Dropdown of tables to show content - php

Question

This script allows the user to choose a table from the dropdown and then displays the contents of the chosen table.
At the moment the DB connection is working.
A list of tables is shown in the dropdown.
Problem: No content from the database table is shown.
I have checked through the code and everything looks OK, but it still only seems to partially work.
<?php
//update this to your DB connection details.
$dbh = "localhost";
$dbn = "dbname";
$dbu = "dbuser";
$dbp = "dbpass";
$conn = mysql_connect($dbh,$dbu,$dbp) or die("Unable to connect do database.");
mysql_select_db($dbn, $conn) or die("Unable to select database.");
//Some vars for Order by and Limit.
if (!isset($ordBy)){
$ordBy = 1;
}
if (!isset($ps)){
$ps = 0;
}
if (!isset($ord)){
$ord = 1;
}
if ($ord == 1){
$tOrder = "ASC";
} else {
$tOrder = "DESC";
}
//Tables drop-down
$result = mysql_query("SHOW TABLES FROM $dbn") or die("Cannot list table names.");
echo "
<form name=\"table_browser\" action=\"".$PHP_SELF."\" method=\"GET\" >
<select name=\"t\" onChange=\"javascript:submit();\">
<option>Select a table</option>
";
while ($row = mysql_fetch_row($result)){
echo " <option value=".$row[0].">".$row[0]."</option>\n";
}
echo " </select>
</form>\n";
if (!isset($t)){
die("Please select a table");
}
//Get number of rows in $t and then select
$result = mysql_query("SELECT COUNT(*) FROM $t") or die("The requested table doesn't exist.");
$total = mysql_result($result,0);
$qry = "SELECT * FROM $t ORDER BY ".$ordBy." ".$tOrder." LIMIT ".($ps*20).",20 ";
if (isset($qry)) {
$result = mysql_query($qry) or die("The requested table doesn't exist.");
$i = 0;
while ($i < mysql_num_fields($result)) {
$meta = mysql_fetch_field($result);
if (!$meta) {
echo "No information available on the table<br />\n";
}
$name[$i] = $meta->name;
$i++;
}
//Display table details
echo "Rows ".($ps*20+1)." to ".((($ps+1)*20 < $total) ? (($ps+1)*20) : ($total))." of $total from table:
<b>$meta->table</b>\n<br /><br />\n";
//Count results
if ($ps > 0) {
echo "20 Previous - ";
} else {
echo "20 Previous - ";
}
if ((($ps+1)*20) < $total ){
echo "Next 20\n";
} else {
echo "Next 20\n";
}
//Column names
echo "<br /><br />\n<table>\n <tr>\n";
for ($j = 0; $j < $i; $j++){
echo " <td><b>$name[$j]</b>";
if ($ordBy == $name[$j]) {
echo "<img src=\"images/arrow$ord.gif\">";
}
echo "</td>\n";
}
echo " </tr>\n";
//Data
while ($row = mysql_fetch_array($result)){
echo " <tr onmouseover=\"this.style.background='#DDDDDD'\" onmouseout=\"this.style.background=''\">";
for ($j = 0; $j < $i; $j++){
echo "<td>".$row[$name[$j]]."</td>";
}
echo "</tr>\n";
}
echo "</table>\n";
}
mysql_close();
?>

Answer 1

#gentlebreeze - my eyes hurt from trying to read that - but I can't see where you are actually setting the variable called $t - which is used to determine the table. You should have
$t = $_GET['t'];
somewhere before the line:
....if (!isset($t)){....
because you need to use it in the line:
....$result = mysql_query("SELECT COUNT(*) FROM $t"....
and you should not be using mysql_query either - old and now deprecated. You should switch to PDO and bound variables.