Question

嘿伙计们，我正在为我的大学做一个小项目。它由一个动态的表（链接到mysql表）组成，每个记录中的一个单元是动态下拉（所有都链接到mysql表）。并且用户必须从生成的每个下拉列表中选择下拉值列表，并且它也可以留空。这是PHP的代码：

    <form class="appnitro"  method="post" action="">
    <div class="form_description">
    <center><h2>NOMINATE ENTRY</h2></center>
    <p><center><font size='3'>
    <?php
    $con=mysqli_connect("localhost","user",pass","db");
    if (mysqli_connect_errno())
    {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
    $dept = $_POST['department'];
    $class = $_POST['class'];
    $result = mysqli_query($con,"SELECT * FROM prizemaster");
    $result1 = mysqli_query($con,"SELECT * FROM studentmaster WHERE dept='$dept' and class='$class'");
    echo "<table border='1'>
    <tr>
    <th>Prize ID &nbsp &nbsp &nbsp &nbsp</th>
    <th>Prize Name &nbsp &nbsp &nbsp &nbsp </th>
    <th>Name &nbsp &nbsp &nbsp &nbsp &nbsp &nbsp </th>
    </tr>";
    while($row = mysqli_fetch_array($result)){
    echo "<tr>";
    echo "<td>" . $row['prizeid'] . "</td>";
    echo "<td>" . $row['name'] . "</td>";
    echo "<td><select name='name'>";
    echo "<option></option>";
    while($drop = mysqli_fetch_array($result1))
    {
    echo "<option value='".$drop['name']."'>" . $drop['name'] . "</option>"; 
    }
    mysqli_data_seek($result1, 0);
    echo "</select></td>";
    echo "</tr>";
    }
    echo "</table>";
    mysqli_close($con);
    ?></center></font></div>
    <p>
    <center><button type="submit" formaction="stnomins.php">Nominate</button></center>
    </form>

以上是表单，当点击提名按钮时，我希望为所选的每个下拉值执行此代码：

<?php
$con=mysqli_connect("localhost","user","pass","db");
$myname = $_POST['name'];
$sql2="SELECT * FROM studentmaster WHERE name='$myname'";
$result = mysqli_query($con,$sql2)or die(mysqli_error());
$row = mysqli_fetch_array($result);
$mydept=$row['dept'];
$myclass=$row['class'];
$myregno=$row['regno'];
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
$sql1="INSERT INTO studenttransaction (`transid`, `date`, `prizeid`, `regno`, `name`, `class`, `department`, `status`) VALUES ('' , CURRENT_TIMESTAMP() , '', '$myregno', '$myname', '$myclass', '$mydept', '1')";
if (!mysqli_query($con,$sql1))
  {
  die('Error: ' . mysqli_error($con));
  }
mysqli_close($con);
?>

请帮助我，我需要在2天内提交。请用代码解释，因为我是php的新手并且不完全知道它

Answer 1

我更新了你的html＆amp; php代码，请看一下：

在表单文件中：

echo "<td><select name='name[]'>"; // added [] here to make it array
echo "<option value=''>Select student</option>"; // added value & option item

上面这些代码行已在相应的

文件中进行了相应的修改

    <form class="appnitro"  method="post" action="">
    <div class="form_description">
    <center><h2>NOMINATE ENTRY</h2></center>
    <p><center><font size='3'>
    <?php
    $con=mysqli_connect("localhost","user",pass","db");
    if (mysqli_connect_errno())
    {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
    $dept = $_POST['department'];
    $class = $_POST['class'];
    $result = mysqli_query($con,"SELECT * FROM prizemaster");
    $result1 = mysqli_query($con,"SELECT * FROM studentmaster WHERE dept='$dept' and class='$class'");
    echo "<table border='1'>
    <tr>
    <th>Prize ID &nbsp &nbsp &nbsp &nbsp</th>
    <th>Prize Name &nbsp &nbsp &nbsp &nbsp </th>
    <th>Name &nbsp &nbsp &nbsp &nbsp &nbsp &nbsp </th>
    </tr>";
    while($row = mysqli_fetch_array($result)){
    echo "<tr>";
    echo "<td>" . $row['prizeid'] . "</td>";
    echo "<td>" . $row['name'] . "</td>";
    echo "<td><select name='name[]'>";
    echo "<option value=''>Select student</option>";
    while($drop = mysqli_fetch_array($result1))
    {
    echo "<option value='".$drop['name']."'>" . $drop['name'] . "</option>"; 
    }
    mysqli_data_seek($result1, 0);
    echo "</select></td>";
    echo "</tr>";
    }
    echo "</table>";
    mysqli_close($con);
    ?></center></font></div>
    <p>
    <center><button type="submit" name="nominate" formaction="stnomins.php">Nominate</button></center>
    </form>

略微更新您的PHP代码。如果点击/提交按钮，则为每个下拉名称启动一个循环。

<?php
$con=mysqli_connect("localhost","user","pass","db");
if (mysqli_connect_errno()){
 echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

if(isset($_POST['nominate'])){
 foreach($_POST['name'] as $key => $myname){
  //$myname = $_POST['name'];
  $sql2="SELECT * FROM studentmaster WHERE name='$myname'";
  $result = mysqli_query($con,$sql2)or die(mysqli_error());
  if($row = mysqli_fetch_array($result)){
   $mydept=$row['dept'];
   $myclass=$row['class'];
   $myregno=$row['regno'];
   $sql1="INSERT INTO studenttransaction (`transid`, `date`, `prizeid`, `regno`, `name`, `class`, `department`, `status`) VALUES ('' , CURRENT_TIMESTAMP() , '', '$myregno', '$myname', '$myclass', '$mydept', '1')";
   if (!mysqli_query($con,$sql1)){
    die('Error: ' . mysqli_error($con));
   }
  }//if mysqli_fetch_array condition closed
 }// for loop closed     
}// if submit button(nominate) closed
mysqli_close($con);
?>

希望这对你有用。

Answer 2

工作正常。我还有一个疑问是，当选择下拉值时，如何回显下拉值。任何知道答案的人，请提供解决方案。

按钮上的多个动态下拉提交单击到mysql表

2 个答案: