PHP从表中选择数据，不工作

好吧，所以我不太了解PHP，当有人提交表单时，它需要检查该datetimestamp是否已经存在于mysql表中，然后执行某些操作。PHP从表中选择数据，不工作

localhost.php：

<?php 
$hostname_localhost = "localhost"; 
$database_localhost = "bookings"; 
$username_localhost = "username"; 
$password_localhost = "password"; 
$localhost = mysql_pconnect($hostname_localhost, $username_localhost, $password_localhost) or trigger_error(mysql_error(),E_USER_ERROR); 
?>

主文件：

<?php require_once('../Connections/localhost.php'); ?> 
<?php 
$date = $_POST['date']; 

$result = mysql_query("SELECT * FROM `bookingtable` WHERE `Date of Session` = '$date'"); 
if(mysql_num_rows($result) > 0) { 
    echo 'the date was found, please try another'; 
} else { 
    echo 'the date wasnt found'; 
}; 
?>

它用于工作，但现在它只是总是说这就是它不是在表的时候它实际上是。

来源

2015-09-06 SadDomino

如果之前stmnt：if（！$ result）{die（mysql_er ROR（））; } – alamnaryab

'echo'找不到日期';'你在这里有一个语法错误，你必须用撇号或者双引号括起来，像这样：'echo'找不到日期';'另外，您使用'$ result'作为查询，但在检查行数时使用'$ result15'。 – Qirel

@Qirel这些只是我输入这个的错误，他们实际上并没有在真正的代码 – SadDomino

写在localhost.php

mysql_select_db($database_localhost);

来源

2015-09-06 20:18:50 alamnaryab

非常感谢，从来没有想过这件事。 – SadDomino

末下面一行试试这个代码：

Localhost.php

<?php 
     $hostname_localhost = "localhost"; 
     $database_localhost = "bookings"; 
     $username_localhost = "username"; 
     $password_localhost = "password"; 
     $tablename_localhost = //your name of your table from database 
     $localhost = mysql_pconnect($hostname_localhost, $username_localhost, $password_localhost, $tablename_localhost) or trigger_error(mysql_error(),E_USER_ERROR); 
    ?>

Main.php

<?php require_once('../Connections/localhost.php'); 


    $date = $_POST['date']; 

    $result = mysql_query($localhost, "SELECT * FROM bookingtable WHERE Date of Session = '$date'"); 

    if(mysql_num_rows($result) > 0) { 
     echo 'the date was found'; 
    } else { 
      echo 'the date wasnt found'; 
    }; 
?>

来源

2015-09-06 20:39:54 aldrin27

PHP从表中选择数据，不工作

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