Question

我正在尝试使用“编辑”页面来获取用户选择的数据并将其显示在表单中。我设法显示数据库中的数据，并允许用户编辑表单中的数据。但我的UPDATE查询将无法在PHP中工作。我尝试回显查询并在xampp中运行它手册，结果证明它可以在xampp中更新手册但不能在php中更新。任何人都可以帮我解决这些问题吗？非常感谢

这是我的php编码

<?php
session_start();
include_once 'dbconnect.php';


 if(isset($_POST['btn-update']))
{
  $ProdCode = mysql_real_escape_string($_POST['productCode']);
 $ProdType = mysql_real_escape_string($_POST['productType']);
 $ProdDes = mysql_real_escape_string($_POST['product_description']);
 $ProdCol = mysql_real_escape_string($_POST['productColour']);
 $ProdPrice = floatval($_POST['productPrice']);
 $XSsize = mysql_real_escape_string($_POST['XSquantity']);
 $Ssize = mysql_real_escape_string($_POST['Squantity']);
 $Msize = mysql_real_escape_string($_POST['Mquantity']);
 $Lsize = mysql_real_escape_string($_POST['Lquantity']);
 $XLsize = mysql_real_escape_string($_POST['XLquantity']);
 $XXLsize = mysql_real_escape_string($_POST['XXLquantity']);
 if(isset($_FILES['productImg'])){

      $file_name = $_FILES['productImg']['name'];
      $file_size = $_FILES['productImg']['size'];
      $file_tmp = $_FILES['productImg']['tmp_name'];
      $file_type = $_FILES['productImg']['type'];
      $file_ext=strtolower(end(explode('.',$_FILES['productImg']['name'])));

      $expensions= array("jpeg","jpg","png");

      if(in_array($file_ext,$expensions)=== false){
         $errors="Please choose JPEG/PNG file.";
         $errorTrigger =true;
      }

      if($file_size > 2097152) {
         $errors='File size must be excately 2 MB';
         $errorTrigger =true;
      }

      if(empty($errors)==true) {
         move_uploaded_file($file_tmp,"images/".$file_name);

               } }
 $query = "UPDATE product SET product_code='$ProdCode', product_type='$ProdType' ,description='$ProdDes' ,colour='$ProdCol',price= '$ProdPrice',size_xs='$XSsize',size_s='$Ssize',size_m='$Msize',size_l='$Lsize',size_xl='$XLsize',size_xxl='$XXLsize' WHERE product_code='%". $ProdCode ."%'";
 echo $query;
if(mysql_query($query))
 {
  echo "<script>
    alert('Product Updated');
        </script>";
 }
 else
 {
     echo mysql_error();
  ?>
        <script>alert('Error while updating');</script>
        <?php
 }
 }

?>

UPDATE SQL查询不会在我的php脚本中更新数据库

0 个答案: