-1

我正在尝试制作一个更新数据库的表单，但它给了我一个错误。你知道它可能来自什么吗？错误：

解析错误：语法错误，第 9 行的 D:\2013.1\xampp\htdocs\ranklist_get.php 中出现意外的 '$Points' (T_VARIABLE)

欢迎.html 

<body>
<form action="ranklist_get.php" method="get">
Skype: <input type="text" id="Skype"><br>
Points: <input type="number" id="Points"><br>
<input type="submit">
</form>
</body>
</html>

ranklist_get.php * Abhik Chakraborty 解决的代码 :)

<?php
$con=mysqli_connect("localhost","root","","my_db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

mysqli_query($con,"UPDATE Persons SET Points='".$Points."' WHERE Skype='".$Skype."'");

mysqli_close($con);
?>
4

2 回答 2

-1

改变

mysqli_query($con,"UPDATE Persons SET Points="$Points";
WHERE Skype="$Skype""); 

to 

mysqli_query($con,"UPDATE Persons SET Points='".$Points."' WHERE Skype='".$Skype."'");
于 2014-01-26T08:31:53.953 回答
-1

使用有效的语法。应该是

mysqli_query($con,"UPDATE Persons SET Points = '" . $Points . "' WHERE Skype = '" . $Skype . "'");
于 2014-01-26T08:36:54.070 回答