具有一種形式的兩個不同的單選按鈕，並使用php發布

Question

我堅持使用此代碼，嘗試發布2個不同的單選組我的問題是我想讓用戶檢查1個單選按鈕以獲取4個結果，我的意思是我如何才能知道用戶何時選擇了其他單選可以選擇多個收音機，有什么幫助嗎？

                $sql = "SELECT * FROM tbl1 WHERE object = '".$obj."' LIMIT 5 ";
                $result2 = mysqli_query($conn,$sql);

              echo "<form action='' method='post'>";
                  while ($row2 = mysqli_fetch_array($result2,MYSQLI_ASSOC)) {

                    echo "<div align='center'>
                    <table class='demo'  dir='rtl'>
                      <tbody>
                      <tr>
                        <td>".$row2['ojs']."</td>
                        <td> </td>
                      </tr>
                      <tr>
                        <td>".$row2['txt1']."</td>
                        <td><input type='radio' name='chk1' value='1' >
                      </tr>
                      <tr>
                        <td>".$row2['txt2']."</td>
                        <td><input type='radio' name='chk2' value='2' >
                      </tr>
                      <tr>
                        <td>".$row2['txt3']."</td>
                        <td><input type='radio' name='chk3' value='3' >
                      </tr>
                      <tr>
                        <td>".$row2['txt4']."</td>
                        <td><input type='radio' name='chk4' value='4' >
                      </tr>
                      </tbody>
                    </table><br>
                    </div>
                    ";
                  }
                  echo "<input type='submit' name='submit'>
                  </form>";
                  if (isset($_POST['submit'])) {
                    $chk1 = $_POST['chk1'];
                    $chk2 = $_POST['chk2'];
                    $chk3 = $_POST['chk3'];
                    $chk4 = $_POST['chk4'];
                    echo $chk4."|".$chk3."|".$chk2."|".$chk1;

//insert into mysql , just the Selected radio button for each different question
$insQry = "insert into tbl2 (id,sel1,sel2,sel3,sel4) VALUES('$chk1','$chk2','$chk3','$chk4')";
if (mysqli_query($conn, $sql)) {
echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
                  }

Answer 1

您需要將整個adio輸入組的名稱設置為一個名稱。 由於您將它們設置為四個不同的名稱，因此將它們視為四個不同的實體。

Answer 2

您需要給它們起相同的名稱，然后在PHP中獲得它，例如

 $sql = "SELECT * FROM tbl1 WHERE object = '".$obj."' LIMIT 5 ";
            $result2 = mysqli_query($conn,$sql);

          echo "<form action='' method='post'>";
              while ($row2 = mysqli_fetch_array($result2,MYSQLI_ASSOC)) {

                echo "<div align='center'>
                <table class='demo'  dir='rtl'>
                  <tbody>
                  <tr>
                    <td>".$row2['ojs']."</td>
                    <td> </td>
                  </tr>
                  <tr>
                    <td>".$row2['txt1']."</td>
                    <td><input type='radio' name='chk' value='1' >
                  </tr>
                  <tr>
                    <td>".$row2['txt2']."</td>
                    <td><input type='radio' name='chk' value='2' >
                  </tr>
                  <tr>
                    <td>".$row2['txt3']."</td>
                    <td><input type='radio' name='chk' value='3' >
                  </tr>
                  <tr>
                    <td>".$row2['txt4']."</td>
                    <td><input type='radio' name='chk' value='4' >
                  </tr>
                  </tbody>
                </table><br>
                </div>
                ";
              }
              echo "<input type='submit' name='submit'>
              </form>";
              if (isset($_POST['submit'])) {
                $chk = $_POST['chk'];
                echo $chk;

然后只需將數據庫中的相鄰一個設置為選定的數字，其他設置為0