更新php mysql

Question

我试图显示和更新一次的所有记录表。我可以成功显示记录。但是我无法更新我的数据库，请帮助我更新表格中的记录  

<?php while($rows=mysqli_fetch_array($result)){ ?>

<tr>
  <td align="center">
    <input name="id[]" type="text" id="name" value="<?php echo $rows['id']; ?>">
  </td>
  <td align="center">
    <input name="name[]" type="text" id="name" value="<?php echo $rows['name']; ?>">
  </td>
  <td align="center">
    <input name="lastname[]" type="text" id="lastname" value="<?php echo $rows['lastname']; ?>">
  </td>
  <td align="center">
    <input name="email[]" type="text" id="email" value="<?php echo $rows['email']; ?>">
  </td>
</tr>

<?php } ?>

<tr>
  <td colspan="4" align="center">
    <input type="submit" name="Submit" value="Submit">
  </td>
</tr>
</table>
</td>
</tr>
</form>
</table>
<?php if(isset($_POST[ 'Submit'])) { $id_array=$_POST[ 'id']; $name_array=$ _POST[ 'name']; $year_array=$_POST[ 'lastname']; $emil_array=$_POST[ 'email']; for($i=0;$i<$count;$i++){ $sql1="UPDATE test_mysql SET name='$name_array[$i]', lastname='$year_array[$i]', email='$emil_array[$i]' WHERE id='id_array[$i]'"
; } mysqli_close($con); } ?>

Answer 1

您的表单没有操作处理程序。

<form name="form1" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">

Answer 2

尝试：

HTML：

<table>
<form name="form1" method="post"> <!-- add form tag -->

<?php while($rows=mysqli_fetch_array($result)){ ?>

<tr>
  <td align="center">
    <input name="id[]" type="text" id="name" value="<?php echo $rows['id']; ?>">
  </td>
  <td align="center">
    <input name="name[]" type="text" id="name" value="<?php echo $rows['name']; ?>">
  </td>
  <td align="center">
    <input name="lastname[]" type="text" id="lastname" value="<?php echo $rows['lastname']; ?>">
  </td>
  <td align="center">
    <input name="email[]" type="text" id="email" value="<?php echo $rows['email']; ?>">
  </td>
</tr>

<?php } ?>

<tr>
  <td colspan="4" align="center">
    <input type="submit" name="Submit" value="Submit">
  </td>
</tr>
</table>

</form>
</table>

PHP：

if(isset($_POST['Submit'])) { 
  $id_array=$_POST['id']; 
  $name_array=$ _POST['name']; 
  $year_array=$_POST['lastname']; 
  $emil_array=$_POST['email']; 
  for($i=0;$i<$count;$i++){ 
    $id = $name_array[$i];
    $name = $year_array[$i]; 
    $year = $emil_array[$i];
    $email = $id_array[$i]; // use $id_array[$i] to access id, as in above code you did not use $ in id_array
    $sql="UPDATE test_mysql SET name='$name', lastname='$year', email='$email' WHERE id='$id'"; 
    if ($conn->query($sql) === TRUE) { // execute query like this
      echo "Record updated successfully";
    } else {
        echo "Error updating record: " . $conn->error;
    }
  } 
  $conn->close();
}

Answer 3

在您的表单中，您没有提及任何操作处理程序。代码必须是

<form name="form1" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">

Answer 4

希望这会对你有所帮助，我没有把mysql_escape_string放在你仍在启动但它有助于逃避mysql攻击。还包装 <form>

上的表单

if(isset($_POST['Submit'])) { 
    $id_array= $_POST['id']; 
    $name_array=$ _POST['name']; 
    $year_array=$_POST['lastname']; 
    $emil_array=$_POST['email']; 


  if($id_array){

     foreach($id_array as $key=>$id_val){
        $name = $name_array[$key]; 
        $year = $year_array[$key];
        $email = $emil_array[$key]; 
        $sql = "update activity_logs set name='{$name}' , lastname = '{$year}' , email='{$email}'  where activity_logs_id = {$id_val}"; 
        $res = mysql_query($sql);

     }

  }

}