Question

我无法理解为什么这不会更新从文本输入中收到的数据：

include "config.php"
printf ("Update PriceList\n");
printf ("<form method=\"post\" action=\"price_red.php\">");
printf ("Select PhoneID: <input type = \"Text\" name = \"PhonID\"><br>");
printf ("PhoneName:<input type = \"Text\" name = \"PhoneName\"><br>");
printf ("PhoneType:<input type = \"Text\" name = \"PhoneType\"><br>");
printf ("ScreenType:<input type = \"Text\" name = \"ScreenType\"><br>");
printf ("Camera: <input type = \"Text\" name = \"Camera\"><br>");
printf ("Quantity: <input type = \"Text\" name = \"Quantity\"><br>");
printf ("Price:<input type = \"Text\" name = \"Price\"><br>");
printf ("Out of stock:<input type = \"Text\" name = \"outofstock\"><br>");
printf ("<input type=\"Submit\" name=\"submit\" value=\"Update\">");
printf ("</form>");

$sql = mysql_query("UPDATE PhonesPriceList SET PhoneName = '$PhoneName', Price = '$Price',     Quantity = '$Quantity', Outofstock = '$outofstock' WHERE PhoneID = '$PhonID';");
$sql1 = mysql_query("UPDATE PhonesDetails SET PhoneType = '$PhoneType', ScreenType = '$ScreenType', Camera ='$Camera' WHERE PhoneID = '$PhonID';");

Answer 1

表单的数据不会自动转换为变量。数据应位于$_POST[ 'name' ]，其中“name”是输入字段的name属性。

请记住在使用它进行任何数据库操作之前清理传入的数据（查找mysql_real_escape_string()）。

还有一个小提示：您可以回退到纯HTML，而不是使用printf()输出大块文本。代码看起来会更清晰。

include "config.php"
// go back to pure HTML
?>
Update PriceList
<form method="post" action="price_red.php">
...
<?php  // back to PHP

Answer 2

尝试这样的事情。只需确保添加适当的卫生设施以防止注射

   include "config.php"

    if(isset('submit'){
 $Phoneid= $_POST['PhoneID'];
  $Phone_name= $_POST['PhoneName'];
  $Phone_type = $_POST['PhoneType'];
  $screen_type= $_POST['ScreenType'];
  $camera = $_POST['Camera'];
   $Quantity = $_POST['Quantity'];
  $price = $_POST['Price'];
 $outofstock = $_POST['outofstock'];

$dbc = mysqli_connect( data info)
 or die ('Error connecting');
   $query = "UPDATE PhonesPriceList SET PhoneName = '$PhoneName', Price = '$price',     
      Quantity =       '$Quantity', Outofstock = '$outofstock' WHERE PhoneID = '$PhonID' ";

    msqli_query( $dbc, $query) or die (' failed to query');

   echo ' Update successful';

     mysqli_close($dbc);
  }

文本类型的输入不会在mysql表中更新

2 个答案: