PHP引用有关问题

WBOY
Release: 2016-06-13 13:23:34
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PHP引用问题
我在A文件中引用B文件的函数cut(),在A页面中写include_once("B.php");然后在下面调用但不知道为什么老是运行报错Fatal error: Call to undefined function cut() in D:\XAMPP1.7.3\xampp\htdocs\web\index.php on line 233
这个是不是说没有定义CUT函数?但我明明定义了啊。然后查看A页面源文件的时候,本该显示结果的那段显示的却是我的CUT函数代码。
CUT函数是
function cut($Str, $Length) {//$Str为截取字符串,$Length为需要截取的长度 

global $s;
$i = 0;
$l = 0;
$ll = strlen($Str);
$s = $Str;
$f = true; 

while ($i if (ord($Str{$i}) $l++; $i++;
} else if (ord($Str{$i}) $l++; $i += 2;
} else if (ord($Str{$i}) $l += 2; $i += 3;
} else if (ord($Str{$i}) $l += 1; $i += 4;
} else if (ord($Str{$i}) $l += 1; $i += 5;
} else if (ord($Str{$i}) $l += 1; $i += 6;


if (($l >= $Length - 1) && $f) {
$s = substr($Str, 0, $i);
$f = false;


if (($l > $Length) && ($i $s = $s . '...'; break; //如果进行了截取,字符串末尾加省略符号“...”
}
}
return $s;
}
?>

不知道我表达的清楚不,希望有人能帮忙解答。

------解决方案--------------------
你开启了php断标签了吗,你A文件的php标签应该是------解决方案--------------------

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你开启了php断标签了吗,你A文件的php标签应该是
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