PHP Multiple Update function not updating and gets

2019-01-29 08:41发布

站内问答 / PHP
1545 1 4

First of all, I use PHP Dreamweaver to make a program with phpMyAdmin Database. This program gets my records in a 1st row blank when I submit the button, and the remaining records in a rows are still there which is 2nd row and below, but the records in a 1st row is still on the database, why should it happen? any idea for me to solve this program, it also won't update my record. this makes me sick, i need help please! :( tnx...

this is the code:

<?php require_once('Connections/tlsc_conn.php');
  mysql_select_db($database_tlsc_conn, $tlsc_conn);
  $query_Recordset1 = "SELECT * FROM tbl_name";
  $Recordset1 = mysql_query($query_Recordset1, $tlsc_conn) or die(mysql_error());
  $row_Recordset1 = mysql_fetch_assoc($Recordset1);
  $totalRows_Recordset1 = mysql_num_rows($Recordset1);

  if(isset($_POST['submit'])) {
      $count = count($_POST['id']);
      $submit = $_GET['id'];

      for($i=0;$i<$count;$i++){
          $sql1="UPDATE $tbl_name SET name='$name[$i]', lastname='$lastname[$i]', email='$email[$i]' WHERE id='$id[$i]'";
          $row_Recordset1=mysql_query($sql1);
       }

    if($row_Recordset1){
            header("location:lulu.php");
            exit;
    }   
 }
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org    /TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
 <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
 <title>Untitled Document</title>
</head>
<body>
  <form name="form2" method="post" action="">
  <table width="634" border="1">
    <tr>
       <td>id</td>
       <td>name</td>
       <td>lastname</td>
       <td>email</td>
    </tr>
   <?php do { ?> 
    <tr>
      <td><?php $id[]=$row_Recordset1['id']; ?><?php echo $row_Recordset1['id']; ?> 
      <input name="id[]" type="hidden" value="<?php echo $row_Recordset1['id'];   ?>" />
      </td>
      <td>
        <input name="name[]" type="text" value="<?php echo $row_Recordset1['name']; ?>">                       
      </td>
      <td>
        <input name="lastname[]" type="text" value="<?php echo $row_Recordset1['lastname']; ?>">
      </td>
      <td>
        <input name="email[]" type="text" value="<?php echo  $row_Recordset1['email']; ?>">       </td>
    </tr>
    <?php } while ($row_Recordset1 = mysql_fetch_assoc($Recordset1)); ?>  
   </table>
    <p>
    <input type="submit" name="submit" value="Submit" />
    </p>
  </form>
   <p>
   </p>
</body>
</html>

1条回答
劳资没心，怎么记你
2楼-- · 2019-01-29 08:55

You need to put:

$name = array_map('mysql_real_escape_string', $_POST['name']);
$lastname = array_map('mysql_real_escape_string', $_POST['lastname']);
$email = array_map('mysql_real_escape_string', $_POST['email']);
$id = array_map('mysql_real_escape_string', $_POST['id']);

before your for loop. You're using these variables but you never filled them in from the form inputs.

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