只能通过引用传递变量-php [英] Only variables can be passed by reference - php

问题描述

我正在尝试此代码，但出现此错误:

i am trying this code, but i get this error:

Only variables can be passed by reference in xxx

脚本

class page {
  function insert($db, $of, $form, &$arr) {

      $i = 0;

      foreach(array_combine($form['value0'], $arr) as $val=>$v){

          $sql->prepare("mysqli query here");
          $sql->bind_param('ssss', $val, $of, $v[$i][0], $v[$i][1]);//error here
          $sql->execute();
          $i++;

      }
      return true;
  }
}

原因是什么，如何解决?谢谢

what is the reason, and how can be solved ? thanks

推荐答案

我假设您正在使用

I assume you're using mysqli::bind_param. All arguments except the first are passed by reference. This means they must be variables, and not strings, array elements, etc. I'm actually not sure why it needs to do this by reference, but never mind. You can fix it pretty easily:

$v0 = $v[$i][0];
$v1 = $v[$i][1];
$sql->bind_param('ssss', $val, $of, $v0, $v1);

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