Question

下面的代码适用于一个if语句，但如果并且它正在显示＆＃39;航班＆＃39;第一个查询中的表但在另一个条件之后没有显示名为＆＃39; isb to muree＆＃39;

的另一个表

<?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

将表名括在反引号（`）中而不是单引号（'）：

SELECT * FROM `isb to murree`

为避免将来出现此问题和其他问题，请始终在反引号中包含模式，表和列名称，以区别于数据和MySQL关键字。考虑使用没有空格的表名，以便于处理。

作为Fred noted，您的代码使用了两个数据库API：mysql_*和mysqli_*。一般来说，两者混合不好。考虑升级所有代码以使用mysqli_*或PDO。有关mysql_*弃用说明，请参阅Deprecated features in PHP 5.5.x和Why are PHP's mysql_ functions deprecated?。

Answer 2

您应该将数据库连接变量移到代码的顶部（因此它们在if语句之外）

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

如何从PHP创建多个MySQL连接

2 个答案: