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我最近从 phpmyadmin 2.6 升级到 5.1，还安装了 wamp...我知道要等很久了。我有一个带有简单变量的网络表单，它们不再插入到我的数据库中

<form method="POST" action="record.php">

 td style="width: 472px">
    <input type=text name=travel size=40 style="color: black;background-color:#FFFF66; width: 270px;"></td>
    </tr>

和插入 php:::

<?
$travel=$_POST['travel'];
$db="xacall";

$link = mysql_connect("localhost", "sql", "sql");
//$link = mysql_connect("localhost",$_POST['username'],$_POST['password']);
if (! $link)
die("Couldn't connect to MySQL");
mysql_select_db($db , $link) or die("Select Error: ".mysql_error());
$result=mysql_query("INSERT INTO ircb (travel) VALUES ('$travel')")or die("Insert Error: ".mysql_error());

//write to file
$fp = fopen("IR_Call_Entries.txt", "a"); 
fwrite($fp, $_POST['travel'] . "\n" . 
close($fp);

?>

我提交，没有收到任何错误消息..但没有任何内容，也没有任何内容...... POST 错误..我应该使用 GET 吗？啊，困惑，谢谢我也回显了变量，但没有显示

4

1 回答 1

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将表格用作：

<form method="post" action="record.php">

<td style="width: 472px">
<input type="text" name="travel" size="40" style="color: black;background-color:#FFFF66; width: 270px;"></td>
</tr>

并使用以下命令调试 post 变量：print_r($_POST)

    <?php #Please use full open tag.
    $travel=$_POST['travel'];
    $db='xacall';

    $link = mysql_connect('localhost', 'sql', 'sql') or die("Couldn't connect to MySQL");
    mysql_select_db($db , $link) or die('Select Error: '.mysql_error());
    $result=mysql_query("INSERT INTO ircb (travel) VALUES ('$travel')")or die('Insert Error: '.mysql_error());


   print_r($_POST);

   #write to file
   $fp = fopen('IR_Call_Entries.txt', 'a'); 
   fwrite($fp, $_POST['travel'] . '\n' ; #you had a concat sign here? 
   close($fp);
   ?>

我也尽可能多地使用单引号。

于 2012-12-11T19:41:52.020 回答