Question

使用html表单填充数据库时遇到问题。一旦我运行代码，网页输出＆＃34;成功完成＆＃34;但记录没有显示。看起来它为每个字段输入空值，因为当我尝试输入另一条记录时，我收到此错误＆＃34;连接成功错误：重复输入＆＃39; 0＆＃39; 0关键＆＃39; PRIMARY＆＃39;＆＃34;我该如何解决这个问题？

<form action="addplayer.php"method "post"/>

     <p> id: <input type="text" name="playerid"/></p>
     <p> Name: <input type="text" name="name"/></p>
     <p> Age: <input type="text" name="age"/></p>
     <p> Position: <input type="text" name="position"/></p>
     <p> Nationality: <input type="text" name="nationality"/></p>
 <input type="submit" value="submit"/>


     ?php
require 'connection.php';


$id = filter_input(INPUT_POST, 'playerid');
$name = filter_input(INPUT_POST, 'name');
$age = filter_input(INPUT_POST, 'age');
$position = filter_input(INPUT_POST, 'position');
$nationality = filter_input(INPUT_POST, 'nationality');

$_id = mysql_real_escape_string( $id );
$_name = mysql_real_escape_string( $name );
$_age = mysql_real_escape_string( $age );
$_position = mysql_real_escape_string( $position );
$_nationality = mysql_real_escape_string( $nationality );
$sql = "INSERT INTO players ( playerid, name, age, position, nationality ) 
         VALUES ( '$_id', '$_name', '$_age', '$_position', '$_nationality' )";

if (!mysql_query($sql)){
    die('Error: ' . mysql_error());
}

Answer 1

您需要在输入中添加id：

<p> id: <input type="text" name="playerid" id="playerid"/></p>

收集如下数据：

$playerID = mysql_real_escape_string($_POST['playerid'])

$sql = "INSERT INTO players
        SET
        playerID = '".$playerID."',
        ........";

使用html表单将记录添加到数据库

1 个答案: