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select box help


techker

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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>

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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?

 

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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">

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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

 

 

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<?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')?

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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

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