PHP Functions

nike_guy_man posted this at 03:01 — 4th May 2002.

Joined: Sep 2000

Hey
Here's what I want to happen:
A form is submitted, with 6 variables... name, description, subject, type, email, and content.
I want it to choose the MySQL table to write to by the type drop down box... I have this set up now, using if/elseif/else loops with the entire code in each. However, that is a real pain to edit, so I want to create a function that inserts the data into the separate tables... Here is the code

<?php
function insert($table) {
$insert = \"INSERT INTO $table (name, description, subject, return, message) VALUES ('$name', 
'$client_message_description', 
'$subject', '$return', '$message') \";
$return = mysql_query($insert);
if ($return) {
print \"Message added to $account<br>\";
echo $client_message_type;
print \"<br>$account<br> $name<br>\";
} else {
print \"Error: It didn't work\";
echo mysql_error();
}
}
?>

Later, it calls this function:

<?php
if ($type == \"New\") {
insert(new);
} elseif ($type == \"Update\") {
insert(update);
}
?>

However, this doesn't add the form data into the database, it just puts empty lines in.
The variables that are in the VALUE() part of the MySQL query are passed from the form.
Why does it do this?
What's wrong?
How can I fix this?
Thanks

Mark Hensler posted this at 23:28 — 4th May 2002.

He has: 4,048 posts

Joined: Aug 2000

hmm...

<?php
function insert($table, $fields) {
    
    for ($i=0; $i<count($fields); $i++) {
        $fields[$i] = key($fields[$i]) . \"='\" . $fields[$i] . \"'\";
    }
    $insert = \"INSERT INTO $table SET \";
    $insert .= implode(\", \", $fields);
    
    $return = mysql_query($insert);
    if ($return) {
        print \"Message added to $account<br>\";
        echo $client_message_type;
        print \"<br>$account<br> $name<br>\";
    } else {
        print \"Error: It didn't work\";
        echo mysql_error();
    }
}

// table  : type
// fields : name, description, subject, email, content
$values = array(
                //field => value
                'name' => $name,
                'description' => $description,
                'subject' => $subject,
                'email' => $email,
                'content' => $content);

if ($type == \"New\") {
    insert('new', $values);
} elseif ($type == \"Update\") {
    insert('update', $values);
}
?>

Mark Hensler
If there is no answer on Google, then there is no question.

nike_guy_man posted this at 01:03 — 5th May 2002.

They have: 840 posts

Joined: Sep 2000

I'm getting this error repeatedly... it never ends lol

Quote:
Warning: Variable passed to key() is not an array or object in /home/parrabal/project/clients_message.php on line 11

Line 11 is

<?php
$fields[$i] = key($fields[$i]) . \"='\" . $fields[$i] . \"'\"; 
?>

Mark Hensler posted this at 06:17 — 5th May 2002.

He has: 4,048 posts

Joined: Aug 2000

Ssorry, I made a boo boo.

<?php
function insert($table, $fields) {
    
    foreach ($fields as $key=>$val) {
        $field[$key] = \"$key='$val'\";
    }
    $insert = \"INSERT INTO $table SET \";
    $insert .= implode(\", \", $fields);

// . . . 
?>

Mark Hensler
If there is no answer on Google, then there is no question.