[SOLVED] $_POST data and PHP code

[Delaran]

Hey guys, I'm a PHP noob.  I am using SQLite and PHP 5.0.5 on my server.  Now that y'all know my background, here's my problem...

 

I'm coding a page with text fields using $_POST data.  I have pageone.php which has the text fields, a submit button, etc. and then I have  pagetwo.php which is my "action" page.  I am attempting to insert the data into my database from pageone's $_POST data, but have run into some problems.  I have checked the forums here as well as googled my problem, but it seems most people are using MySQL on this site.  I hope I'm not out of line to ask this question and if I am -- I'll understand if this thread is locked/deleted.

 

My code is as follows (shortened to the problem area):

 

// set path of database file

$db = $_SERVER['DOCUMENT_ROOT']."/../wantdb.sqlite2";

 

// open database file

$handle = sqlite_open($db) or die("Could not open database");

 

// check to see if the form was submitted with a new record

if (isset($_POST['submit'])) {

 

// make sure all fields are present

if (!empty($_POST['name']) && !empty($_POST['item']) && !empty($_POST['phone']) && !empty($_POST['email'])) {

        // generate and execute query

    $name = sqlite_escape_string($_POST['name']);

    $item = sqlite_escape_string($_POST['item']);

    $phone = sqlite_escape_string($_POST['phone']);

    $email = sqlite_escape_string($_POST['email']);

    $query = "INSERT INTO wantdb (name, item, phone, email) VALUES ('".$name."', '".$item."', '".$phone."', '".$email."')";

    $result = sqlite_query($handle, $query) or die("ERROR: $query. ".sqlite_error_string(sqlite_last_error($handle)));

        echo "<i>Record successfully inserted with ID ".sqlite_last_insert_rowid($handle)."!</i><p />";

    }

    else {

        // missing data

        // display error message

        echo "<i>Incomplete form input. Record not inserted!</i><p />";

 

---------------------------------------

 

As you can see, I have four text fields I want inserted into the db with $_POST data.  I get no error messages so I assume that the database has received the information, but when I use "select * from wantdb" it retrieves only the two I manually entered beforehand.  Nothing from the web form shows up.  So my question(s) are:

 

1) Should I place this code onto pageone.php, or leave it on pagetwo.php?

 

2) Is this code correct in the syntax, or have I added/omitted something necessary?

 

I greatly appreciate anyone's help or if you could link another thread where this has been solved.  I looked but could not find anything related to an SQLite topic; most of the threads revolved around MySQL.

 

~delaran

[mpharo]

in your insert statement you do not need the periods...try this line....

 

$query = "INSERT INTO wantdb (name, item, phone, email) VALUES ('$name', '$item', '$phone', '$email')";

 

[Delaran]

Tried that just now -- still nothing.  Gives me no errors and when I check the database it still has nothing in it.  Could there be something else?  ???

[bobleny]

Try this:

$query = "INSERT INTO wantdb (name, item, phone, email) VALUES ('".$name."', '".$item."', '".$phone."', '".$email."')";

echo $query;
die();

[Delaran]

Still nothing, no echo of the query either.  I appreciate all the help, thanks guys.

[mpharo]

 

$query = "INSERT INTO wantdb (name, item, phone, email) VALUES (\''.$name.'\'', \''.$item.'\', \''.$phone.'\', \''.$email.'\')";

 

[Delaran]

Copied and pasted in, used scp to transfer my file from my computer to the server, tested it via the webpage, then queried the database with select * from wantdb, and still nothing...

 

This one is a tough one, I think I tried some of those code snippets before and got nothing.. 

[bobleny]

If you used this:

$query = "INSERT INTO wantdb (name, item, phone, email) VALUES ('".$name."', '".$item."', '".$phone."', '".$email."')";

echo $query;
die();

 

And it didn't echo the query, I think you have another issue at play.

 

That should have echoed the query...

[Delaran]

Alright.. I inserted that code again and got no echo.  Below is the -entire- code of the page.  I hope I'm not breaking any rules.  Now, should I pull anything out of this code or add something in? Maybe a username/pw?

 

<html>

<head>

<title>Web-based customer requests</title>

</head>

<center>

 

<br>

<br>

Thank you <?php echo htmlspecialchars($_POST['name']); ?>, we look forward to helping you find the item you are looking for.

<br>

Upon finding the item, we will contact you at <?php echo (int)($_POST['phone']); ?>, or by email at <?php echo htmlspecialchars($_POST['email']); ?>.

 

</center>

 

<?php

 

// set path of database file

$db = $_SERVER['DOCUMENT_ROOT']."/../wantdb.sqlite2";

 

// open database file

$handle = sqlite_open($db) or die("Could not open database");

 

// check to see if the form was submitted with a new record

if (isset($_POST['submit'])) {

 

// make sure all fields are present

if (!empty($_POST['name']) && !empty($_POST['item']) && !empty($_POST['phone']) && !empty($_POST['email'])) {

        // generate and execute query

    $name = sqlite_escape_string($_POST['name']);

    $item = sqlite_escape_string($_POST['item']);

    $phone = sqlite_escape_string($_POST['phone']);

    $email = sqlite_escape_string($_POST['email']);

    $query = "INSERT INTO wantdb (name, item, phone, email) VALUES (\''.$name.'\'', \''.$item.'\', \''.$phone.'\', \''.$email.'\')";

    echo $query;

    die();

    $result = sqlite_query($handle, $query) or die("ERROR: $query. ".sqlite_error_string(sqlite_last_error($handle)));

        echo "<i>Record successfully inserted with ID ".sqlite_last_insert_rowid($handle)."!</i><p />";

    }

    else {

        // missing data

        // display error message

        echo "<i>Incomplete form input. Record not inserted!</i><p />";

    }

}

 

 

// all done

// close database file

sqlite_close($handle);

?>

 

[mpharo]

You are using die() before the query is executed...remove that outta there and try it....die should be used for error handling...or forcibly terminating a script...

[Delaran]

Alright, so this thread can be "SOLVED", here is what was wrong with my code, I modified it so just take a look at the above code and match it to the one below to see what I removed (it appears to be the code that was supposed to check if the text fields were submitted with something in them, or blank fields):

 

<html>

<head>

<title>Web-based customer requests</title>

</head>

<center>

 

<br>

<br>

Thank you <?php echo htmlspecialchars($_POST['name']); ?>, we look forward to helping you find the item you are looking for.

<br>

Upon finding the item, we will contact you at <?php echo (int)($_POST['phone']); ?>, or by email at <?php echo htmlspecialchars($_POST['email']); ?>.

 

</center>

</html>

 

<?php

 

// set path of database file

$db = $_SERVER['DOCUMENT_ROOT']."/../wantdb.sqlite2";

 

// open database file

$handle = sqlite_open($db) or die("Could not open database");

 

{

 

          // generate and execute query

    $name = sqlite_escape_string($_POST['name']);

    $item = sqlite_escape_string($_POST['item']);

    $phone = sqlite_escape_string($_POST['phone']);

    $email = sqlite_escape_string($_POST['email']);

    $query = "INSERT INTO wantdb (name, item, phone, email) VALUES ('".$name."', '".$item."', '".$phone."', '".$email."')";

        $result = sqlite_query($handle, $query) or die("ERROR: $query. ".sqlite_error_string(sqlite_last_error($handle)));

        echo "<i>Record successfully inserted with ID ".sqlite_last_insert_rowid($handle)."!</i><p />";

 

        }

 

 

// all done

// close database file

sqlite_close($handle);

?>

 

[bobleny]

You are using die() before the query is executed...remove that outta there and try it....die should be used for error handling...or forcibly terminating a script...

 

To clarify, that was to test the $query line. I find it helpful to kill it after such a thing to prevent the rest of the scrip from interfering.

 

We thought the problem was the $query, so if you check to see what exactly is being queryed, you might discover a problem.

 

He was not getting anything echoed, which means there is another problem. Because of the "die();", you know the problem is above the "die();".

 

Just a small little check I leaned from Wildteen I think...