我的ask.php语法有什么问题？

So I modified the script as I was instructed earlier, but I am still getting a blank page when I run the HTML form. What is wrong here?

所以我按照之前的指示修改了脚本，但是当我运行HTML表单时，我仍然得到一个空白页面。这有什么不对？

UPDATE 2: I get the following error now.

更新2：我现在收到以下错误。

Parse error: syntax error, unexpected T_VARIABLE in /home/content/o/m/o/omorgan/html/dimephysics/adviseme/ask.php on line 32

解析错误：语法错误，第32行/home/content/o/m/o/omorgan/html/dimephysics/adviseme/ask.php中的意外T_VARIABLE

<?php

ini_set('display_errors',1);
error_reporting(E_ALL|E_STRICT);

//include('config.php');
//include('open_connection.php');



if (!isset($_POST['name']) || !isset($_POST['question']))
    {
        header ("Location: ask.html");
        exit;
    }

    // Database parameters
    $dbhost = '...';
    $dbuser = 'questionme';
    $dbpass = 'Question_2011';
    $dbname = 'questionme';

    $db_name = "questionme";
    $table_name = "questions";

    //Open connection
    $connection = mysql_connect("", "questionme", "Question_2011")
        or die(mysql_error());

    $db = mysql_select_db($db_name, $connection) or die(mysql_error())

    $name = mysql_escape_string($_POST[name]);
    $question = mysql_escape_string($_POST[question]);

    //Insert data into database
    $sql = "INSERT INTO questions
    (name, question) VALUES
    ('$name', '$question')";

?>

<html>
<head>
<title>Ask</title>
<head>
<body>

<h1>Question Submitted</h1>

<p><strong>Name:</strong>
<?php echo $_POST['name']; ?></p>

<p><strong>Question:</strong>
<?php echo $_POST['question']; ?></p>

</body>
</html>

7 个解决方案

#1

At a bare minimum you need quotes around your array indexes on this line:

至少，您需要在此行上的数组索引周围引用：

if (!isset($_POST[name]) || !isset($_POST[question]))

Make it

做了

if (!isset($_POST['name']) || !isset($_POST['question']))

Also, check your error level (can't give you a link at the moment, in a bit of a rush), PHP should be warning you about this.

另外，检查你的错误级别（暂时不能给你一个链接，有点急），PHP应该警告你这个。

#2

try changing

尝试改变

if (!isset($_POST[name]) || !isset($_POST[question]))

至

if (!isset($_POST['name']) || !isset($_POST['question']))

#3

Place this at the top of the script:

将它放在脚本的顶部：

ini_set('display_errors',1);
error_reporting(E_ALL|E_STRICT);

This should let you know whats going on.

这应该让你知道发生了什么。

#4

Given your error message, it's obvious that your query is failing. You've supressed errors with @ on it, so the or die(...) never kicks in.

鉴于您的错误消息，很明显您的查询失败。你已经用@来抑制错误，所以或者死（......）永远不会开始。

Your $table_name in the query is undefined, so the query looks like

您的$ table_name在查询中是未定义的，因此查询看起来像

INSERT INTO (name, question) ...

which is incorrect SQL.

这是不正确的SQL。

The two major fixes you need:

您需要的两个主要修复：

Remove the @ supression on mysql_query(). It is almost NEVER acceptable to supress errors, particularly when dealing with a database.
删除mysql_query（）上的@ supression。压制错误几乎是不可接受的，特别是在处理数据库时。
Define $table_name in your script, or change the variable inside the query string to a proper table name.
在脚本中定义$ table_name，或将查询字符串中的变量更改为正确的表名。

#5

First of all, don't check with isset. Since $_POST is always generated you should be using !empty().

首先，不要用isset检查。由于$ _POST总是生成，所以你应该使用！empty（）。

Remove @'s before mysql_* commands, this makes your script slow and suppresses helpful errors.

在mysql_ *命令之前删除@，这会使您的脚本变慢并抑制有用的错误。

And you are having issues because you don't have table variable set, $table_name needs to be defined.

而且由于您没有设置表变量，因此需要定义$ table_name。

If you are inserting questions to a table named 'questions' simply change your SQL to:

如果要将问题插入名为“questions”的表中，只需将SQL更改为：

//Insert data into database
$sql = "INSERT INTO `questions` (name, question)
             VALUES ('$name', '$question')";

#6

You forgot add a semicolon *;* in line no. 30

你忘了在行号中加一个分号*; *三十

it should be

它应该是

   $db = mysql_select_db($db_name, $connection) or die(mysql_error());

instead of

代替

  $db = mysql_select_db($db_name, $connection) or die(mysql_error())

#7

You need to end this line:

你需要结束这一行：

$db = mysql_select_db($db_name, $connection) or die(mysql_error())

change it to:

将其更改为：

$db = mysql_select_db($db_name, $connection) or die(mysql_error());

#1