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Hi I have a simple login (user and password) form.
This form also includes this multiple menu:
How can I do that each option will redirect the user to another url?PHP Code:<select name="selectgrade" id="selectgrade">
<option>A</option>
<option>B</option>
<option>C</option>
<option>D</option>
</select>
Thanks
Use theonclickattribute of each<option>to setwindow.location.href.
but I should redirect the user only when he clicked the button (and he passes the validation..)
Code:<select name="selectgrade" id="selectgrade" onchange="window.location.href=this['options'][this['selectedIndex']]['value']">
<option></option>
<option value="http://www.google.ca/">A</option>
<option value="http://www.yahoo.com/">B</option>
<option>C</option>
<option>D</option>
</select>
Actually, after re-reading your post, why don't you just do this with $_POST["selectgrade"] on the validation page? That is, assuming you are using a form.
This is exatcly what I've tried to do, I just don't know how to get the information from the select menu
This is working:
Now, what's the problem on it? What validation are you expecting?PHP Code:<?php echo $_POST['selectgrade']; ?>
<form method="post" action="<?php $_SERVER['PHP_SELF']?>">
<select name="selectgrade" id="selectgrade">
<option>A</option>
<option>B</option>
<option>C</option>
<option>D</option>
</select>
<input type="submit" value="Submit">
</form>
Hope that keeps you going.
Yeah, sorry I didn't look at the "PHP" at the top of the page. I was trying to answer a JavaScript question you didn't have.
The $_POST data will echo the value of the selected index, so you need a value tag in each option.
PHP Code:<?php echo $_POST['selectgrade']; ?>
<form method="post" action="<?php $_SERVER['PHP_SELF']?>">
<select name="selectgrade" id="selectgrade">
<option value="option_a">A</option>
<option value="option_b">B</option>
<option value="option_c">C</option>
<option value="option_d">D</option>
</select>
<input type="submit" value="Submit">
</form>
Hi
can someone tell what's wrong in here:
?PHP Code:$sql="INSERT INTO grades (selectgrade)
VALUES
('$_POST['selectgrade']')";
It wrote me that there is some syntax error there.
What's the exact error message? The parens look wrong to me, but I can't tell you for sure.
MY FORM:
add2.php:PHP Code:<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>database first try</title>
</head>
<body>
<form method="post" action="add2.php">
<table>
<tr>
<td>link url:</td>
<td><input name="link_url" type="text" /></td>
</tr>
<tr>
<td>link name:</td>
<td><input name="link_name" type="text" /></td>
</tr>
<tr>
<td>choose:</td>
<td>
<select name="selectgrade">
<option value="option_a">A</option>
<option value="option_b">B</option>
<option value="option_c">C</option>
<option value="option_d">D</option>
</select>
</td>
</tr>
</table>
<input type="submit" value="sumbit" />
</form>
</body>
</html>
What I've tried to do is to get the selected value and insert it into a database.PHP Code:<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>database first try</title>
</head>
<body>
<?php
$connection = mysql_connect("localhost",
"XXX",
"XXX");
if (!$connection)
die('Could not connect: ' . mysql_error());
mysql_select_db("XXX", $connection);
$sql="INSERT INTO links (link_url, link_name, selectgrade)
VALUES
('$_POST[link_url]','$_POST[link_name]','$_POST['selectgrade']')";
if (!mysql_query($sql,$connection))
die('Error: ' . mysql_error());
echo "1 record added";
mysql_close($connection);
?>
</body>
</html>
Thank you for posting the entire source, but what's the error message?
Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING
The code you originally said was causing the problem wasBut if that line became thisCode:$sql="INSERT INTO grades (selectgrade)
VALUES
('$_POST['selectgrade']')";
then change it to thisCode:$sql="INSERT INTO links (link_url, link_name, selectgrade)
VALUES
('$_POST[link_url]','$_POST[link_name]','$_POST['selectgrade']')";
See the second code-section at http://www.php.net/manual/en/languag...string.parsing.Code:$sql="INSERT INTO links (link_url, link_name, selectgrade)
VALUES
('{$_POST['link_url']}','{$_POST['link_name']}','{$_POST['selectgrade']}')";
Hi thank you,
now there are no errors, but still it doesn't insert it in the right way. All the values from the select menu turned into 0 in the database.
Try this anywhere in that script and report the output:Code:var_dump(isset($_POST['selectgrade']));
var_dump($_POST['selectgrade']);
I think it might have something to do with your single quotes. Here is the way it would be saved.
Try this:Code:$sql="INSERT INTO grades (selectgrade) VALUES ('$_POST['selectgrade']')";
Also as to the problem with changing to 0, make sure that the field in the DB is set to text/varchar and not int.Code:$grade = $_POST['selectgrade'];
$sql="INSERT INTO grades (selectgrade) VALUES ('{$grade}')";
Thank you very much!!
Now it is working :)