Question

我该怎么做？我试过各种不同的方法，但我不能这样做！

基本上，我试图从“初学者表”中提取信息并仅显示登录的用户数据。

这是代码，它给出了一条我无法解决的错误消息：

 <?php
session_start();
require_once '../database.php';
if (isset($_SESSION['myusername'])){
echo "Welcome ".$_SESSION['myusername'];
}
?>

<?php
include '../database.php';

$userid = $_SESSION["myusername"];

#the where clause is where im stuck at the moment!

Line 50:
$result = mysql_query("SELECT Recipename, Ingredients, Method, Time FROM starters WHERE username = $_SESSION['myusername']");

echo "<table border='0'><table border width=65%> <tr><th>Recipie Name</th><th>Ingredients</th><th>Method</th><th>Time</th></tr>";

 while($row = mysql_fetch_array($result))
{
        echo "<tr>";
        echo "<td>" . $row['Recipename']. "</td>";
        echo "<td>" . $row['Ingredients']. "</td>";
        echo "<td>" . $row['Method']. "</td>";
        echo "<td>" . $row['Time']. 'minutes'."</td>";

echo "</tr>";
}
echo "</table>";

        ?>
</table>

我得到的错误信息如下：

    Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in /home/jahedhus/public_html/cook/usersloggedin/starters.php on line 50

第50行是select语句！我将衷心感谢您的帮助，非常感谢！

Answer 1

您需要将用户名包装在引号中：

$result = mysql_query("SELECT Recipename, Ingredients, Method, Time FROM starters WHERE username = '{$_SESSION['myusername']}'");

此外，您相信$_SESSION['myusername']是有效的用户名（不是某些附加SQL字符串）。如果您不知道数据是安全的，则至少需要escape the data。

Answer 2

更改行：

$result = mysql_query("SELECT Recipename, Ingredients, Method, Time FROM starters WHERE username = $_SESSION['myusername']");

这一个：

$result = mysql_query("SELECT Recipename, Ingredients, Method, Time FROM starters WHERE username = '" . $_SESSION['myusername'] . "'");

Answer 3

引号的问题在于它们会全部搞砸了，我会将它们分开：

$result = mysql_query("SELECT Recipename, Ingredients, Method, Time FROM starters WHERE username = ".$_SESSION['myusername']);

OR

$result = mysql_query("SELECT Recipename, Ingredients, Method, Time FROM starters WHERE username = '$_SESSION[myusername]'");

匹配用户名（mysql表中的字段）到PHP中的会话ID

3 个答案: