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问PHP从表中选择数据,但无法工作
EN

Stack Overflow用户
提问于 2015-09-06 19:43:59
回答 2查看 176关注 0票数 0

好的,所以我并不真正了解PHP,当有人提交一个表单时,它需要检查这个日期时间戳是否已经在mysql表中了,然后再做一些事情。

localhost.php:

代码语言:javascript
复制
<?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);
?>

主档案:

代码语言:javascript
复制
<?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';
};
?>

它曾经起作用,但现在它总是说,它不是在桌子上,当它真的是。

EN

回答 2

Stack Overflow用户

回答已采纳

发布于 2015-09-06 20:18:50

在localhost.php的末尾写下一行

代码语言:javascript
复制
mysql_select_db($database_localhost);
票数 0
EN

Stack Overflow用户

发布于 2015-09-06 20:39:54

试试下面的代码:

Localhost.php

代码语言:javascript
复制
   <?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

代码语言:javascript
复制
    <?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';
     };
 ?>
票数 0
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/32427581

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