数据库错误没有插入值 phpMyAdmin答案

【问题标题】：Database error doesn't get values inserted phpMyAdmin数据库错误没有插入值 phpMyAdmin
【发布时间】：2013-06-07 05:10:51
【问题描述】：

ctimings.html

    <html>
        <head>
            <link rel="stylesheet" type="text/css" href="cstyle.css">
            //  <script type="text/javascript">
            //      function validate(form)
            //       { 
            //          if(document.myForm.sno.value == "")
            //          {
            //              alert("Please Complete Serial NUmber");
            //              return false;
            //          }
            //          if(document.myForm.ssign.value == "")
            //          {
            //              alert("Please Complete Signature");
            //              return false;
            //          }
            //          if(document.myForm.empName.value == "")
            //          {
            //              alert("Please Complete Employee Name");
            //              return false;
            //          }
            //          if(document.myForm.empSign.value == "")
            //          {
            //              alert("Please Complete Employee Signature");
            //              return false;
            //          }
            //          if(document.myForm.empCode.value == "")
            //          {
            //              alert("Please Complete Employee Code");
            //              return false;
            //          }
            //      }
            // </script>
            <title>Conference Room Timings</title>
        </head>
        <body>
            <form  action = "ctimings.php" method = "post" name = "myForm" onsubmit="return validate(this);">
                <legend>ADD DETAILS</legend>
                <fieldset>
                    <label>Serial Number <input type = "int" name = "sno" size="20" class = "required" /></label><br/>
                    <label>Security Signature <input type = "text" name = "ssign" size="20" class = "required" /></label><br/>
                    <legend>Employee Details</legend>
                    <fieldset>
                        <label>Employee Name <input type = "text" name = "empName" size="20" class = "required" /></label><br/>
                        <label>Employee Signature <input type = "text" name = "empSign" size="20" class = "required"/></label><br/>
                        <label>Employee Code <input type = "int" name = "empCode" size="20" class = "required" /></label><br/>
                    </fieldset>


                    <legend>Date & Time</legend>
                    <fieldset>
                        <label>Date <input type = "int" name = "date" size="20" class = "required"/></label><br/>
                        <label>In - Time <input type = "int" name = "inTime" size="20" class = "required"/></label><br/>
                        <label>In - Time <input type = "int" name = "outTime" size="20" class = "required"/></label><br/>
                    </fieldset>

                    <legend>Usage</legend>
                    <fieldset>
                        <label>
                        Phone Set Usage
                        <input type = "radio" name = "radio1" value = "YES" />
                        <input type = "radio" name = "radio1" value = "NO" /><br/>
                    </label>
                    <label>
                        Projector Usage
                        <input type = "radio" name = "radio2" value = "YES" />
                        <input type = "radio" name = "radio2" value = "NO" /><br/>
                    </label>
                    </fieldset>

                    <input type = "submit" value = "SUBMIT" class = "" name =  "SUBMIT"/>

                </fieldset>

            </form>
        </body>


    </html>

ctimings.php

<html>
    <body>
    <?php


        if(isset($_POST['SUBMIT']))
        {
            echo "pressed";
            $db_host = "localhost";
            $db_username = "root";
            $db_pass = "";
            $db_name = "ibm";
            $con=mysql_connect("localhost","root","","ibm");

        // Check connection
            $link = mysql_connect('localhost', 'root', '');
            if (!$link) {
                die('Not connected : ' . mysql_error());
            }
            else
            {
                echo 'Connection was OK';
            }
            $db_selected = mysql_select_db($db_name, $link);
            if (!$db_selected) {
                die ('Can\'t use ibm : ' . mysql_error());
            }
            else
            { 
                echo "Database was OK!\n";
            }
            $sql = "INSERT INTO ctimings(sno,ssign,empName,empSign,date,intime,outtime,empCode,Proj.Use,Ph.Use) VALUES('$_POST[sno]','$_POST[ssign]','$_POST[empName]','$_POST[empSign]','$_POST[date]','$_POST[inTime]','$_POST[outTime]','$_POST[empCode]','$_POST[radio1]','$_POST[radio2]')";
            mysql_query($sql,$link);
            mysql_close($link);
          }
          else
            echo 'error';

    ?>
    </body>
</html>

这是一个简单的过程，但我无法缩小为什么没有将数据发送到数据库，因为 phpmyadmin 表“ctimings”显示它没有记录。按下提交按钮后应该将数据post到数据库中，但是没有成功。数据库中的数据库名和表名是正确的。

【问题讨论】：

标签： php mysql phpmyadmin xampp database-connection

【解决方案1】：

$_POST 是一个全局数组，你必须像这样使用它 $_POST["field_name"];

【讨论】：

【解决方案2】：

你的查询会是这样的

$sql = "INSERT INTO ctimings(sno,ssign,empName,empSign,date,intime,outtime,empCode,Proj.Use,Ph.Use) VALUES
('".$_POST[sno]."','".$_POST[ssign]."','".$_POST[empName]."','".$_POST[empSign]."','".$_POST[date]."',
 '".$_POST[inTime]."','".$_POST[outTime]."','".$_POST[empCode]."','".$_POST[radio1]."'
,'".$_POST[radio2]."')";

您需要使用 mysql_real_escape_string

转义 post 变量

主要原因是由于 Proj.Use,Ph.Use 我认为您一次使用插入两个表，

【讨论】：

【解决方案3】：

你的sql字符串不正确，应该是；

$sql = "INSERT INTO ctimings(sno,ssign,empName,empSign,date,intime,outtime,empCode,Proj.Use,Ph.Use) VALUES('".$_POST[sno]."','".$_POST[ssign]."','".$_POST[empName]."','".$_POST[empSign]."','".$_POST[date]."','".$_POST[inTime]."','".$_POST[outTime]."','".$_POST[empCode]."','".$_POST[radio1]."','".$_POST[radio2]."')";

【讨论】：

【解决方案4】：

要添加到上面的答案，清理数据可能是一个想法。

$_POST = array_map('mysql_real_escape_string', $_POST);

【讨论】：