如何从PHP建立多个MySQL连接

Question

下面的代码适用于一个if语句，但没有给出另一个if语句的结果，它在第一个查询中显示了'flights'表，但在另一个条件之后，没有显示另一个名为'ISB to Muree'的表。

<?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

您应该将数据库连接变量移到代码的顶部（这样它们就在if语句之外）

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