“ php不在xampp中插入mysql服務器”

Question

“我已經閱讀了很多類似於我的問題在stackoverflow中解決的問題，並且看到了很多示例，但是我的代碼仍未插入mysql。但是，如果我硬喂它會插入php。我的信息是來自html帖子的提交。我有良好的服務器連接以及與數據庫的連接，如果我錯過任何事情，任何人都可以幫助我。這是下面的代碼。”

<?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的輸出為2.5”

Answer 1

輸入提交按鈕的名稱

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

然后在PHP文件

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

}

這個if語句將運行

Answer 2

您沒有為button指定name屬性，因此請提供name="submit" ，如果要上傳文件，請更改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

您正在檢查isset($_POST['submit'])但是沒有輸入字段帶有提交名稱。.您需要在submit按鈕中添加名稱屬性。 你也沒有在mysqli_query傳遞$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>

還有一個建議總是在代碼中使用PDO來防止SQL注入。 您的代碼容易受到sql注入的攻擊。