0

我有一个从员工列表中填充的更新表单。这些值正在传递但未在数据库中更新。这是我的代码以及我显示的已通过。

<?php

$con = mysql_connect("localhost","root","*******");
 if (!$con)
 {
  die('Could not connect: ' . mysql_error());
  }


$query = mysql_query("select * from backup");

 if(isset($_POST['update']))
 $id = $_POST['id'];
 $first = $_POST['first'];
$last = $_POST['last']; 
$store  = $_POST['store']; 
 $title  = $_POST['title']; 
 $title2  = $_POST['other']; 
 $phone  = $_POST['phone']; 
 $email  = $_POST['email'];
 $dept = $_POST['dept'];
 $bio   = $_POST['bio']; 

 $query="UPDATE backup SET first='$first', last='$last', store='$store',   title='$title', title2='$title2', phone='$phone', email='$email', bio='$bio' WHERE id='$id'";
 mysql_query($query);
 echo "Record Updated";
 mysql_close();
 print_r($_POST)




?>

这是结果

Record UpdatedArray ( [id] => 1396 [first] => Charles [last] => Adams [store] => [dept] => Accounting [title] => Accounting Clerk [other] => [phone] => 410 -555-1212[email] => [email2] => [bio] => Charlie 从 2009 年 8 月开始。这是一个测试.... [Submit] => Submit )

有人可以帮我解决我可能做错的事情吗？至于注射，我将在完成测试后修复它。我知道这可能听起来倒退，但我需要先找出为什么这不起作用。

感谢您对此的任何帮助

4

4 回答 4

1

您还应该选择数据库：

// make foo the current db
$db_selected = mysql_select_db('foo', $link);
if (!$db_selected) {
    die ('Can\'t use foo : ' . mysql_error());
}

但最重要的是，您应该使用 MySQLi 或 PDO_MySQL。

于 2013-10-01T14:28:39.280 回答
1

这是因为，你忘记了{之后if (isset(...))。

此外，没有选择数据库。

更正后的代码如下：

<?php

$con = mysql_connect("localhost","root","*******");
mysql_select_db('DB_NAME');
 if (!$con)
 {
  die('Could not connect: ' . mysql_error());
  }


$query = mysql_query("select * from backup");

 if(isset($_POST['update'])) {
 $id = $_POST['id'];
 $first = $_POST['first'];
$last = $_POST['last']; 
$store  = $_POST['store']; 
 $title  = $_POST['title']; 
 $title2  = $_POST['other']; 
 $phone  = $_POST['phone']; 
 $email  = $_POST['email'];
 $dept = $_POST['dept'];
 $bio   = $_POST['bio']; 

 $query="UPDATE backup SET first='$first', last='$last', store='$store',   title='$title', title2='$title2', phone='$phone', email='$email', bio='$bio' WHERE id='$id'";
 mysql_query($query);
 echo "Record Updated";
 mysql_close();
 print_r($_POST)
}

?>
于 2013-10-01T14:28:47.673 回答
0

您尚未在连接中定义数据库。

连接到数据库后执行此操作

于 2013-10-01T14:30:09.123 回答
0

您的代码可以更正为：

<?php
  $con = mysqli_connect("localhost","root","**","db");
  if (!$con)
  {
   die('Could not connect: ' . mysqli_error());
   }
  //$query = mysqli_query($con,"select * from backup");
  if(isset($_POST['update'])):
  $id = $_POST['id'];
  $first = $_POST['first'];
  //other stuffs
  $query="UPDATE backup SET first='$first', last='$last', store='$store',   title='$title', title2='$title2', phone='$phone', email='$email', bio='$bio' WHERE id='$id'";
  $res = mysqli_query($con,$query);
  if(!$res)
  die("could not update records. Error = ".mysqli_error($con)); 
  echo "Record Updated";
  mysqli_close($con);
  print_r($_POST);
  endif;
  ?>
于 2013-10-01T14:40:33.680 回答