【发布时间】:2017-08-15 03:16:01
【问题描述】:
所以我试图从我的表单提交中获取我的数据以放入 mysql 数据库,但每当我提交表单时,它都会给我这个错误:错误:INSERT INTO form_submissions(ID, first, @ 987654329@, phone, class) 值 ([value-1],[value-2],[value-3],[value-4],[value-5])
现在这是我的 PHP 代码:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "form_database";
$value = $_POST['first'];
$value1 = $_POST['last'];
$value2 = $_POST['phone'];
$value3 = $_POST['class'];
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error){
die("connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO `form_submissions`(`ID`, `first`, `last`, `phone`,
`class`) VALUES ([value-1],[value-2],[value-3],[value-4],[value-5])";
if ($conn->query($sql) === TRUE) {
echo "Submitted Successfully";
} else {``
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
【问题讨论】:
-
使用引号而不是括号作为字符串分隔符
-
你所拥有的
[value-x]是 mssql/oracle 语法,而不是 mysql -
并确保在看到您"Table 'form_submissions' is read only" 的此评论后,您拥有执行该操作的正确权限 - 您在这里阅读 cmets,对吧?
-
我猜你不是。
标签: php mysql forms mysqli xampp