[RESOLVED] PHP Beginner Help

[hab_786]

Hello All,

Firstly I would like to let you know that I'm new to this forum and PHP/MySQL. I've been passing PHP/MySQL work to freelancers for years and finally thought to learn it myself. I am familiar with HTML and PHP concepts but am a total beginner.

Anyway, I am making my first programme in PHP. User simply uses a form to add some text (a comment) into the MySQL database. I have set-up the database on my localhost using Wampserver. I managed to create a setup.php file:

<?php
include("dbinfo.inc.php");
mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query="CREATE TABLE comments (id int(6) NOT NULL auto_increment,date varchar(10) NOT NULL,comment varchar(200) NOT NULL,PRIMARY KEY (id),UNIQUE id (id),KEY id_2 (id))";
mysql_query($query);
mysql_close(); 
echo "Database created";
?>

The above code worked and I now have a table in my database. And I also managed to create an input data form:

<form action="insert.php" method="post">
Todays Date: <input type="text" name="date"><br>
Your Comment: <input type="text" name="comment"><br>
<input type="Submit">
</form>

Which should work with this insert.php file:

<?php
include("dbinfo.inc.php");
mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database"); 

$query = "INSERT INTO comments VALUES ('','$date','$comment')";
mysql_query($query);
mysql_close();
?> 

The problem occurs when I try to submit the form. The browser just shows me the source for insert.php, and adds nothing into the database.

My question is, what am I doing wrong. Thank You in advance for all your help!

Hab.786

[bradgrafelman]

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hab.786;11001852 wrote:

I managed to create a setup.php file:

<?php
include("dbinfo.inc.php");
mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query="CREATE TABLE comments (id int(6) NOT NULL auto_increment,date varchar(10) NOT NULL,comment varchar(200) NOT NULL,PRIMARY KEY (id),UNIQUE id (id),KEY id_2 (id))";
mysql_query($query);
mysql_close(); 
echo "Database created";
?>

Few comments:

1. You should ensure that display_errors is set to On and error_reporting is set to E_ALL, because this:

   mysql_connect(localhost,$username,$password);

   should be generating an error message (an 'undefined constant' one) since localhost is first being treated as if it were a [man]constant[/man] rather than a [man]string[/man]; the latter must be delimited in some way (such as with single or double quotes).

2. Don't use the '@' error suppressor. Ever. (It doesn't do anything helpful; what it does do is simply prevent you from knowing about potential problems with your code.)

3. The entire [man]mysql[/man] extension (e.g. any functions that start with 'mysql_') has been outdated for some time and is deprecated in favor of [man]MySQLi[/man] (or even [man]PDO[/man]). See [man]mysqlinfo.api.choosing[/man] for more info.

4. You should always check the return value of [man]mysql_query/man to see if it returned boolean FALSE (indicating an error has occurred) and, if so, either output or log the MySQL error message (and perhaps the SQL query as well to aid in debugging).

5. Any reason why you chose to create the table via PHP rather than the MySQL CLI or some GUI front end such as phpMyAdmin?

6. Couple comments about your table schema:

   1. If you've got a both KEY and UNIQUE are a subset of the PRIMARY KEY type of index, so specifying all three is redundantly redundant, repetitive, and redundant. (😉)

   2. Why is the date column a VARCHAR column and not a... well, a DATE column? (Unless by dateyou're referring to the tasty fruit rather than something you'd find on a calendar.)

hab.786;11001852 wrote:

Which should work with this insert.php file:

<?php
include("dbinfo.inc.php");
mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database"); 

$query = "INSERT INTO comments VALUES ('','$date','$comment')";
mysql_query($query);
mysql_close();
?> 

Few comments:

1. Same comments as #1-4 above.

2. If I showed you this INSERT query:

   INSERT INTO myTable VALUES (123, 'foo bar', NULL)

   could you tell me about the columns in this table? Or perhaps just explain which column 'foo bar' is being INSERT'ed in?

   The answer is you can't, because I omitted the column list and instead just threw a bunch of values in a SQL query. Furthermore, if the table structure ever changes, that query could even become broken and/or not behave properly.

   Same problem with your code. IMHO you should always list the columns in an INSERT INTO query, e.g.:

   INSERT INTO myTable (id, name, favorite_color) VALUES (123, 'foo bar', NULL)

3. Assuming you take my above advice, note that AUTO_INCREMENT columns should be omitted altogether from INSERT queries - just leave it out of your column list and VALUES section and MySQL will take care of it on its own. Otherwise, note you should be using NULL - not an empty string (the two are not the same).

4. Where do you define the variables $date and $comment? Perhaps you meant $POST['date'] and $POST['comment'] instead? See the manual page [man]variables.external[/man] to learn more about how to access data from external sources (e.g. that which is POST'ed from a form or appears in the query string of the URL).

5. Note that user-supplied data should never be placed directly into a SQL query string, else your code will be vulnerable to SQL injection attacks and/or just plain SQL errors. Instead, you must first sanitize the data such as with a function like [man]mysql_real_escape_string/man (for string data) or by using prepared statements. See [man]security.database.sql-injection[/man] for an introduction to the topic of SQL injections if you haven't heard of it before.

hab.786;11001852 wrote:

The browser just shows me the source for insert.php

This suggests a much bigger problem; either you aren't accessing the pages via your webserver, or else your webserver isn't passing the requests along to the PHP interpreter and is instead treating the PHP script as if it were any other plain text document (such as HTML) to be sent directly to the user.

So... how are you accessing the HTML form in your browser?

[hab_786]

Wow... so much wrong... :-)

*Thank you soooooo much for your help!

Ok, so I was following this manual: http://www.freewebmasterhelp.com/tutorials/phpmysql/3 which was most likely the problem. So I switched to this:

<?php

// CONNECTION TO THE MYSQL DATABASE
$hostname = "localhost"; // change to your hostname
$db_user = "root"; // change to your database username
$db_password = ""; // change to your database password
$database = "internet1"; // provide your database name
$db_table = "comments"; // leave this as is

// THIS CODE IS USED TO CONNECT TO THE MYSQL DATABASE
$db = mysql_connect($hostname, $db_user, $db_password);
mysql_select_db($database,$db);
?>

<html>
<head>
<title>ADD COMMENTS</title>
</head>
<body>

<?php
if (isset($REQUEST['Submit'])) {
// THIS CODE TELL MYSQL TO INSERT THE DATA FROM THE FORM INTO YOUR MYSQL TABLE
$sql = "INSERT INTO $db_table(date,comment) values ('".mysql_real_escape_string(stripslashes($REQUEST['date']))."','".mysql_real_escape_string(stripslashes($_REQUEST['comment']))."')";
if($result = mysql_query($sql ,$db)) {
echo 'Thank you. Your comment has been added.';
} else {
echo "ERROR: ".mysql_error();
}
} else {
?>
<h1>Add a comment here:</h1>
<form method="post" action="">
Date (MM.DD.YY):<br>
<input type="text" name="date">
<br>
Comment: <br>
<input type="text" name="comment">
<br><br>
<input type="submit" name="Submit" value="Submit">
</form>
<?php
}
?>
</body>
</html>

And now its working like a charm. Now the issue is, how to export the data saved onto a table in HTML. So for that, I have used this manual: http://www.phpbuilder.com/board/archive/index.php/t-10207538.html

But the following code is not working:

<?php

$hostname = "localhost"; // change to your hostname
$db_user = "root"; // change to your database username
$db_password = ""; // change to your database password
$database = "internet1"; // provide your database name
$db_table = "comments"; // leave this as is

mysql_connect($hostname, $db_user, $db_password)
mysql_select_db($database);

print "<TABLE BORDER=0 CELLPADDING=2>\n";
$fields = mysql_list_fields($database, $db_table);
$columns = mysql_num_fields($fields);
print "<TR>";
for ($x = 0; $x < $columns;$x++) {
print "<TH>" . mysql_field_name($fields,$x) . "</TH>\n";
}
print "</TR>\n";

$sql = "SELECT * from YOURSQLTABLENAME";
$result = mysql_query($sql);
while ($obj = mysql_fetch_object($result,MYSQL_ASSOC)) {
print "<TR>";
foreach($obj as $val) {
print "<TD>" . $val . "</TD>";
}
print "</TR>\n";
}
echo "</TABLE>\n";

?>

Any help would be appreciated. Thanks again for your above comment.

[hab_786]

Ok, I get an error on Line 22:

<?php

$hostname = "localhost"; // change to your hostname
$db_user = "root"; // change to your database username
$db_password = ""; // change to your database password
$database = "internet1"; // provide your database name
$db_table = "comments"; // leave this as is

mysql_connect($hostname, $db_user, $db_password);

print "<TABLE BORDER=0 CELLPADDING=2>\n";
$fields = mysql_list_fields($database, $db_table);
$columns = mysql_num_fields($fields);
print "<TR>";
for ($x = 0; $x < $columns;$x++) {
print "<TH>" . mysql_field_name($fields,$x) . "</TH>\n";
}
print "</TR>\n";

$sql = "SELECT * from $db_table";
$result = mysql_query($sql);
while ($obj = mysql_fetch_object($result,MYSQL_ASSOC)) {
print "<TR>";
foreach($obj as $val) {
print "<TD>" . $val . "</TD>";
}
print "</TR>\n";
}
echo "</TABLE>\n";

?>

[hab_786]

Sorry, one of my messages above was not posted...

The first-issue has been resolved with thanks.

[hab_786]

All resolved... Solution:

<?php

// CONNECTION TO THE MYSQL DATABASE
$hostname = "localhost"; // change to your hostname
$db_user = "root"; // change to your database username
$db_password = ""; // change to your database password
$database = "internet1"; // provide your database name
$db_table = "comments"; // leave this as is

// THIS CODE IS USED TO CONNECT TO THE MYSQL DATABASE
$db = mysql_connect($hostname, $db_user, $db_password);
mysql_select_db($database,$db);
?>

<html>
<head>
<title>ADD COMMENTS</title>
</head>
<body>

<?php
if (isset($REQUEST['Submit'])) {
// THIS CODE TELL MYSQL TO INSERT THE DATA FROM THE FORM INTO YOUR MYSQL TABLE
$sql = "INSERT INTO $db_table(date,comment) values ('".mysql_real_escape_string(stripslashes($REQUEST['date']))."','".mysql_real_escape_string(stripslashes($_REQUEST['comment']))."')";
if($result = mysql_query($sql ,$db)) {
echo 'Thank you. Your comment has been added.';
} else {
echo "ERROR: ".mysql_error();
}
} else {
?>
<h1>Add a comment here:</h1>
<form method="post" action="">
Date (MM.DD.YY):<br>
<input type="text" name="date">
<br>
Comment: <br>
<input type="text" name="comment">
<br><br>
<input type="submit" name="Submit" value="Submit">
</form>
<?php
}
?>
</body>
</html>

[Weedpacket]

When posting PHP or HTML code, please use the board's [noparse]

..

[/noparse] or [noparse]

..

[/noparse] bbcode tags (respectively) as they make your code much easier to read and analyze.