使用PHP和變量更新sql查詢

Question

當我運行此頁面時，它表示You couldnt execute query 。 我認為存在語法問題，但我看不出問題。

當我回應它們時，變量看起來很好。 我添加了列名作為代碼注釋。 所有列都是VARCHAR 。

我不相信這是一個連接問題，因為我可以從其他php頁面執行SELECT AND DELETE操作。

<?php 

$id = $_POST["id"];
$name = $_POST["name"];
$surname = $_POST["surname"];
$phone = $_POST["phone"];

/* Table in the DDBB(all are VARCHAR): CUST_ID  CUST_FORENAME   CUST_PHONE  CUST_SURNAME*/

$query = "UPDATE customers SET CUST_ID='$id',$CUST_FORENAME='$name',
            CUST_PHONE='$phone', CUST_SURNAME='$surname' WHERE CUST_ID=$id";

$result = mysqli_query ($connection,$query)
    or die ("You couldn’t execute query");

echo "<br /><br />User $id has been updated.";

?>
</div>
</body>
</html>

Answer 1

如下更改您的查詢，它將工作

$query = "UPDATE customers SET CUST_ID='$id', CUST_FORENAME='$name',
                 CUST_PHONE='$phone', CUST_SURNAME='$surname' WHERE CUST_ID='$id'";

Answer 2

謝謝大家，這只是其中一個列中的$符號，並在$ id>'$ id'中再次放回引號。

$id = $_POST["id"];
$name = $_POST["name"];
$surname = $_POST["surname"];
$phone = $_POST["phone"];

/*CUST_ID   CUST_FORENAME   CUST_PHONE  CUST_SURNAME*/

$query = "UPDATE customers SET CUST_ID='$id',CUST_FORENAME='$name',
            CUST_PHONE='$phone', CUST_SURNAME='$surname' WHERE CUST_ID='$id'";

$result = mysqli_query ($connection,$query)
    or die(mysqli_error($connection));

echo "<br /><br />User $id has been updated.";

?>