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Asked: 2026-05-31T03:21:13+00:00

I’ve been trying to fix this problem forever now and its bugging me. the

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I’ve been trying to fix this problem forever now and its bugging me. the code is to display a table selected from the dropbox which works fine..but when i put it in with my layout/template it throws an error & i have no idea why!

Here’s the code:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta name="keywords" content="" >
<meta name="description" content="" >
<meta http-equiv="content-type" content="text/html; charset=utf-8" >
<title>SNYSB Archive</title>
<link href="style.css" rel="stylesheet" type="text/css" media="screen" >
<!-- Location of javascript. -->
<script language="javascript" type="text/javascript" src="swfobject.js" ></script>
</head>
    <div id="wrapper">
    <div id="header">
    <!-- KEEP THIS BIT [ITS FORMATTING] --> 
    </div>
    <!-- end #header -->
    <div id="menu">
            <ul>
            <li><a href="Hpage.php">Home</a></li>
            <li><a href="Register.php">Register</a></li>
        </ul>
    </div>
    <!-- end #menu -->
    <div id="page">
    <div id="page-bgtop">
    <div id="page-bgbtm">
    <div id="content">
            <div class="post">
            <div class="post-bgtop">
            <div class="post-bgbtm">
<h1 class="title">PUT HEADING HERE!</h1>
                <div class="entry">
                    <p class="Body">
<?php
$dbname = 'snysbarchive';
$conn= mysql_connect('localhost', 'root', 'usbw');
if (!$conn) {
    echo 'Could not connect to mysql';
    exit;
}

$sql = "SHOW TABLES FROM $dbname";
$result = mysql_query($sql);

if (!$result) {
    echo "DB Error, could not list tables\n";
    echo 'MySQL Error: ' . mysql_error();
    exit;
}
if (mysql_select_db($dbname, $conn))
{
?>
    <form method="post" action="new 2.php">
    <select name="tables">
    <?php
    while ($row = mysql_fetch_row($result)) {
    ?>    
    <?php
        echo '<option value="'.$row[0].'">'.$row[0].'</option>';
    }
    ?>
    </select>
    <input type="submit" value="Show">
</form>
<?php
//mysql_free_result($result);
if (isset($_POST) && isset($_POST['tables']))
{
    $tbl=$_POST['tables'];
    //echo $_POST['tables']."<br />";
    $query="SELECT * from $tbl";
    $res=mysql_query($query);
    echo $query;
    if ($res)
    {
    ?>
    <table border="1">
    <?php
        while ( $row = mysql_fetch_array($res))
        {
            echo "<tr>";
            echo "<td>".$row[0]."</td>";
            echo "<td>".$row[1]."</td>";
            echo "<td>".$row[2]."</td>";
            echo "<td>".$row[3]."</td>";        
            echo "</tr>";
        } ?>
    </table>
    <?php
    }
}
?>
</div>
</div>
            </div>
            </div>

        <div style="clear: both;">&nbsp;</div>
        </div>
        <!-- end #content -->
        <div id="sidebar">
            <ul>
                <li>
                    <h2>Welcome!</h2>
                    <p>Welcome to SNYSBs archive!
                       </p>
                </li>
                <li>
                    <h2>SNYSB</h2>
                    <p>
            <a href="Contact.php">Contact Us!</a>
                    </p>
                </li>
            </ul>
        </div>
        <!-- end #sidebar -->
        <div style="clear: both;">&nbsp;</div>
    </div>
    </div>
    </div>
    <!-- end #page -->
    <div id="footer">
        <p>Copyright (c) 2008 Sitename.com. All rights reserved. Design by <a href="http://www.freecsstemplates.org/">Free CSS Templates</a>.</p>
    </div>
    <!-- end #footer -->
</div>
</body>
</html>

(Just pasted the entire page’s code from the original with my template at the start and bottom)

This is the line it doesnt like:

if (isset($_POST['tables']))

Thanks

Edit: New error message:
Parse error: syntax error, unexpected $end in FILENAME on line 128

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1 Answer

  1. Editorial Team

    The error you’re seeing is likely because $_POST doesn’t have a key called ‘tables’.

    echo"<form method="post" action="show_tabs.php">";
echo"<select name="tables">";

    both produce errors when I run them. Notice the syntax highlighting color changes on SO, indicating that not all is being interpreted as a string literal. You need to either escape the inner quotes or switch the outer ones to single quotes:

    echo '<form method="post" action="show_tabs.php">';
echo '<select name="tables">';

    or

    echo "<form method=\"post\" action=\"show_tabs.php\">";
echo "<select name=\"tables\">";

    The same goes for

    echo"<input type="submit" value="Show">";
...
echo"<table border="1">";

    Also,

    if ($res)
}

    should read

    if ($res)
{

    Furthermore, you have an error in your usage of the mysql_* functions. I can see you’re selecting your db only after having run a query, and I don’t see a call to mysql_connect. The order is connect, select_db, query, fetch, free, close as seen in http://www.php.net/manual/en/mysql.examples-basic.php

    (After above code was updated)
    Wrapping the whole thing in a heredoc is clearly the cause, and is very unnecessary. Simply use php open and closing tags to switch between text and code. For instance instead of

    echo <<< TEMPLATE
<h1 class="title">PUT HEADING HERE!</h1>
<div class="entry">
<p class="Body">

    use

    ?>
<h1 class="title">PUT HEADING HERE!</h1>
<div class="entry">
<p class="Body">
<?php

    Code inside of a heredoc doesn’t get executed (although variables are substituted), so none of the code from <<< TEMPLATE to TEMPLATE; is being run. Only variables should go inside a heredoc.

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