PHP，MySQL表查詢語法錯誤？

Question

我希望有人可以幫助您查看這里出了什么問題：我有一個包含兩個字段EMAIL和PASSWORD的表單，該表單打開一個php頁面，我打算在該頁面上對表運行簡單查詢。 我收到一條毫無意義的錯誤消息：
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '@gmail.com' at line 1 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '@gmail.com' at line 1 。
我在這種情況下輸入的電子郵件地址確實以“ @ gmail.com”結尾
這是代碼：

<?php
$dbhost = 'somewhere.net';
$dbuser = 'someUser';
$dbpass = 'pass';

$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql');

$dbname = 'medreunten_db1';
mysql_select_db($dbname) or die(mysql_error($conn)); 

$email = mysql_real_escape_string($_POST['email']);

$query = "SELECT * FROM employee WHERE email = $email";
$result = mysql_query($query, $conn) or die (mysql_error($conn));
extract(mysql_fetch_assoc($result));  


while ($row = mysql_fetch_array($result)) {
extract($row);
echo $row['name'];
echo $row['surname'];
echo $row['age'];
}
?>

任何意見，將不勝感激。

Answer 1

您缺少字符串字段周圍的引號：

$query = "SELECT * FROM employee WHERE email = '$email'";

另外，

extract(mysql_fetch_assoc($result));  

將從數據庫中獲取第一行，因此while循環將從第二行開始。

Answer 2

您必須將值放在SQL字符串內的引號中。

$email = mysql_real_escape_string($_POST['email']);

$query = "SELECT * FROM employee WHERE email = '$email'";

（介意額外''大約$電子郵件）

Answer 3

您的查詢翻譯為：

SELECT * FROM emloyee WHERE email = foo@bar.com

這不起作用，您必須將字符串放在引號中。 將您的代碼更改為以下代碼，它將起作用：

$query = "SELECT * FROM employee WHERE email = '$email'";

Answer 4

僅將變量'$ email'單引號，因為它是varchar類型的value和field。 如所寫，達哈澤:)

Answer 5

完整的固定代碼：

<?php
$dbhost = 'somewhere.net';
$dbuser = 'someUser';
$dbpass = 'pass';

$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql');

$dbname = 'medreunten_db1';
mysql_select_db($dbname) or die(mysql_error($conn)); 

$email = mysql_real_escape_string($_POST['email']);

$query = "SELECT * FROM employee WHERE email = '$email'";
$result = mysql_query($query, $conn) or die (mysql_error($conn));
extract(mysql_fetch_assoc($result));  


while ($row = mysql_fetch_array($result)) {
extract($row);
echo $row['name'];
echo $row['surname'];
echo $row['age'];
}
?>