Question

当我运行此页面时，它会显示You couldnt execute query。我认为存在语法问题，但我看不出问题。

当我回应它们时，变量看起来很好。我添加了列名作为代码注释。所有列都是VARCHAR。

我不相信这是一个连接问题，因为我可以从其他SELECT AND DELETE页面执行php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

按如下所示更改您的查询，它将起作用

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

Answer 2

谢谢大家，这只是其中一个专栏中的$符号，并在$ id＆gt;中再次放回引号＆＃39; $ ID＆＃39;

$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.";

?>

使用PHP和变量更新sql查询

2 个答案: