Question

请帮我这个。我认为我的查询是正确的，但它不会插入到我的表中。

$dbhost = "localhost";
$dbuser = "root";
$dbpass = "root";
$dbname = "test";

$connection = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);

if (!empty($_POST['name']) AND !empty($_POST['info'])) {
    $name = mysql_real_escape_string($_POST['name']);
    $info = mysql_real_escape_string($_POST['info']);

    $query = "INSERT INTO comment (name, info) VALUES('$name', '$info') ";
    if(mysqli_query($connection,$query)){
        $print = "success";
    }
    else{
        $print = "not success";
    }
}

Answer 1

弗雷德试图说的是你应该调整你的代码

$dbhost = "localhost";
$dbuser = "root";
$dbpass = "root";
$dbname = "test";

$con= mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);

if (!empty($_POST['name']) AND !empty($_POST['info'])) {
$name = $con->real_escape_string($_POST['name']);
$info = $con->real_escape_string($_POST['info']);

$query = "INSERT INTO comment (name, info) VALUES('$name', '$info') ";
if(mysqli_query($con,$query)){
    $print = "success";
}
else{
    $print = "not success";
}
}

应该有效

参考文献：php MYSQLI

Answer 2

一旦您尝试仅指定值，即

$query = "INSERT INTO comment VALUES('$name', '$info') ";

某些时候它可以运作

Answer 3

试试这个：

$query = "INSERT INTO comment (name, info) VALUES('".$name."', '".$info."') ";

Answer 4

我的兄弟......

你用过这个：$ query =＆＃34; INSERT INTO评论（姓名，信息）价值观（＆＃39; $ name＆＃39;，＆＃39; $ info＆＃39;）＆＃34 ;; < / p>

但是你忘记了SET：使用这个：

$query = "INSERT INTO comment SET (name, info) VALUES('$name', '$info') ";

mySQL插入失败

4 个答案: