插入用戶數據從PHP到用戶輸入形式

Question

我想弄清楚如何從用戶輸入從使用PHP的HTML表單插入信息到我的數據庫，但我不是100％確定如何。

這是我目前的。

HTML

<form id="playerForm" action="insert.php" method="post"> 
     Player Name: <input type="text" value="playerName"> 
     Player Race: <input type="text" value="playerRace"> 
     Player Class: <input type="text" value="playerClass"> 
     <button id="submit">Submit</button> 
    </form> 

PHP

<?php 
    $servername = "localhost"; 
    $username = "username"; 
    $password = "password"; 
    $dbname = "database"; 

    //create connection 
    $conn = mysqli_connect($servername, $username, $password, $dbname); 

    //check connection 
    if(!$conn){ 
    die("Unable to connect to MySQL: " . mysqli_connect_error()); 
    } 

    echo "connected successfully"."<br><br>"; 

    //selects the different data types from the professions table 
    $sql = "INSERT INTO Player (PlayerName, PlayerRace, PlayerClass) 
    VALUES()"; 

    $result = $conn->query($sql);     

    $conn->close(); 

Answer 1

需要分配形式name attribute，如果你不使用的名字，那麼你無法獲得數據，如果你想設置的輸入字段爲空，因此在這案例不設定值

<form id="playerForm" action="insert.php" method="post"> 
     Player Name: <input type="text" name="playerName" value="playerName"> 
     Player Race: <input type="text" name="playerRace" value="playerRace"> 
     Player Class: <input type="text" name="playerClass" value="playerClass"> 
     <input type="submit" value="submit"> 
    </form> 

如果您要設置輸入字段爲空，那麼在這種情況下不要設置值

<form id="playerForm" action="insert.php" method="post"> 
     Player Name: <input type="text" name="playerName" value=""> 
     Player Race: <input type="text" name="playerRace" value=""> 
     Player Class: <input type="text" name="playerClass" value=""> 
     <input type="submit" value="submit"> 
    </form> 

，這是你insert.php文件，重要的是你的表結構必須確定，使用自動增量也和主鍵，幫助您

<?php 

    $link = mysqli_connect("localhost", "jgbydtoz_root", "123warcraft", "jgbydtoz_warcraft"); 


    if($link === false){ 

     die("ERROR: Could not connect. " . mysqli_connect_error()); 

    } 


    $playerName = mysqli_real_escape_string($link, $_POST['playerName']); 

    $playerRace = mysqli_real_escape_string($link, $_POST['playerRace']); 

    $playerClass = mysqli_real_escape_string($link, $_POST['playerClass']); 

    $sql = "INSERT INTO Player (`PlayerName`, `PlayerRace`, `PlayerClass`) VALUES ('$playerName', '$playerRace', '$playerClass')"; 

    if(mysqli_query($link, $sql)){ 

     echo "added successfully."; 

    } else{ 

     echo "ERROR: . " . mysqli_error($link); 

    } 



    mysqli_close($link); 

    ?> 

Answer 2

第一件事情你需要這樣做是html form tutorial

你的表單輸入沒有名稱值可言，因此，你將無法在PHP中提取任何價值和存儲

這是怎麼了你的表單應該是：

<form id="playerForm" action="insert.php" method="post"> 
     Player Name: <input type="text" name="playerName"> 
     Player Race: <input type="text" name="playerRace"> 
     Player Class: <input type="text" name="playerClass"> 
     <button id="submit" type="submit">Submit</button> 
</form> 

然後在insert.php檢查，如果輸入設置，並不proccesing前空。那麼如果所有設置你的好去然後用mysqli prepared statements將數據插入/ pdo prepared statements

<?php 

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

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

$errors = ""; 
$fields = array("playerName","playerRace","playerClass"); 

foreach ($fields as $fieldName) { 

    if (!isset($_POST[$fieldName]) && !empty($_POST[$fieldName])) { 

     $errors++; 
     die("Fill in all the fields"); 
    } 
} 

if ($errors <= 0) { 

    $playername = $_POST['playerName']; 
    $playerrace = $_POST['playerRace']; 
    $playerclass = $_POST['playerClass']; 

    $sql = $conn->prepare("INSERT INTO Player (PlayerName,PlayerRace,PlayerClass) VALUES(?,?,?"); 
    $sql->bind_param("sss", $playername, $playerrace, $playerclass); 

    if ($sql->execute()) { 

     echo "data inserted"; 
    } else { 

     echo "Error : " . $conn->error; // in live site send this to error log not browser 
    } 
} 


?> 

Answer 3

試試下面code.and必須在「球員」表有一個AUTO_INCREMENT場。

<?php 
    $servername = "localhost"; 
    $username = "username"; 
    $password = "password"; 
    $dbname = "database"; 

    //create connection 
    $conn = mysqli_connect($servername, $username, $password, $dbname); 

    //check connection 
    if(!$conn){ 
     die("Unable to connect to MySQL: " . mysqli_connect_error()); 
    } 

    echo "connected successfully"."<br><br>"; 
    if(isset($_REQUEST["submit"])){ 
     $p_name = $_POST['playerName']; 
     $p_race = $_POST['playerRace']; 
     $p_class = $_POST['playerClass']; 
     //selects the different data types from the professions table 
     $sql = mysqli_query($conn,"INSERT INTO Player (PlayerName, PlayerRace, PlayerClass) 
      VALUES('$p_name','$p_race','$p_class')"); 


    }    

    $conn->close(); 
    ?> 


<form id="playerForm" method="post"> 
      Player Name: <input type="text" name="playerName"> 
      Player Race: <input type="text" name="playerRace"> 
      Player Class: <input type="text" name="playerClass"> 
      <input type="submit" name="submit" value="SUBMIT"/> 
     </form>