警告：在 PHP 中除以零

Question

我收到错误，但我不明白为什么。我有一种感觉，它是处理 $bank 变量的东西，但我对 PHP 很陌生，我不太理解所有代码：

PHP:

<?php

  $thisgame = fetchinfo("value","info","name","current_game");
  $bank = fetchinfo("cost","games","id",$thisgame);

  $thisuserid=$rsx['uid'];

  $result = mysql_query("SELECT SUM(value) AS value FROM `game$thisgame` WHERE `userid`='$thisuserid'");
  $getrowval = mysql_fetch_assoc($result);
  $generatechance=round($getrowval["value"]*100/$bank,1);
  $howmanyitems=mysql_num_rows(mysql_query('SELECT * FROM `game'.$thisgame.'` WHERE   `userid`="'.$thisuserid.'"'));

  $profilele='http://steamcommunity.com/profiles/'.$thisuserid;
?>

那些是第 77 行到第 90 行......有人有想法吗？

Answer 1

将 $generatechance=round($getrowval["value"]*100/$bank,1); 替换为

并告诉我它是否有效

if( $bank > 0 ){
    $generatechance=round($getrowval["value"]*100/$bank,1);
}else{
    $generatechance = <YOU NEED TO DO THIS PART>
}

更好的主意是 $bank = fetchinfo("cost","games","id",$thisgame); 在这行之后检查您是否收到 $bank。

检查 $result = mysql_query("SELECT SUM(value) AS value FROMgame$thisgameWHEREuserid='$thisuserid'"); 查询是否正确 game$thisgame 。

警告：在 PHP 中除以零

Warning: Division by Zero in PHP

php

warnings

division

zero