How to make multiple MySQL connections from PHP

Question

The code below is working for one if statement but not giving results for another else if and it is showing the 'flights' table in first query but after another condition not display the other table named 'isb to muree'

<?php
$from = isset($_POST['from'])?$_POST['from']:'';
 $To = isset($_POST['To'])?$_POST['To']:'';
 if( $from =='Islamabad'){
 if($To == 'Lahore'){
$db_host = 'localhost';
$db_user = 'root';
$database = 'homedb';
//$table = 'flights';
if (!mysql_connect($db_host, $db_user))
 die("Can't connect to database");
if (!mysql_select_db($database))
 die("Can't select database");
$result = mysql_query("SELECT * FROM flights");
if (!$result) {
 die("Query to show fields from table failed");
}
$fields_num = mysql_num_fields($result);
echo "<h1>Table: 'flights'</h1>";
echo "<table border='1'><tr>";
while($row = mysql_fetch_row($result))
{
 echo "<tr>";
 // $row is array... foreach( .. ) puts every element
 // of $row to $cell variable
 foreach($row as $cell)
 echo "<td>$cell</td>";
 echo "</tr>\n";
}
}
else if( $from =='Islamabad'){
 if($To == 'murree'){
if (!mysql_connect($db_host, $db_user))
 die("Can't connect to database");
if (!mysql_select_db($database))
 die("Can't select database");
$result = mysql_query("SELECT * FROM 'isb to murree'");
if (!$result) {
 die("Query to show fields from table failed");
}
$fields_num = mysql_num_fields($result);
echo "<h1>Table: 'isb to murree'</h1>";
echo "<table border='1'><tr>";
while($row = mysql_fetch_row($result))
{
 echo "<tr>";
 // $row is array... foreach( .. ) puts every element
 // of $row to $cell variable
 foreach($row as $cell)
 echo "<td>$cell</td>";
 echo "</tr>\n";
}
}
}
}
mysqli_close($con);
?>

Answer 1

Enclose the table name in backticks (`) instead of single quotes ('):

SELECT * FROM `isb to murree`

To avoid this and other problems in the future, always enclose schema, table, and column names in backticks to distinquish them from data and MySQL keywords. Consider using table names without spaces for easier handling.

As Fred noted, your code is using two database APIs: mysql_* and mysqli_* . In general, the two do not mix well. Consider upgrading all of your code to use mysqli_* or PDO. See Deprecated features in PHP 5.5.x and Why are PHP's mysql_ functions deprecated? for mysql_* deprecation notes.

• Funk Forty Niner: +1 Yes, indeed. Just an added note, OP is mixing APIs with mysqli_close($con);

Answer 2

You should move the database connection variables to the top of your code (so they are outside of the if statement)

$db_host = 'localhost';
$db_user = 'root';
$database = 'homedb';

• user3909877: yes now its working for the three elseif conditions but not for the 4th one..syntax is same . what should be the problem?? @billy
• Billy: Double check your if statements.
• Billy: And check your inputs, murree is lowercase where Lahore is title case.
• user3909877: yess they are okay.. elseif ( $from =='Lahore'){ if($To == 'Islamabad'){ @billy
• user3909877: in this part its diverted now from Lahore to Islamabad... @billy