trouble with inserting

alijsh

Hello.
I've been posting on this subject in the newbie forum, now I'm gonna try my luck here.

I've got apache 2 with php4 and mysql installed on FreeBSD.
I installed everything from the FreeBSD ports-collection and all went well.

Now the problem:
I snagged two scripts from a tutorial, one for displaying what's in a table, and the other one to insert things into a database, it's the latter line that's giving me a headake.

It looks like this

<html>
<head><title>Web Database Sample Inserting</title>
</head>

<body bgcolor=#ffffff>
<?
    if (isset($name) && isset($phone)) {
        mysql_connect("localhost", "webuser", "");
        $query = "INSERT INTO mytable VALUES ($name, $phone)";
        $result = mysql_db_query("example", $query);

if (isset($result)) {
    echo "<p>$name was added to the database</p>";
}
}
?>


if (isset($result)) {
    echo "<p>$name was added to the database</p>";
}
}
?>

The user, doesn't have a password, and it's suppose to be that way (it's just for testing, I'll change it later).

The thing is that the script won't work, I've tried modifying it after the tips I got in the newbie forum, but that didn't really do it.

I can display what's in the database without any problems, it's the inserting bit that doesn't work.

I'm starting to suspect that it's not the script-part that is causing this.

The <host>.err log doesn't say much about this.
I tried inserting som "or die()" in the script, but they didn't give anything.

If someones had a similar problem, please help me with this one.

nnichols

Try adding the following line after your query -

if (!$result) {
echo 'DB error: ' . mysql_error();
}

It should tell you what is failing. When you post code in the future, PLEASE put it in

 open and close tags.  It makes it much easier to read.

Hope this helps ;)

alijsh

nope, it didn't do anything

Thanks anyways.

nnichols

Please post the structure of the table in your db and the full code for your page and I'll have a look.

alijsh

Alright, here it is.

mysql> use example
Database changed
mysql> select * from mytable;
+----------------+----------+
| name           | phone    |
+----------------+----------+
| Homer Simpson  | 555-1234 |
| Bart Simpson   | 555-4321 |
| Ned Flanders   | 555-3294 |
| Lisa Simpson   | 555-3214 |
| Marge Simpson  | 555-2314 |
| Maggie Simpson | 555-3142 |
+----------------+----------+
6 rows in set (0.01 sec)

and the insert script:

<html>
<head><title>Web Database Sample Inserting</title>
</head>

<body bgcolor=#ffffff>
<?
if (isset($name) && isset($phone)) {
        mysql_connect("localhost", "webuser", "");
        $query = "INSERT INTO mytable VALUES ('$name', '$phone')";
        $result = mysql_db_query("example", $query);

    if (!$result) {
            echo 'DB error: ' . mysql_error();
    }

    if ($result) {
            echo "<p>$name was added to the database</p>";
    }
}
?>

<h1>Add an entry</h1>
<form>
Name: <input type=text name='name'><br>
Phone: <input type=text name='phone'><br>
<input type=submit>
</form>

<p><a href="index.php">Back to index</a>
</body>
</html>

As I said, the insert script doesn't work, or there's something wrong with my other settings. But as I said earlier, the script listing the contents of a table works. It looks like this:

<html>
<head>
</head>

<body bgcolor=#ffffff>
<h1>Data from mytable</h1>
<?
    mysql_connect("localhost", "webuser", "");
    $query = "SELECT name, phone FROM mytable";
    $result = mysql_db_query("example", $query);

if ($result) {
    echo "Found these entries in the database:<ul>";
    while ($r = mysql_fetch_array($result)) {
        $name = $r["name"];
        $phone = $r["phone"];
        echo "<li>$name, $phone";
    }
    echo "</ul>";
} else {
    echo "No data.";
}
mysql_free_result($result);
?>
</body>
</html>

The webuser does't have a password set, he has these permissions set SELECT, INSERT, DELETE and USAGE.

Thanks

xblue

Do I understand it correctly that neither the DB error nor the success message is displayed?

If so, this would make me suspect that the whole block of code is not executed.

Try echoing $name and $phone outside the block to make sure you can access the variables this way, or search for the register_global value in your php.ini file.

If this is set to off (as per default since PHP 4.2), you need to access the variables in a different way, use $GET[$name] or $POST[$name] etc. with PHP 4.2.

Hope this help...

alijsh

Yes, that's it. I completely forgot about that, was using php on the webserver at my university, and it has register globals set to on. I didn't do that in the install.

Thanks a lot.

Now the thing is that $GET[] and $POST[] don't do it either.
Think I'm gonna recompile php with register globals set to on.

nnichols

Try -

$POST['name']
$POST['phone']

or the GET equivalent

alijsh

thanks, I just figured it out myself and was going to delete the post..

Tanks a bunch, I've been cursing over this problem for the last few days, and all thta just because I forgot about the globals...
Thanks a lot again.