從php的結果下拉列表中獲取選定的值在html html

我已經從sql結果php中填充下拉列表，現在我試圖在同一頁面中將選定的值變爲php變量，但它不是加工。你能幫我嗎。以下是代碼。從php的結果下拉列表中獲取選定的值在html html

<?php 
    $mid="mario"; 
    $sql = "SELECT * FROM tbl_prdy" ; 
    $result = mysqli_query($conn,$sql); 
    echo "<select name='list'>"; 
    while ($row = mysqli_fetch_array($result,MYSQLI_ASSOC)) { 
    echo "<option value='" . $row['col_of_fac'] . "'>" . $row['col_of_fac'] . " 
    </option>"; 
    } 
    echo "</select>"; 
    $varsel = $_POST['list']; 
    echo "hai"; 
    echo $varsel; 

?>

$varsel = $_POST['list'];不工作。

來源

2017-07-21 djac

請添加您的表格，以便我們可以幫助您更好 –

請在插入查詢中發佈您的選擇名稱.. – user3663

你應該

1-應用的形式

2-發送與方法「POST」變量相同的文件PHP

<?php 
    // display the errors 
    error_reporting(E_ALL); 
    ini_set("display_errors", 1); 
?> 

<?php 
    // when the form submitted 
    if(!empty($_POST['list'])){ 
     echo $_POST['list']; 
    } 
?> 

<?php 
    // connection test 
    $conn=mysqli_connect("localhost","user","pass","db"); //replace the (user, pass, db) with your parameters 
    if (mysqli_connect_errno()) 
    { 
     echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
     die(); 
    } 

    //if your connection succeded 
    $sql = "SELECT * FROM tbl_prdy" ; 
    $result = mysqli_query($conn,$sql); 
?> 

<form method="post" action="<?= $_SERVER['PHP_SELF']; ?>"> 
    <select name="list"> 
    <?php while ($row = mysqli_fetch_array($result,MYSQLI_ASSOC)): ?> 
     <option value="<?= $row['col_of_fac']; ?>"><?= $row['col_of_fac']; ?></option> 
    <?php endwhile; ?> 
    </select> 
    <input type="submit" value="valider"> 
</form>

，如果你想打印選定的價值和不是沒有形式

，那麼你必須使用jquery

<?php 
     // display the errors 
     error_reporting(E_ALL); 
     ini_set("display_errors", 1); 
    ?> 

    <?php 
     // when the form submitted 
     if(!empty($_POST['list'])){ 
     echo $_POST['list']; 
     } 
    ?> 

    <?php 
     // connection test 
     $conn=mysqli_connect("localhost","user","pass","db"); //replace the (user, pass, db) with your parameters 
     if (mysqli_connect_errno()) 
     { 
     echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
     die(); 
     } 

     //if your connection succeded 
     $sql = "SELECT * FROM tbl_prdy" ; 
     $result = mysqli_query($conn,$sql); 
    ?> 

    <select name="list"> 
    <?php while ($row = mysqli_fetch_array($result,MYSQLI_ASSOC)): ?> 
     <option value="<?= $row['col_of_fac']; ?>"><?= $row['col_of_fac']; ?></option> 
    <?php endwhile; ?> 
    </select> 

    <p id="value-selected"></p> 

    <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script> 
    <script> 
     $(function(){ 
      $("select[name=list]").on("change", function() { 
       var valueSelected = $(this).val(); 
       $("#value-selected").html(valueSelected); 
      }); 
     }); 
    </script>

來源

2017-07-21 06:58:06 bfahmi

嘗試了代碼。但它不起作用。 – djac

@djac;你有什麼結果？ – bfahmi

「海」和$ varsel的價值沒有得到顯示。我認爲，即使條件本身不起作用。 FYI-下拉列表已填充。 – djac

$ varsel = $ _POST ['list'];不管用。：當表單POST方法提交它只會工作，

（與當前代碼選擇框打印，並且您無需提交形式獲取後）的邏輯應該是這樣形式選擇框是由使用下面的代碼和用戶選擇並點擊提交：

<?php 
    $mid="mario"; 
    $sql = "SELECT * FROM tbl_prdy" ; 
    $result = mysqli_query($conn,$sql); 
    echo "<select name='list'>"; 
    while ($row = mysqli_fetch_array($result,MYSQLI_ASSOC)) { 
    echo "<option value='" . $row['col_of_fac'] . "'>" . $row['col_of_fac'] . " 
    </option>"; 
    } 
    echo "</select>"; 
?>

後提交下面的代碼形式應執行並在網絡控制檯，在頁頭，你可以，如果需要看數據傳輸到請求的頁面

來源

2017-07-21 06:59:27 LogicBlower

從php的結果下拉列表中獲取選定的值在html html

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