELECT name FROM authuser ".
"ORDER BY name";

$rs = mysql_query($sql);

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

[Lamez]

you mean like it shows suggestions and you click on it, and it does not add it to the input box?

 

If so look at Ajax or JS

[techker]

i just need to retreive the user from my database

[Lamez]

ah!

 

<?php
$username = "john_awesome68";
$q = mysql_query("SELECT * FROM `users` WHERE `username`='".$username."'")
$r = mysql_fetch_array($q);
echo $r['username'];
?>

 

mess around with that!

[techker]

its good but i need the part were it gets the select name?

[grim1208]

its good but i need the part were it gets the select name?

You need to be more clear on what it is you are trying to do.

 

a) are you trying to pass the variable from the page where you created the select box to the next page?

b) user selects option then in a random spot populates what they selected (alert w/ js) or what?

 

[techker]

in the select box it is populated with user name in my db.now i want to when i select a user it show his profile.

[Lamez]

would a IF statement do the job?

[techker]

i guess?but where i bug is how would i do it so it can get the selected user

[CroNiX]

Well, your select isn't in a form (that I can see from the code you posted) and you need a submit button.  Then the person selects the username and clicks submit to send it somewhere for processing.  During that processing you select the user info from the db.

 

<form name="myform" action="<?php echo $_SERVER['php_self']; ?>" method="post">
<?php
// Connect database
mysql_connect("localhost","techker_","o");
mysql_select_db("o_test");
$username = isset($_POST['categoryID']) ? mysql_real_escape_string($_POST['categoryID']) : "";
?>
<select name="categoryID">
<?php
$sql = "SELECT name FROM authuser ".
"ORDER BY name";

$rs = mysql_query($sql);

while($row = mysql_fetch_array($rs))
{
  echo "<option value=\"".$row['name']."\">".$row['name']."\n  ";
}
?>
</select>
<input type="submit" />
</form>
<?php
if(!empty($username))
{
    $q = mysql_query("SELECT * FROM `users` WHERE `username`='".$username."'")
    $r = mysql_fetch_array($q);
    echo $r['username'];
    //rest of your user data....
}
?>

Didn't test it but hopefully you can see whats going on...

Also, if you are selecting the username, Im not sure why you are calling it categoryID in this line: <select name="categoryID">

[techker]

hey thanks dude i just got to debug it..the select box part works fine but the las part

 

<?php

if(!empty($username))

{

    $q = mysql_query("SELECT * FROM `authuser` WHERE `username`='".$username."'")

    $r = mysql_fetch_array($q);

    echo $r['username'];

    //rest of your user data....

}

?>

 

gives me errors

 

 

[Lamez]

if the topic is solved then press the Topic Solved button

[techker]

not really its not working ..

[grim1208]

<?php

if(!empty($username))

{

    $q = mysql_query("SELECT * FROM `authuser` WHERE `username`='".$username."'")

    $r = mysql_fetch_array($q);

    echo $r['username'];

    //rest of your user data....

}

?>

why would you want to echo $r['username'] from the db when you already have the usersname from you select box ('categoryID')?

[grim1208]

not sure if this will help, but if you want to see the value you are getting from the select box put in a javascript function & call like this:

 

<form name="myform" action="<?php echo $_SERVER['php_self']; ?>" method="post">
<?php
// Connect database
mysql_connect("localhost","techker_","o");
mysql_select_db("o_test");
$username = isset($_POST['categoryID']) ? mysql_real_escape_string($_POST['categoryID']) : "";
?>
<select name="categoryID" id="categoryID" onChange="testSelect();">
<?php
$sql = "SELECT name FROM authuser ".
"ORDER BY name";

$rs = mysql_query($sql);

while($row = mysql_fetch_array($rs))
{
  echo "<option value=\"".$row['name']."\">".$row['name']."\n  ";
}
?>
</select>
<input type="submit" />
</form>
<script>
function testSelect() {
  alert(document.getElementById('categoryID').value);
}
</script>

 

Notice I added id="categoryID" to you select and onChange="testSelect();"  I use js to help debug a lot to make sure I am getting the output I want

[techker]

it works a pop up apears and show the user i select.

 

but what i need is to grab the user selected profile in my database.

[CroNiX]

So adjust your query and do something like you did at the top of your code using a where statement.  Are you trying to get us to do it all for you?

[techker]

lol im just asking for help.cause i freez on that part.thx

[grim1208]

GOOGLE is your best friend