“php没有插入到xampp中的mysql服务器”

Question

“我已经阅读了很多在 stackoverflow 中解决的与我的问题类似的问题，并且已经看到了很多示例，但我的代码仍然没有插入到 mysql 中。但是，如果我硬输入 php，它会插入。 “

<?php
$servername = "localhost";
$username = "root";
$password = "";
$db="image";


// Create connection

$connection = mysqli_connect($servername, $username, $password, $db); // Establishing Connection with Server
if (!$connection) {
      die("Connection failed: " . mysqli_connect_error());
    }
else{
      echo "Connected successfully"; 
    }


if(isset($_POST['submit'])){ // Fetching variables of the form which travels in URL
$name = $_POST['name'];
$image = $_POST['image'];

echo $name;
echo $image;

if($name !=''||$image !=''){
//Insert Query of SQL
$query = mysqli_query("INSERT INTO image (id, name, imagename) VALUES ('NULL', '$name', '$image')");
echo "Data Inserted successfully...!!";
}
else{
echo "Insertion Failed <br/> Some Fields are Blank....!!";
}
}
mysqli_close($connection); // Closing Connection with Server
?>


<form action = "test2.php" method="POST" enctype="multipart/form-data">
    <label>name: </label><input type="text" name="name" />
    <label>File: </label><input type="text" name="image" />
    <input type="submit" />
</form>
</body>
</html>

" 我希望 5/2 的输出为 2.5"

Answer 1

写下提交按钮的名称

<input type="submit" name="submit" />

然后在php文件中

if(isset($_POST['submit'])){ // Fetching variables of the form which travels in URL

}

这个 if 语句将运行

Answer 2

你没有给name属性button所以给name="submit"如果你想上传文件然后改变type="file"

<?php
$servername = "localhost";
$username = "root";
$password = "";
$db="image";


// Create connection

$connection = mysqli_connect($servername, $username, $password, $db); // Establishing Connection with Server
if (!$connection) {
      die("Connection failed: " . mysqli_connect_error());
    }
else{
      echo "Connected successfully"; 
    }


if(isset($_POST['submit'])){ // Fetching variables of the form which travels in URL
$name = $_POST['name'];
$image = $_POST['image'];

echo $name;
echo $image;

if($name !=''||$image !=''){
//Insert Query of SQL
$query = mysqli_query("INSERT INTO image (id, name, imagename) VALUES ('NULL', '$name', '$image')");
echo "Data Inserted successfully...!!";
}
else{
echo "Insertion Failed <br/> Some Fields are Blank....!!";
}
}
mysqli_close($connection); // Closing Connection with Server
?>


<form action = "test2.php" method="POST" enctype="multipart/form-data">
    <label>name: </label><input type="text" name="name" />
    <label>File: </label><input type="file" name="image" />
    <input type="submit" name="submit" />
</form>
</body>
</html>

Answer 3

您正在检查isset($_POST['submit'])，但没有使用提交名称发布的输入字段。您需要在submit 按钮中添加名称属性。你也没有通过mysqli_query中的$connection。

    $servername = "localhost";
    $username = "root";
    $password = "";
    $db="image";


    // Create connection

    $connection = mysqli_connect($servername, $username, $password, $db); // Establishing Connection with Server
    if (!$connection) {
          die("Connection failed: " . mysqli_connect_error());
        }
    else{
          echo "Connected successfully"; 
        }


    if(isset($_POST['submit'])){ // Fetching variables of the form which travels in URL
    $name = $_POST['name'];
    $image = $_POST['image'];

    echo $name;
    echo $image;

    if($name !=''||$image !=''){
    //Insert Query of SQL
    $query = mysqli_query($connection, "INSERT INTO image (id, name, imagename) VALUES ('NULL', '$name', '$image')");
    if($query !== false){
        echo "Data Inserted successfully...!!";
    }
    else{
        echo "Query failed";
    }
    }
    else{
    echo "Insertion Failed <br/> Some Fields are Blank....!!";
    }
    }
    mysqli_close($connection); // Closing Connection with Server
    ?>


    <form action = "test2.php" method="POST" enctype="multipart/form-data">
        <label>name: </label><input type="text" name="name" />
        <label>File: </label><input type="text" name="image" />
        <input type="submit" name = "submit" />
    </form>
    </body>
    </html>

还有一个建议总是在代码中使用 PDO 来防止 SQL 注入。您的代码容易受到 sql 注入的攻击。