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Unexpected $end

colleyboy

By colleyboy
in PHP Coding Help

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colleyboy Member

colleyboy

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Just trying to do a basic query... not working and don't know why:

 

here is inserts.php:

 

<?php

$username="wormste1_barry";
$password="barry";
$database="wormste1_barry";

$CarName=$_POST['CarName'];
$CarTitle=$_POST['CarTitle'];
$CarPrice=$_POST['CarPrice'];
$CarMiles=$_POST['CarMiles'];
$CarDescription=$_POST['CarDescription'];

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

$query = "INSERT INTO tablename VALUES
('','$CarName','$CarTitle','$CarPrice','$CarMiles','$CarDescription');

mysql_query($query);

mysql_close();

?>

 

That parses the simple form of :

 

form.html:

 

<HTML>

<HEAD>

</HEAD>

<BODY>


<form action="inserts.php" method="post">
car Name: <input type="text" name="CarName"><br>
Car Title: <input type="text" name="CarTitle"><br>
Car Price: <input type="text" name="CarPrice"><br>
Car Miles: <input type="text" name="CarMiles"><br>
Car Description: <input type="text" name="CarDescription"><br>
<input type="Submit">
</form>

</BODY>

</HTML>

 

I get the error: Parse error: syntax error, unexpected $end in /home/wormste1/public_html/tilburywebdesign/shop/FTPServers/barryottley/showroom/inserts.php on line 23

 

Don't know whats wrong?

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colleyboy Member

colleyboy

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  • Author
Posted

That has got rid of that problem but now I get:

 

Parse error: syntax error, unexpected '(' in /home/wormste1/public_html/tilburywebdesign/shop/FTPServers/barryottley/showroom/inserts.php on line 17

 

<?php

$username="wormste1_barry";
$password="barry";
$database="wormste1_barry";

$CarName=$_POST['CarName'];
$CarTitle=$_POST['CarTitle'];
$CarPrice=$_POST['CarPrice'];
$CarMiles=$_POST['CarMiles'];
$CarDescription=$_POST['CarDescription'];

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

$query = "INSERT INTO tablename VALUES"
('','$CarName','$CarTitle','$CarPrice','$CarMiles','$CarDescription');

mysql_query($query);

mysql_close();

?>

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Psycho Prolific Member

Psycho

Posted
Posted 

<?php

$username="wormste1_barry";
$password="barry";
$database="wormste1_barry";

$CarName        = mysql_real_escape_string(trim($_POST['CarName']));
$CarTitle       = mysql_real_escape_string(trim($_POST['CarTitle']));
$CarPrice       = mysql_real_escape_string(trim($_POST['CarPrice']));
$CarMiles       = mysql_real_escape_string(trim($_POST['CarMiles']));
$CarDescription = mysql_real_escape_string(trim($_POST['CarDescription']));

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

$query = "INSERT INTO tablename
          VALUES ('','{$CarName}','{$CarTitle}','{$CarPrice}','{$CarMiles}','{$CarDescription}')";
mysql_query($query);
mysql_close();

?>

 

Note: moved the variable creation for the field values to after the database connection since mysql_real_escape_string() requires a db connection and it makes no sense defining the variables if the db connection would fail.

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colleyboy Member

colleyboy

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  • Author
Posted 

<?php

$username="wormste1_barry";
$password="barry";
$database="wormste1_barry";

$CarName=$_POST['CarName'];
$CarTitle=$_POST['CarTitle'];
$CarPrice=$_POST['CarPrice'];
$CarMiles=$_POST['CarMiles'];
$CarDescription=$_POST['CarDescription'];

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

$query = "INSERT INTO cars VALUES"
('','$CarName','$CarTitle','$CarPrice','$CarMiles','$CarDescription');

mysql_query($query);

mysql_close();

php?>

 

 

Basically I created a database called "wormste1_barry"

 

I have made a table called "cars"

 

and I have 5 fields:

 

CarName

CarTitle

CarPrice

CarMiles

CarDescription

 

all are varchar (15) except CarTitle which is varchar (50)

 

have I got the correct query then?? :S

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colleyboy Member

colleyboy

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Tried the code you have written:

 

I get the following errors appear:

 

Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: Access denied for user 'wormste1'@'localhost' (using password: NO) in /home/wormste1/public_html/tilburywebdesign/shop/FTPServers/barryottley/showroom/inserts.php on line 7

 

Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: A link to the server could not be established in /home/wormste1/public_html/tilburywebdesign/shop/FTPServers/barryottley/showroom/inserts.php on line 7

 

Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: Access denied for user 'wormste1'@'localhost' (using password: NO) in /home/wormste1/public_html/tilburywebdesign/shop/FTPServers/barryottley/showroom/inserts.php on line 8

 

Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: A link to the server could not be established in /home/wormste1/public_html/tilburywebdesign/shop/FTPServers/barryottley/showroom/inserts.php on line 8

 

Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: Access denied for user 'wormste1'@'localhost' (using password: NO) in /home/wormste1/public_html/tilburywebdesign/shop/FTPServers/barryottley/showroom/inserts.php on line 9

 

Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: A link to the server could not be established in /home/wormste1/public_html/tilburywebdesign/shop/FTPServers/barryottley/showroom/inserts.php on line 9

 

Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: Access denied for user 'wormste1'@'localhost' (using password: NO) in /home/wormste1/public_html/tilburywebdesign/shop/FTPServers/barryottley/showroom/inserts.php on line 10

 

Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: A link to the server could not be established in /home/wormste1/public_html/tilburywebdesign/shop/FTPServers/barryottley/showroom/inserts.php on line 10

 

Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: Access denied for user 'wormste1'@'localhost' (using password: NO) in /home/wormste1/public_html/tilburywebdesign/shop/FTPServers/barryottley/showroom/inserts.php on line 11

 

Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: A link to the server could not be established in /home/wormste1/public_html/tilburywebdesign/shop/FTPServers/barryottley/showroom/inserts.php on line 11

 

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Psycho Prolific Member

Psycho

Posted
Posted

have I got the correct query then?? :S

 

As I stated, the string variable $query is malformed.

$query = "INSERT INTO cars VALUES"
('','$CarName','$CarTitle','$CarPrice','$CarMiles','$CarDescription');

 

The value of $query is "INSERT INTO cars VALUES" because you end the query with the double quote at the end of that line. The second line with the values is not in the string and is what is causing the error. I provided a properly created sting in my sample code above.

 

But, now you are stating that the table only has five fields. I assumed tha the first field was an ID field because in the values you have 6 different values, the first being an empty string. Personally, I prefer to explicitly state the fields to insert the values into.

 

$query = "INSERT INTO cars (`CarName`, `CarTitle`, `CarPrice`, `CarMiles`, `CarDescription`)
         VALUES"('{$CarName}','{$CarTitle}','{$CarPrice}','{$CarMiles}','{$CarDescription}');

 

 

EDIT: Regarding the erros you are getting, most likely your database connection is not being made. Change the connection lines to the following:

mysql_connect(localhost,$username,$password) or die(mysql_error());
mysql_select_db($database) or die(mysql_error());

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colleyboy Member

colleyboy

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  • Author
Posted

Dude, your gonna hate me :(

 

I still get the error: Parse error: syntax error, unexpected '(' in /home/wormste1/public_html/tilburywebdesign/shop/FTPServers/barryottley/showroom/inserts.php on line 16

 

I will just run over the details exactly as they stand.

 

I have a database called: wormste1_barry

the table in it is called: cars

it has 5 fields:

 

CarName

CarTitle

CarPrice

CarMiles

CarDescription

 

theres no id tag...

 

the html file called addcar.html looks as follows:

 

<HTML>

<HEAD>

</HEAD>

<BODY>


<form action="inserts.php" method="post">
car Name: <input type="text" name="CarName"><br>
Car Title: <input type="text" name="CarTitle"><br>
Car Price: <input type="text" name="CarPrice"><br>
Car Miles: <input type="text" name="CarMiles"><br>
Car Description: <input type="text" name="CarDescription"><br>
<input type="Submit">
</form>

</BODY>

</HTML>

 

 

the php file looks as follows:

 

inserts.php:

<?php

$username="wormste1_barry";
$password="barry";
$database="wormste1_barry";

$CarName = mysql_real_escape_string(trim($_POST['CarName']));
$CarTitle = mysql_real_escape_string(trim($_POST['CarTitle']));
$CarPrice = mysql_real_escape_string(trim($_POST['CarPrice']));
$CarMiles = mysql_real_escape_string(trim($_POST['CarMiles']));
$CarDescription = mysql_real_escape_string(trim($_POST['CarDescription']));

mysql_connect(localhost,$username,$password) or die(mysql_error());
mysql_select_db($database) or die(mysql_error());

$query = "INSERT INTO cars (`CarName`, `CarTitle`, `CarPrice`, `CarMiles`, `CarDescription`)         VALUES"('{$CarName}','{$CarTitle}','{$CarPrice}','{$CarMiles}','{$CarDescription}');
mysql_query($query);
mysql_close();

?>

 

 

I am well confused but cant see why it keeps sending an error back

 

many thanks for the help... much appreciated!!!

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Psycho Prolific Member

Psycho

Posted
Posted

Ah, it looks like I added the error in the last edit I did. Ending double quote was in wrong place.

 

Use this:

$query = "INSERT INTO cars (`CarName`, `CarTitle`, `CarPrice`, `CarMiles`, `CarDescription`)
          VALUES('{$CarName}','{$CarTitle}','{$CarPrice}','{$CarMiles}','{$CarDescription}')";

 

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