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cronジョブに入れるphpスクリプトを作成しました。深夜に実行する必要がありますが、これは問題ではありません。問題は、cronが私のコードを実行すると、データベースからすべてのレコードを削除しますが、CSVファイルから新しいレコードを挿入しないことです。ヒントはありますか？

PHPコードは以下のとおりです。

<?php 

# I run this script from a cron job every night to update 
# the mysql database I use with my employee web site 
# so it matches my local database every day.  Feel free to 
# modify it to meet your specific needs.  If you find it 
# usefull, drop me an email and let me know. 

# edit the follow six items to use the script 

# first connect to your mysql database 
# i have my connection settings in a diferent file 
# so i just include that file in all my scripts 
include("db.php"); 

# assign the tables that you want to import to to the table array 
$table = array( 
                'table1', 
                'table2', 
                'table3', 
                'table4', 
                'table5', 
        ); 

# if the first row of your csv file contains column headings: 
# $columnheadings=1 
# if the first row does not contain column headings and should be imported: 
# $columnheadings=0 
$columnheadings = 0; 

# contains the email address you want the results sent to 
$emailaddress = "user@domain.com"; 

# contains the subject you want the message to have 
$subject = "Enter Subject Here"; 

# contains the email address that will show in the from line 
$emailfrom = "user@domain.com"; 

# perform the required operations for every table listed in the table array 
foreach ($table as $tablename) { 

    # empty the table of its current records 
    $deleterecords = "TRUNCATE TABLE `$tablename`"; 
    mysql_query($deleterecords); 

    # intialize your counters for successful and failed record imports 
    $pass = 0; 
    $fail = 0; 

    # the csv file needs to be the same name as the table, 
    # comma seperated with the columns in the same order as the table, 
    # and in the same dir as this script 
    $filecontents = file ("$tablename.csv"); # .csv is added to the table name to get the name of the csv file 

    # every record in the csv file will be inserted into the table unless an error occurs with that record 
    for($i=$columnheadings; $i<sizeof($filecontents); $i++) { 
        $insertrecord = "Insert Into `$tablename` Values ($filecontents[$i])"; 
        mysql_query($insertrecord); 
        if(mysql_error()) { 
            $fail += 1;     # increments if there was an error importing the record 
        } 
        else 
        { 
            $pass += 1;        # increments if the record was successfully imported 
        } 
    } 

    # adds a line to the email message we will send stating how many records were imported 
    # and how many records failed for each table 
    $message .= "Table $tablename: Success=$pass  Failure=$fail \n"; 
} 

# set to the date and time the script was run 
$runtime = (date("d M Y H:i")); 

# add the run time to the body of the email message 
$message .= "\nTime of the message: $runtime (server time zone)\n\n"; 

# Send the email message 
mail($emailaddress, $subject, $message, "From: '$emailfrom'"); 

?>
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2 に答える 2

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あなたのコードはファイルの存在をチェックしません-それで、あなたがそれを探していると思っている場所にない場合はどうなりますか？これをcronから実行する方法によっては、Webから実行した場合とは異なるディレクトリから実行される場合があります。

于 2012-06-21T12:07:42.400 に答える
0

テーブルのスキーマは何$messageですか、そして最後に何を取得していますか？すべての行を失敗として取得していますか？

追加した：

すべての行が失敗しているため、挿入ステートメントに問題があるか、csvとターゲットテーブルのスキーマが一致していない可能性があります。これを試して

    # and in the same dir as this script 
    $filecontents = file ("tablename.csv"); # .csv is added to the table name to get the name of the csv file 

    # every record in the csv file will be inserted into the table unless an error occurs with that record 
    for($i=$columnheadings; $i<sizeof($filecontents); $i++) { 
        $insertrecord = "Insert Into `$tablename` Values ($filecontents[$i])"; 
        mysql_query($insertrecord); 
        if(mysql_error()) { 
            $fail += 1;     # increments if there was an error importing the record
             //this is include exact error in email, and will help you to identify the
             //problem very easily.
             $message .= mysql_errno() . ": " . mysql_error() . "\n"; 
        } 
        else 
        { 
            $pass += 1;        # increments if the record was successfully imported 
        } 
    }

これを試して、あなたの発見で私に知らせてください？

于 2012-06-21T12:07:37.103 に答える