Question

我们有一个VOIP服务器，用于存储用于呼叫录音的wav文件。我的目的是整理一个PHP文件，我可以在文件的URL中传递calldate和uniqueid值（因为服务器将这些作为文件名的一部分存储）以检索文件的链接。

但是，尝试这样做时会出现以下错误：

“解析错误：语法错误，第42行的/var/www/html/maint/helloworld.php意外$结束”

<?php 

function getRecordingLink($callDate, $uniqueId){

$callDate_arr = explode(' ', $callDate);

$removeChar = array('-',':');
foreach($callDate_arr as $value){
    $callDate_arr_adj[] = str_replace($removeChar,'',$value);
}

$audioDirectory_array = array();
if ($handle = opendir('/var/spool/asterisk/monitor/')) {
    while (false != ($file = readdir($handle))) { 
        if ($file != "." && $file != "..") { 
            $audioDirectory_array[] = $file; 
        } 
    }
    closedir($handle); 
}

foreach($audioDirectory_array as $key => $value){

    if (preg_match ("/".$uniqueId."/i", $value)) {
        if(file_exists('/var/spool/asterisk/monitor/'.$value)){
            $wavFile = $value;
            //return '<a href="/maint/modules/cdrreport/monitor/'.$wavFile.'" target="_blank">Recorded File</a>';
            return '/maint/cache/monitor/' . $wavFile;
        }
    }else{
        //return "Not Recorded";
    }

getRecordingLink ('20110513','1305274000.2'); //sample calldate and uniqueid values for testing

＆GT？;

以下是MySQL表的描述，我将从中获取值以传递给此calidate和uniqueid`的URL

    mysql> describe cdr
    -> ;
+-------------+--------------+------+-----+---------------------+-------+
| Field       | Type         | Null | Key | Default             | Extra |
+-------------+--------------+------+-----+---------------------+-------+
| calldate    | datetime     | NO   |     | 0000-00-00 00:00:00 |       |
| clid        | varchar(80)  | NO   |     |                     |       |
| src         | varchar(80)  | NO   |     |                     |       |
| dst         | varchar(80)  | NO   |     |                     |       |
| dcontext    | varchar(80)  | NO   |     |                     |       |
| channel     | varchar(80)  | NO   |     |                     |       |
| dstchannel  | varchar(80)  | NO   |     |                     |       |
| lastapp     | varchar(80)  | NO   |     |                     |       |
| lastdata    | varchar(80)  | NO   |     |                     |       |
| duration    | int(11)      | NO   |     | 0                   |       |
| billsec     | int(11)      | NO   |     | 0                   |       |
| disposition | varchar(45)  | NO   |     |                     |       |
| amaflags    | int(11)      | NO   |     | 0                   |       |
| accountcode | varchar(20)  | NO   |     |                     |       |
| uniqueid    | varchar(32)  | NO   |     |                     |       |
| userfield   | varchar(255) | NO   |     |                     |       |
+-------------+--------------+------+-----+---------------------+-------+

Answer 1

是的，还有另外一个很难看的:-p

您在getRecordingLink()上方文件底部缺少两个大括号：

function getRecordingLink($callDate, $uniqueId){

    $callDate_arr = explode(' ', $callDate);

    $removeChar = array('-',':');
    foreach($callDate_arr as $value){
        $callDate_arr_adj[] = str_replace($removeChar,'',$value);
    }

    $audioDirectory_array = array();
    if ($handle = opendir('/var/spool/asterisk/monitor/')) {
        while (false != ($file = readdir($handle))) { 
            if ($file != "." && $file != "..") { 
                $audioDirectory_array[] = $file; 
            } 
        }
        closedir($handle); 
    }

    foreach($audioDirectory_array as $key => $value){
        if (preg_match ("/".$uniqueId."/i", $value)) {
            if(file_exists('/var/spool/asterisk/monitor/'.$value)){
                $wavFile = $value;
                //return '<a href="/maint/modules/cdrreport/monitor/'.$wavFile.'" target="_blank">Recorded File</a>';
                return '/maint/cache/monitor/' . $wavFile;
            }
        }else{
            //return "Not Recorded";
        }
    }
}
getRecordingLink ('20110513','1305274000.2');

我从here得到的错误是：

解析错误：语法错误，意外结束/homepages/26/d94605010/htdocs/lz/writecodeonline.com/php/index.php(98）：第32行的eval（）代码

该文件现在为我解析，但我无法测试它实际上做了什么。如果这不起作用，请在评论中遵循我的建议。

获取PHP中的文件链接

1 个答案: