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将数据存储到mysql数据库的PHP语法错误_Php_Mysql - Fatal编程技术网

将数据存储到mysql数据库的PHP语法错误

将数据存储到mysql数据库的PHP语法错误,php,mysql,Php,Mysql,我已经解决了这个问题。下面的代码现在可以完美地工作。 谢谢大家 请访问dbcontroller.php文件的相关部分,如下所示: <?php class DBController { function runQuery2($query) { $result = mysql_query($query); return $result; } } 您的代码: <?php session_start(); require_on

我已经解决了这个问题。下面的代码现在可以完美地工作。 谢谢大家

请访问dbcontroller.php文件的相关部分,如下所示:

<?php
class DBController {


function runQuery2($query) {
    $result = mysql_query($query);
            return $result;
    }

 }
您的代码:

<?php
    session_start();
    require_once("dbcontroller.php");
    $db_handle = new DBController();

    if(!empty($_POST["submit"])) {

        $sql = "INSERT INTO cquestionstable (postid, ccode, nick, queries) VALUES ( 1,'cc-001', 'james', 'what could be the problem?')";
        if ($db_handle->runQuery($sql) === TRUE) {
            echo "New record created successfully";
        } else {
            echo "Error: " . $**sql . "<br>" . $db_handle->error;
        }
    }
?>
dbcontroller.php

<?php

$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

?>

如果您仍然不明白为什么它不工作,请查看


希望这就是我们正在寻找的答案。

您遇到了什么错误?也可以这样更改行:-
echo“Error:”$sql。“
”$db_句柄->错误
remove
**
如果postid设置为自动增量,我不会将其包含在
insert语句中。可能问题出在您试图插入的数据中<代码>变量转储
数据。显示dbHandler类。原因:请按要求查找dbHandler类。
<?php
    // session_start(); You do not need this when inserting into database
    include "dbcontroller.php";

    if (isset($_POST["submit"])) {

        $sql = "INSERT INTO cquestionstable (postid, ccode, nick, queries) VALUES (1, 'cc-001', 'james', 'what could be the problem?')";
        if ($conn->query($sql) === TRUE) {
            echo "New record created successfully";
        } else {
            echo "Error: " . $sql . "<br>" . $connn->error;
        }
    }
?>
<?php

$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

?>