PHP question

Divine · 25-02-2004 6:36pm #1

Hey can anybody tell me whats wrong with this code. Whenever i press the submit button it does not seem to execute anything. I am using foxserve as my localhost and i am logged into MYSQLadmin as sleekfx. Can anybody shed some much needed light on the matter.

Thanks.

<HTML>
<HEAD>
<TITLE>Putting data</TITLE>
</HEAD>
<BODY>
<?php
/* This program enters news items into a database */

if(isset($submit)) {

$db = mysql_connect ("localhost", "sleekfx");

mysql_select_db ("DBNAME", $db);

$date = date("Y-m-d");

$sql = "INSERT INTO news

VALUES (NULL, '$heading','$body', '$date',
'$auth', '$auth_email')";

mysql_query($sql);

print("<h2>The data has been entered.</h2>");
print("<b>You can add another news story below</b>");

}
?>

<h1>Enter your news item into the database</h1>
<form action="data_in.php" method="post"><br>
Name:<br> <input type ="text" name="auth"> <br>
Email:<br> <input type ="text" name="auth_email"><br>
Heading:<br> <input type ="text" name="auth"> <br>
News:<br> <input type ="text" name="body"> <br>

<input type="submit" name="submit" value="Submit News!">
</form>
</body>
</html>

aidan_walsh · 25-02-2004 6:45pm

My guess would be that you have to remove the ' from the variable names in the SQL statement. It could think that you are trying to insert literals and not variable values...

dahamsta · 25-02-2004 7:42pm

Is DBNAME the actual name of the database? Are you getting any output at all?

Try turning error reporting up full by adding this at the top of your script:

[PHP]error_reporting(E_ALL);[/php]adam

chakotha · 25-02-2004 8:04pm

Put

if (!mysql_query($sql)) {
die("Problems with the query: ".mysql_error());
}

instead of

mysql_query($sql);

Divine · 25-02-2004 9:01pm

no dooddle_sketch that is not the problem, i considered that ages ago.

I have tried the other suggestions but nothing seems to be happening at all? Can anyone put it through their own localhost and see what happens. yes the name of the database is DBNAME but nothings ****ING HAPPENING AND IVE BEEN AT IT FOR THE LAST 6 ****ING HOURS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

chakotha · 25-02-2004 9:18pm

You haven't provided a password. Try

[PHP]
$dbcnx = @mysql_connect ("localhost", "sleekfx", "your_mysql_password");

if (!$dbcnx){
echo("<P>Unable to connect to the server at this time.</P>");
exit();
}
//select the database
if ( ! @mysql_select_db("DBNAME")){
echo("<P>Unable to locate the database at this time.</P>");
exit();
}
[/PHP]

If your password is "" it might be worth trying anyway to see if you can connect to MySQL

Divine · 25-02-2004 10:43pm

ok i just tried that but i am getting no sort of response at all. I give up i dont know what to do. This is for a dating website so i was wondering could anyone recommend a book that would teach you PHP and include functions you would be likely to find included in a dating site.

Thanks,

The Corinthian · 25-02-2004 11:10pm

In fairness, you seem to have bitten off more than you can chew. Either lower your targets or farm it out.

Ph3n0m · 26-02-2004 12:46pm

Originally posted by Divine
<HTML>
<HEAD>
<TITLE>Putting data</TITLE>
</HEAD>
<BODY>
<?php
/* This program enters news items into a database */

if(isset($submit)) {

$db = mysql_connect ("localhost", "sleekfx");

mysql_select_db ("DBNAME", $db);

$date = date("Y-m-d");

$sql = "INSERT INTO news

VALUES (NULL, '$heading','$body', '$date',
'$auth', '$auth_email')";

mysql_query($sql);

print("<h2>The data has been entered.</h2>");
print("<b>You can add another news story below</b>");

}
?>

<h1>Enter your news item into the database</h1>
<form action="data_in.php" method="post"><br>
Name:<br> <input type ="text" name="auth"> <br>
Email:<br> <input type ="text" name="auth_email"><br>
Heading:<br> <input type ="text" name="auth"> <br>
News:<br> <input type ="text" name="body"> <br>

<input type="submit" name="submit" value="Submit News!">
</form>
</body>
</html>

OK had a quick look and the thing that strikes me are these:

1.The variable you are passing, would be called $auth, $auth_email, $auth and $body
So first of all you got two fields of the form named the same - that is a problem

2. The php code for feeding information for the database uses variables called:
$heading, $body, $date, $auth, $auth_email

Where is the $heading variable in your form??

[CrimsonGhost] · 26-02-2004 12:49pm

To start with for each of your mysql_<functionname> calls do something like
mysql_query($sql) or die(mysql_error());
This will hopefully show up any problems you have connecting/talking to the database.

Are you getting any error messages at all?

oq4v3ht0u76kf2 · 26-02-2004 10:29pm

[PHP]
if(isset($submit)) {
[/PHP]

Are you sure register_globals is turned on? If not, you may have to specify $_POST rather than just using $submit... try changing that before you mess with anything else. All modern builds come with register_globals turned off by default... you'll also have to change all other references to variables to $_POST.

oq4v3ht0u76kf2 · 26-02-2004 10:51pm

Here, this code should work for you. It is basically the same as your code except slightly cleaner and, um, works. I tested it myself. Make sure you set the action value in the form tag to point to the file itself. Good luck.

[PHP]<html>
<?php
/* news.php, for Divine by Bob Burke
You must define these variables */
$dbHost = ""; // Database host, e.g. "localhost"
$dbUser = ""; // Database username, e.g. "bob"
$dbPass = ""; // Database password
$dbName = ""; // Database name, e.g. "products"
?>

<html>
<head>
<title>Putting data</title>
</head>
<body>

<?php
if(isset($_POST)) {

mysql_connect($dbHost, $dbUser, $dbPass) or die("Database error: ".mysql_error());
mysql_select_db($dbName) or die("Database error: ".mysql_error());

$date = date("Y-m-d");

$sql = "INSERT INTO `news` "
. "VALUES (NULL, '".$_POST."', "
. "'".$_POST."', '".$date."', "
. "'".$_POST."', '".$_POST."');";

$query = mysql_query($sql) or die("Database error: ".mysql_error());

print("<h2>The data has been entered.</h2><br />");
print("<b>You can add another news story below</b><br />");

}
?>

<h1>Enter your news item into the database</h1>
<form action="./news.php" method="post"><br>
Name:<br> <input type ="text" name="auth"> <br>
Email:<br> <input type ="text" name="auth_email"><br>
Heading:<br> <input type ="text" name="heading"> <br>
News:<br> <input type ="text" name="body"> <br>

<input type="submit" name="submit" value="Submit News!">
</form>
</body>
</html>[/PHP]

Divine · 27-02-2004 12:49am

ok Orange_Rhino i have tried that but there still does not seem to be anything happening. What i did was:

Log into phpmyAdmin as "sleekfx" with a password of "password1".

I set up a database called "des" that only i have access to when i am logged in as "sleekfx".

I set up a table called news and inserted the required fields.

I have made alterations to the code you have given me but still nothing seems to be happening. When i press the submit button the information just vanishes but nothing actually happens. I do not get the message saying "The data has been entered". In fact nothing happens at all. I do not get any error messages or anything as [CrimsonGhost] mentioned. My screen just stays the same except the information disappears.

Oh and also the register_globals is in fact turned off whether or not that actually makes a differance with the new code you have given me. Below is the code you gave me and the changes i have made to it. Any more ideas id really appreciate it.

Thanks,

Des.

<html>
<head>
<title>Putting data</title>
</head>
<body>

<?php
if(isset($_POST)) {

mysql_connect($localhost, $sleekfx, $password1) or die("Database error: ".mysql_error());
mysql_select_db($des) or die("Database error: ".mysql_error());

$date = date("Y-m-d");

$sql = "INSERT INTO `news` "
. "VALUES (NULL, '".$_POST."', "
. "'".$_POST."', '".$date."', "
. "'".$_POST."', '".$_POST."');";

$query = mysql_query($sql) or die("Database error: ".mysql_error());

print("<h2>The data has been entered.</h2><br />");
print("<b>You can add another news story below</b><br />");

}
?>

<h1>Enter your news item into the database</h1>
<form action="data_in.php" method="post"><br>
Name:<br> <input type ="text" name="auth"> <br>
Email:<br> <input type ="text" name="auth_email"><br>
Heading:<br> <input type ="text" name="heading"> <br>
News:<br> <input type ="text" name="body"> <br>

<input type="submit" name="submit" value="Submit News!">
</form>
</body>
</html>

oq4v3ht0u76kf2 · 27-02-2004 2:10pm

Try:

mysql_connect("localhost", "sleekfx", "password1") or die("Database error: ".mysql_error());
mysql_select_db("des") or die("Database error: ".mysql_error());

Instead of your corresponding lines.

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