MySQLi更新准备语句不更新数据库 [英] MySQLi update prepared statement not updating database

问题描述

所以我有这个更新语句，当我转储的$ _POST变量，我得到我想要的输出。

So I have this update statement which when I dump the $_POST variables., I get the outputs I want.

 $stmt = $dbConnectionW->prepare("UPDATE members SET 
                          fname='". mysqli_real_escape_string($dbConnectionW, $_POST['fname']) ."',
                          sname='". mysqli_real_escape_string($dbConnectionW, $_POST['sname']) ."',
                          gender='". mysqli_real_escape_string($dbConnectionW, $_POST['gender']) ."',
                          nationality='". mysqli_real_escape_string($dbConnectionW, $_POST['nation']) ."',
                          year='". mysqli_real_escape_string($dbConnectionW, $_POST['year']) ."',
                          dep1='". mysqli_real_escape_string($dbConnectionW, $_POST['dep1']) ."',
                          dep2='". mysqli_real_escape_string($dbConnectionW, $_POST['dep2']) ."',
                          f_pos='". mysqli_real_escape_string($dbConnectionW, $_POST['f_pos']) ."',
                          f_region='". mysqli_real_escape_string($dbConnectionW, $_POST['f_region']) ."',
                          exp_comp='".$comp."',
                          exp_dep='".$comp_dep."',
                          shareinfo='".$shareinfo."',
                          interest='".$interest."',
                          userconfirm = '1'
                              WHERE confirmcode = '".$passkey."';");
              $stmt->execute(); 
              if (!$stmt)
              {
              die('Error: ' . mysqli_error($dbConnectionW));
              }
              $smst-> close(); }}} mysqli_close($dbConnectionW);
      }

基本上，问题是它不会更新数据库！它工作没有错误，但数据库不会得到更新后这个sql / php尝试。

Basically the issue is that it won't update the database! It works with no errors, but the database does not get updated after this sql/php attempt.

任何人都看到我的代码有什么问题吗？什么是一些可能的原因，为什么我的数据库不会更新？

Can anyone see anything wrong with my code? What are some possible causes for why my would my database not be updated? I've been starting at this for the past hour.

推荐答案

您不需要在准备语句中转义变量，而是应该在执行语句之前绑定变量。

You don't need to escape your variables in a prepared statement, instead you should bind your variables before executing the statement. Also the column names should be inside ` marks.

$stmt = $dbConnectionW->prepare("UPDATE members SET 
                      `fname`=?,
                      `sname`=?,
                      `gender`=?,
                      `nationality`=?,
                      `year`=?,
                      `dep1`=?,
                      `dep2`=?,
                      `f_pos`=?,
                      `f_region`=?,
                      `exp_comp`=?,
                      `exp_dep`=?,
                      `shareinfo`=?,
                      `interest`=?,
                      `userconfirm`=?
                          WHERE `confirmcode`=?");
$stmt->bind_param('ssssissssssssis',$_POST['fname'],$_POST['sname'],$_POST['gender'],...);          
$stmt->execute();

为简洁起见，我没有包括所有的绑定参数。

I haven't included all the bound parameters for brevity.

希望这有助于。

这篇关于MySQLi更新准备语句不更新数据库的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！