Cron作业不要运行php代码

Question

我尝试运行cron job以对数据库进行备份。

cron job正在执行的问题，我收到了cron job正在执行的电子邮件，但问题是我没有备份文件。 但是，如果我在没有cron job情况下在浏览器上运行该页面，那么我会有一个备份文件。 我也有一个按钮，如果我单击此按钮，我会收到备份文件。 那意味着我的代码工作。

Cron作业文件：

<?php
include("includes/connect.php");
include("includes/functions.php");
include("includes/backup.php");
?>

备份功能，此功能在functions.php中

<?php
function Export_Database($host,$user,$pass,$name,$tables=false,$backup_name=false)
    {
        $mysqli = new mysqli($host,$user,$pass,$name); 
        $mysqli->select_db($name); 
        $mysqli->query("SET NAMES 'utf8'");

        $queryTables    = $mysqli->query('SHOW TABLES'); 
        while($row = $queryTables->fetch_row()) 
        { 
            $target_tables[] = $row[0]; 
        }   
        if($tables !== false) 
        { 
            $target_tables = array_intersect( $target_tables, $tables); 
        }
        foreach($target_tables as $table)
        {
            $result         =   $mysqli->query('SELECT * FROM '.$table);  
            $fields_amount  =   $result->field_count;  
            $rows_num=$mysqli->affected_rows;     
            $res            =   $mysqli->query('SHOW CREATE TABLE '.$table); 
            $TableMLine     =   $res->fetch_row();
            $content        = (!isset($content) ?  '' : $content) . "\n\n".$TableMLine[1].";\n\n";

            for ($i = 0, $st_counter = 0; $i < $fields_amount;   $i++, $st_counter=0) 
            {
                while($row = $result->fetch_row())  
                { //when started (and every after 100 command cycle):
                    if ($st_counter%100 == 0 || $st_counter == 0 )  
                    {
                            $content .= "\nINSERT INTO ".$table." VALUES";
                    }
                    $content .= "\n(";
                    for($j=0; $j<$fields_amount; $j++)  
                    { 
                        $row[$j] = str_replace("\n","\\n", addslashes($row[$j]) ); 
                        if (isset($row[$j]))
                        {
                            $content .= '"'.$row[$j].'"' ; 
                        }
                        else 
                        {   
                            $content .= '""';
                        }     
                        if ($j<($fields_amount-1))
                        {
                                $content.= ',';
                        }      
                    }
                    $content .=")";
                    //every after 100 command cycle [or at last line] ....p.s. but should be inserted 1 cycle eariler
                    if ( (($st_counter+1)%100==0 && $st_counter!=0) || $st_counter+1==$rows_num) 
                    {   
                        $content .= ";";
                    } 
                    else 
                    {
                        $content .= ",";
                    } 
                    $st_counter=$st_counter+1;
                }
            } $content .="\n\n\n";
        }

     $fp = fopen($_SERVER['DOCUMENT_ROOT']."/backup"."/mybackup-".date('d-m-Y')."-".date('H:i:s').".sql","wb");
fwrite($fp,$content);
fclose($fp);
 exit();
    }?>

备用页面：

<?php
    $mysqlUserName      = "***";
    $mysqlPassword      = "";
    $mysqlHostName      = "****";
    $DbName             = "****";
    $backup_name        = "mybackup.sql";

     $tables = array();
$showTable = "SHOW TABLES from $DbName";
$getData = mysqli_query($conn, $showTable);
while ($row = mysqli_fetch_row($getData)) {
   $tables[] = $row;
}
    Export_Database($mysqlHostName,$mysqlUserName,$mysqlPassword,$DbName,  $tables=false, $backup_name=false );
?>

当执行cron作业时，如何解决此问题以运行代码？

Answer 1

为什么不使用mysqldump ？

就像文档中所说的那样：
4.5.4 mysqldump —数据库备份程序

[cron-times] mysqldump -u <dbuser> --password <pw> [--host <host] --all-databases > /opt/mysql.bkp/`date`