只有变量应该通过引用传递

// Other variables
$MAX_FILENAME_LENGTH = 260;
$file_name = $_FILES[$upload_name]['name'];
//echo "testing-".$file_name."<br>";
//$file_name = strtolower($file_name);
$file_extension = end(explode('.', $file_name)); //ERROR ON THIS LINE
$uploadErrors = array(
    0=>'There is no error, the file uploaded with success',
    1=>'The uploaded file exceeds the upload max filesize allowed.',
    2=>'The uploaded file exceeds the MAX_FILE_SIZE directive that was specified in the HTML form',
    3=>'The uploaded file was only partially uploaded',
    4=>'No file was uploaded',
    6=>'Missing a temporary folder'
);

什么好主意吗?过了2天，还是卡住了。

当前回答

结束(…[爆炸”。'， $file_name)])从PHP 5.6开始工作。这在RFC中有记录，但PHP文档本身没有记录。

2019-04-16 00:52:31

其他回答

Php 7兼容的正确用法:

$fileName      = 'long.file.name.jpg';
$tmp           = explode('.', $fileName);
$fileExtension = end($tmp);

echo $fileExtension;
// jpg

2014-01-09 09:35:06

其他所有人都已经给出了出错的原因，但这里是完成您想要做的事情的最佳方式: $file_extension = pathinfo($file_name, PATHINFO_EXTENSION);

2011-01-08 21:21:01

首先，您必须像这样将值存储在变量中

$value = explode("/", $string);

然后你可以使用end函数从这样的数组中获取最后一个索引

echo end($value);

我希望这对你有用。

2019-05-03 09:53:53

就像你不能立即索引数组一样，你也不能对它调用end。首先将它赋值给一个变量，然后调用end。

$basenameAndExtension = explode('.', $file_name);
$ext = end($basenameAndExtension);

2011-01-08 21:17:55

由于它在超过10年的时间里升起了一个旗帜，但工作得很好，并返回预期的值，一个小stfu操作符是你都在寻找的最好的坏习惯:

$file_extension = @end(explode('.', $file_name));

但是警告，不要在循环中使用，因为会影响性能。最新版本的php 7.3+提供了array_key_last()和array_key_first()方法。

https://www.php.net/manual/en/function.array-key-last.php

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2019-11-03 19:04:56

只有变量应该通过引用传递

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