为什么这个PHP脚本不会将表单数据提交给mysql？

问题描述：

This script is supposed to get the content of a text area and submit it to mysql, but it isnt can anyone see why?

    if ($_SERVER["REQUEST_METHOD"] == "POST") 
{
$error = '';
$like = mysql_real_escape_string($_POST['like_box']);

mysql_query("INSERT INTO likes (like) VALUES ($like)");

$id = mysql_query("SELECT id FROM likes WHERE like=$like");
header('Location:like.php?id='.$id.'');
}?>



<form method="post" action="post.php">
                <textarea name="like_box" id="like_box" style="border-style: none; border-color: inherit; border-width: 0; width: 458px; height: 65px" class="style11120"></textarea>

            <tr>
                <td style="height: 53px">
                <div class="style11116" style="width: 417px">
                    <input name="Submit" type="submit" value="submit" />
                    </form>

这个脚本应该获取文本区域的内容并将其提交给mysql，但它不是任何人都能看到的为什么？ p>

  if（$ _SERVER [“REQUEST_METHOD”] ==“POST”）
 {
 $ error =''; 
 $ like = mysql_real_escape_string（$  _POST ['like_box']）; 
 
mysql_query（“INSERT INTO like（like）VALUES（$ like）”）; 
 
 $ id = mysql_query（“SELECT id FROM like WHERE like = $ like”）;  
header（'Location：like.php？id ='。$ id。''）; 
}？＆gt; 
 
 
 
＆lt; form method =“post”action =“post.php”＆gt  ; 
＆lt; textarea name =“like_box”id =“like_box”style =“border-style：none; border-color：inherit; border-width：0; width：458px; height：65px”class =“style11120”  ＆gt;＆lt; / textarea＆gt; 
 
＆lt; tr＆gt; 
＆lt; td style =“height：53px”＆gt; 
＆lt; div class =“style11116”style =“width：417px”＆gt; 
  ＆lt; input name =“Submit”type =“submit”value =“submit”/＆gt; 
＆lt; / form＆gt; 
  code>  pre> 
   DIV>

答

Having some error reporting would tell you that you need ' around the $like and you need ` around like in the columns section, since like is a reserved word, inside the insert.

mysql_query("INSERT INTO likes (`like`) VALUES ('$like')") or trigger_error('Query Error: ' . mysql_error());

Should work.

Also you will need to enclose the like in ` for the select:

$id = mysql_query("SELECT id FROM likes WHERE `like`=$like") or trigger_error('Query Failed: ' . mysql_error());

答

Your first SQL query is wrong. You have to enclose string values with quotation marks in SQL:

mysql_query("INSERT INTO likes (`like`) VALUES ('$like')");

The rest of your PHP part will not perform as you wish either. $id will be a resultset, not a value. You'll have to fetch the row like this:

$result = mysql_query("SELECT id FROM likes WHERE `like`='$like'");
$row = mysql_fetch_assoc($result);
$id = $row['id'];

Or even better, just replace the second SQL query with a call to mysql_insert_id:

$id = mysql_insert_id();

Still more possibilities for improvement:

Don't forget to call exit after you've done a header('Location: ...'); call, otherwise you may get unexpected results since the script will continue running.
Add some error handling of your insert. At least check with an if statement that it succeeded.

为什么这个PHP脚本不会将表单数据提交给mysql？

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