Another Parse Error

gman225566

I'm working on a very simple (at least I think it should be) php file to insert something into a database. I'm not a program but I've learned enough to get by.

Below is my code. I am getting the error:

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in i................... input.php on line 22

I am using a host gator account which I believe is php5

HTML FORM FILE:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 
Transitional//EN" 
"http://www.w3.org/TR/html4/loose.dtd">
<html>

<head>
<title>Form Input Data</title>
</head>

<body>
<table border="1">
  <tr>
    <td align="center">Form Input Employees Data</td>
  </tr>
  <tr>
    <td>
      <table>
        <form method="post" action="input.php">
        <tr>
          <td>Name</td>
          <td><input type="text" name="name" size="60">
          </td>
        </tr>
        <tr>
          <td>Address</td>
          <td><input type="text" name="address" size="80">
          </td>
        </tr>
		<tr>
          <td>Description</td>
          <td><input type="text" name="description" size="100">
          </td>
        </tr><tr>
          <td>Link</td>
          <td><input type="text" name="link" size="180">
          </td>
        </tr><tr>
          <td>Lat</td>
          <td><input type="text" name="lat" size="40">
          </td>
        </tr><tr>
          <td>Long</td>
          <td><input type="text" name="lng" size="40">
          </td>
        </tr>
		<tr>
          <td>Type</td>
          <td><input type="text" name="type" size="30">
          </td>
        </tr>
        <tr>
          <td></td>
          <td align="right"><input type="submit" 
          name="submit" value="Send"></td>
        </tr>
        </table> 
      </td>
    </tr>
</table>
</body>  

</html>

Input.php File

<?php
//the example of inserting data with variable from HTML form
//input.php

require("sqlconnect.php");

// Opens a connection to a MySQL server

$connection=mysql_connect (localhost, $username, $password);
if (!$connection) {  die('Not connected : ' . mysql_error());} 

// Set the active MySQL database

$db_selected = mysql_select_db($database, $connection);
if (!$db_selected) {
  die ('Can\'t use db : ' . mysql_error());
} 

//inserting data marker
//possible INSERT INTO 'meingh_canadamap`.`markers` (
$marker = "INSERT INTO markers (id, name, address, description, link, lat, lng, type, approved)
		VALUES(NULL, $_POST['address'], $_POST['description'], $_POST['link'], $_POST['lat'], $_POST['lng'], $_POST['type'], '0')";

//declare in the marker variable
$result = mysql_query($marker,$connection);	//marker executes

if (!mysql_query($markers,$connection))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";


mysql_close($connection);
?>

bradgrafelman

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

..

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

Your parse error is the result of using quoted array indexes inside of a double-quoted string. See this manual page for the proper syntax of how to reference such array items.

Also, your code is vulnerable to SQL injection attacks (or just plain SQL errors); user-supplied data should never be placed directly into a SQL query string. Instead, it must first be sanitized with a function such as [man]mysqli_real_escape_string/man.

EDIT: Also note that you've got two separate calls to mysql_query() with the latter referencing an undefined variable.

gman225566

Thanks

I have someone who is going to help me with the rest of this, I just needed to get the basic functionality working for now.

I will read through the links you posted!

Thanks again.