如果PHP中的语句结构 [英] If statement structure in PHP

问题描述

我在以下代码中遇到错误。这可能是一些小事，但我看不出有什么问题。

I keep getting an error with the following bit of code. It is probably some small thing but I don't see what is wrong.

while($row = mysql_fetch_array($result))
  {
   $varp = $row['ustk_retail'];
   if ($varp<80000) { $o1 = 1; }
   if (($varp=>80000) && ($varp<100000)) { $o2 = "1"; }
   if (($varp=>100000) && ($varp<120000)) { $o3 = "1"; }
   if (($varp=>120000) && ($varp<140000)) { $o4 = "1"; }
   if (($varp=>140000) && ($varp<160000)) { $o5 = "1"; }
   if (($varp=>160000) && ($varp<180000)) { $o6 = "1"; }
   if (($varp=>180000) && ($varp<200000)) { $o7 = "1"; }
   if (($varp=>200000) && ($varp<220000)) { $o8 = "1"; }
   if (($varp=>220000) && ($varp<240000)) { $o9 = "1"; }
   if (($varp=>240000) && ($varp<260000)) { $o10 = "1"; }
   if (($varp=>260000) && ($varp<280000)) { $o11 = "1"; }
   if (($varp=>280000) && ($varp<300000)) { $o12 = "1"; }
   if ($varp>=300000) { $o13 = "1"; }
  }

推荐答案

大于或等于 > = 签收，而非 =>

Greater than or equal to is >= sign, not =>

更新：

你是对的。这很小但很难找到错误。

我把整行分成几块来看问题所在：

Update:
You are right. It's small but hard to find mistake.
It took me to split whole line into pieces to see where the problem is:

<?php
if 
(
$varp
=>
80000
)

所以，它说第5行的解析错误，我不得不重新检查这个运算符。

当然，起初我将问题行与其余代码分开以确定。

So, it says parse error on line 5 and I had to doublecheck this operator.
Of course, at first I separated the problem line from the rest of the code to be certain.

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