Php 当我提交数据时，它不会显示在表中，并显示错误。刷新后正确显示。我使用ajax来实现这一点吗？

我想从数据库中赋值

HTML

---

如果点击submit，无论是否附加了文件，都可以关闭第66行的数据库连接

然后尝试从关闭的连接获取新数据，这是不可能的

您必须将
mysql\u关闭（$conn）到脚本中的其他位置

<?php

$dbhost = 'localhost:3306';
$dbuser = 'root';
$dbpass = 'root';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
mysql_select_db("test_db", $conn);



if(isset($_POST['submit']))
{
if(!empty($_FILES))
  {
$allowedExts = array("gif", "jpeg", "jpg", "png","txt");
$temp = explode(".", $_FILES["resume"]["name"]);
$extension = end($temp);

if (($_FILES["resume"]["size"] < 50000)
&& in_array($extension, $allowedExts))
  {
  if ($_FILES["resume"]["error"] > 0)
    {
    echo "Error: " . $_FILES["resume"]["error"] . "<br>";
    }
  else
    {
    echo "Upload: " . $_FILES["resume"]["name"] . "<br>";
    echo "Type: " . $_FILES["resume"]["type"] . "<br>";
    echo "Size: " . ($_FILES["resume"]["size"] / 1024) . " kB<br>";
    echo "Stored in: " . $_FILES["resume"]["tmp_name"];
    }
  }
else
  {
  echo "Invalid file\n";
  }

$name = $_POST['name'];
$age = $_POST['age'];
$quali = $_POST['quali'];
$state = $_POST['state'];
$country = $_POST['country'];
$msg = $_POST['msg'];
$resume = $_FILES["resume"]["name"];



$sql = "INSERT INTO form ".
       "(name,age,quali,state,country,msg,resume) ".
       "VALUES('$name','$age', '$quali', '$state','$country','$msg','$resume')";

mysql_select_db('test_db');

$retval = mysql_query( $sql, $conn );
if(! $retval )
{
  die('Could not enter data: ' . mysql_error());
}

echo "Entered data successfully\n";

move_uploaded_file( $_FILES["resume"]["tmp_name"], "files/".$_FILES["resume"]["name"]);
}

mysql_close($conn);

}
?>
<div>
    <form action="<?php $_PHP_SELF ?>" method="POST" enctype="multipart/form-data">
        <div>Name:<input type="text" name="name" /></div>
        <div style="height:10px"></div>
        <div>Age:<input type="text" name="age" /></div>
        <div style="height:10px"></div>
        <div>Qualification:<input type="text" name="quali" /></div>
        <div style="height:10px"></div>
        <div>State:
        <select name="state">
          <option value="tn">TamilNadu</option>
          <option value="kl">Kerala</option>
          <option value="ka">Karnataka</option>
          <option value="ani">Andhara</option>
        </select>
</div>
        <div style="height:10px"></div>
        <div>Country:<input type="text" name="country" /></div>
        <div style="height:10px"></div>
        <div>Resume:<input type="file" name="resume" /></div>
        <div style="height:10px"></div>
        <div>Message:<textarea cols="10" rows="5" name="msg"></textarea></div>
        <div style="height:10px"></div>
        <div><input type="Submit" name="submit" value="Submit" /></div>
    </form>
</div>
<div style="height:30px"></div>

<div>
<table class="table">
    <tr>
        <th>Name</th>
        <th>Age</th>
        <th>Qualification</th>
        <th>State</th>
        <th>Country</th>
        <th>Resume</th>
        <th>Message</th>
    </tr>
    <?php
$sql1 = "SELECT * FROM form";
$retval1 = mysql_query($sql1,$conn);
while($row = mysql_fetch_array($retval1))
{
$name_s = $row['name'];
$age_s = $row['age'];
$quali_s = $row['quali'];
$state_s = $row['state'];
$country_s = $row['country'];
$resume_s = $row['resume'];
$msg_s = $row['msg'];
?>
<tr>
    <td><label name="name_s"><?php echo $name_s; ?></label></td>
    <td><label name="age_s"><?php echo $age_s; ?></label></td>
    <td><label name="quali_s"><?php echo $quali_s; ?></label></td>
    <td><label name="state_s"><?php echo $state_s; ?></label></td>
    <td><label name="country_s"><?php echo $country_s; ?></label></td>
    <td><label name="resume_s"><?php echo $resume_s; ?></label></td>
    <td><label name="msg_s"><?php echo $msg_s; ?></label></td>
</tr>
<?php
}
?>
</table>
</div>