Member Avatar for andyseaton2003

Hi all,

This is my first post and first attempt at PHP so sorry if this sounds daft. I'm about a fifth of the way through the php bible and a quarter through php/mysql for dummies, so it can only get better.

Anyway I'm getting this error when I try to run a form which takes the data from the form fields and inserts it into a mysql database I set up on my hosting account.

Parse error: parse error, unexpected T_STRING, expecting ']' in /home/peilife/public_html/phpform.php on line 32

Here's the code:

<HTML>
<HEAD>
<TITLE>PHP Form submit</TITLE>
<META NAME="GENERATOR" CONTENT="MAX's HTML Beauty++ 2004">
</HEAD>

<BODY>
<form action="<?php echo $SERVER?>" method="post">
">

<p> Please enter the name of the business: <BR>
<input type="text" name="$busname" size=40 value="<?php echo $PHP_SELF?>">
<p> Please enter the phone number of the business: <BR>
<input type="text" name="$busphone" size=40 value="<?php echo $PHP_SELF?>">
<p> Please enter the primary key: <BR>
<input type="text" name="$primkey" size=40 value="<?php echo $PHP_SELF?>">
<p> Please enter the image location where the business image is stored: <BR>
<input type="text" name="$imageloc" size=40 value="<?php echo $PHP_SELF?>"> <BR>

<input type="submit" value="submit"><input type="reset" value="clear values">

<?php
$user = "peilife_admin";
$host = "localhost";
$password = "*******";
$database = "peilife_busdir"


$connection = mysql_connect($host, $user, $password) or die ("Could not make connection to server");
**line 32**mysql_select_db ('peilife_admin') or die ("Couldn't select database");

$query1 = "INSERT INTO name(busname) VALUES ($busname)"
$query2 = "INSERT INTO phone(busphone) VALUES ($busphone)"
$query3 = "INSERT INTO primary(primkey) VALUES ($primkey)"
$query4 = "INSERT INTO images(imageloc) VALUES ($imageloc)"

$result1 = mysql_query($query1);
$result2 = mysql_query($query2);
$result3 = mysql_query($query3);
$result4 = mysql_query($query4);

echo "Record updated";

?>


</BODY>
</HTML>

I just added line 32 there to show where its going wrong. I've been googling for a while trying to find a solution but I can't figure it out. I'd be really grateful if anyone could help.

Thanks

Andy Seaton

  • 4 Contributors
  • 6 Replies
  • 115 Views
  • 4 Days Discussion Span
  • Latest Post Latest Post by billah_norm

Recommended Answers

All 6 Replies

Member Avatar for ReDuX

$database = "peilife_busdir"

That line requires an ; (semicolon) at the end of it.

As do these lines..

$query1 = "INSERT INTO name(busname) VALUES ($busname)"
$query2 = "INSERT INTO phone(busphone) VALUES ($busphone)"
$query3 = "INSERT INTO primary(primkey) VALUES ($primkey)"
$query4 = "INSERT INTO images(imageloc) VALUES ($imageloc)"

Also line:
<form action="<?php echo $SERVER?>" method="post">
">

shouldnt have the additional "> at the end of it...

Member Avatar for andyseaton2003

Thanks for the help mate, but that didn't seem to fix it. Any other suggestions? I'm still checking on it.

Thanks

Andy

Member Avatar for ReDuX

In the line:

mysql_select_db ('peilife_admin') or die ("Couldn't select database");

You are selecting a database with named after your username, not the database name..

Should it not be:

mysql_select_db ('peilife_busdir') or die ("Couldn't select database");

?

Member Avatar for PoA 
$query1 = "INSERT INTO name(busname) VALUES ($busname)";
$query2 = "INSERT INTO phone(busphone) VALUES ($busphone)";
$query3 = "INSERT INTO primary(primkey) VALUES ($primkey)";
$query4 = "INSERT INTO images(imageloc) VALUES ($imageloc)";

Make sure to put single quote if $busname, $busphone, $primkey and $imageloc are varchar type in mysql.

Hope it helps.

Member Avatar for andyseaton2003

Sorry guys, none of these suggestions has helped - if I comment out the offending line it just tells me the first line of php code is wrong - where I set the user. I can't figure this - its just not willing to run it.

Thanks for any help.

Andy

Member Avatar for billah_norm

You should have to change the extension from *.html to *.php to execute the code that I have edited.....

You forgot to put semicolones. fix it.
Just Folow this code..
I think using PHP extension is a good approach. thank you

<HTML>
<HEAD>
<TITLE>PHP Form submit</TITLE>
<META NAME="GENERATOR" CONTENT="MAX's HTML Beauty++ 2004">
</HEAD>
<BODY>

<?php

echo "
<form action=\"$SERVER[PHP_SELF]\" method=post>  
<p>Please enter the name of the business: <BR>
<input type=text name=busname size=40 value=$PHP_SELF>
<p> Please enter the phone number of the business: <BR>
<input type=text name=busphone size=40 value=$PHP_SELF>
<p> Please enter the primary key: <BR>
<input type=text name=primkey size=40 value=$PHP_SELF>
<p> Please enter the image location where the business image is stored: <BR>
<input type=text name=imageloc size=40 value=$PHP_SELF><BR>
<input type=submit value=submit name =submit><input type=reset value=clear name=reset>
</form>";

if ($submit=="submit"){  //$reset = reset type name and $ submit = submit type name

	$user = "peilife_admin";
	$host = "localhost";
	$password = "*******";
	$database = "peilife_busdir";
	//if (!($connection = mysql_connect("localhost","","")))
	if (!($connection = mysql_connect($user,$host,$password)))
		{
			print "<h3>could not connect to database</h3>\n";
			exit;
		}

	//$connection = mysql_connect($host, $user, $password) or die ("Could not make connection to server");
	mysql_select_db ('peilife_admin') or die ("Couldn't select database");
 

	$query1 = "INSERT INTO name(busname) VALUES ('$busname')";
	$query2 = "INSERT INTO phone(busphone) VALUES ('$busphone')";
	$query3 = "INSERT INTO primary(primkey) VALUES ('$primkey')";
	$query4 = "INSERT INTO images(imageloc) VALUES ('$imageloc')";

	$result1 = mysql_query($query1);
	$result2 = mysql_query($query2);
	$result3 = mysql_query($query3);
	$result4 = mysql_query($query4);

	echo "Record updated";
}
?>

</BODY>
</HTML>
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