php - VAR 数据未推送至 SQL

Question

我遇到了一个小问题，但我不知道为什么。也许有人可以帮助我。首先是一些免责声明；我仍在学习 PHP，我知道 mysqli 或 pdo，但是它将运行的服务器运行的是旧版本 4 的 php。

好的，现在开始解决问题。

我有一个表单传递到我的 post-data.php 表单以推送到 SQL 数据库。

但是，当它推送数据时，它只是推送变量，而不是表单后操作中的变量内的数据。

Screenshot of submitted data in PHPmyadmin

我的代码如下:

<?php
    $hostname = "localhost"; $username = "goldme_owner";
    $dbName = "goldme_dealer_meeting";

    $connect = mysql_connect($hostname, $username);
    if (!$connect) { 
        echo "Please try later."; 
    } 
    else { 
        mysql_select_db($dbName, $connect); 
        $checkboxA1 = isset($_POST['checkboxA1']) ? $_POST['checkboxA1'] : 'No';
        $checkboxE1 = isset($_POST['checkboxE1']) ? $_POST['checkboxE1'] : 'No';
        $checkboxF1 = isset($_POST['checkboxF1']) ? $_POST['checkboxF1'] : 'No';
        $checkboxG1 = isset($_POST['checkboxG1']) ? $_POST['checkboxG1'] : 'No';
        $checkboxH1 = isset($_POST['checkboxH1']) ? $_POST['checkboxH1'] : 'No';
        $checkboxI1 = isset($_POST['checkboxI1']) ? $_POST['checkboxI1'] : 'No';
        $checkboxJ1 = isset($_POST['checkboxJ1']) ? $_POST['checkboxJ1'] : 'No';
        $checkboxK1 = isset($_POST['checkboxK1']) ? $_POST['checkboxK1'] : 'No';
    }

    echo "$checkboxA1"; //just want to make sure checkbox vars are passing //will delete in final code
    echo "$checkboxE1"; //just want to make sure checkbox vars are passing //will delete in final code
    echo "$checkboxF1"; //just want to make sure checkbox vars are passing //will delete in final code
    echo "$checkboxG1"; //just want to make sure checkbox vars are passing //will delete in final code
    echo "$checkboxH1"; //just want to make sure checkbox vars are passing //will delete in final code
    echo "$checkboxI1"; //just want to make sure checkbox vars are passing //will delete in final code
    echo "$checkboxJ1"; //just want to make sure checkbox vars are passing //will delete in final code
    echo "$checkboxK1"; //just want to make sure checkbox vars are passing //will delete in final code
    echo "$_POST[confirm]"; //just want to make sure confirm code generated //will delete in final code

        $sql_statement = 'INSERT INTO 2014_registrations'. 
        '(confirm_number,timecode,company_name,country,address1,address2,city'.
        ',state,zip,phone,fax,email,zone_manager,transport,first_name,'.
        'last_name,tuesday_tours,tuesday_meat,wed_lunch,wed_dinner,'.
        'wed_pool_tourney,thurs_lunch,thurs_dinner,fri_shop,fri_tour,dietary)'.
        'VALUES ("$_POST[confirm]","$_POST[timecode]","$_POST[company_name]",'.
        '"$_POST[address]","$_POST[address2]","$_POST[city]","$_POST[state]",'.
        '"$_POST[zip]","$_POST[country]","$_POST[phone]","$_POST[fax]",'.
        '"$_POST[email]","$_POST[zonemanager]","$_POST[transport]",'.
        '"$_POST[fattendee1]","$_POST[lattendee1]","$_POST[checkboxA1]",'.
        '"$_POST[radio1]","$_POST[checkboxE1]","$_POST[checkboxF1]",'.
        '"$_POST[checkboxG1]","$_POST[checkboxH1]","$_POST[checkboxI1]",'.
        '"$_POST[checkboxJ1]","$_POST[checkboxK1]","$_POST[dietary1]")';


    $rec_insert = mysql_query($sql_statement);
    if(! $rec_insert ){
        die('Could not enter data: ' . mysql_error());
    }
    echo "Entered data successfully\n";
    mysql_close($connect);
?> 

Answer 1

单引号内的变量不会被读取为变量。例如echo '$a'将打印$a但是echo "$a"将打印 $a 的值.

在您的代码中，您尝试使用类似于 '$a' 的内容，或者换句话说，您在变量周围使用单引号而不是双引号。

尝试将其写为

"VALUES ('$_POST[confirm]','$_POST[timecode]','$_POST[company_name]',".

或

'VALUES ("'.$_POST[confirm].'","'.$_POST[timecode].'","'.$_POST[company_name].'",'.

等等..

请注意使用$_POST[confirm]发出通知。正确的使用方法是$_POST['confirm'] .

另请注意，您的代码容易受到 SQL 注入(inject)攻击。考虑使用准备好的语句。