PHP基礎有關問題

PHP基礎問題

PHP code

";    $sql = "show databases";  //尋找某個庫是否存在    print("$sql
");    $sql_return = mysql_query($sql);    echo "$sql_return 
";    while($row = mysql_fetch_array($sql_return))    {        for($i = 0; $i != count($row); ++$i)        {            printf("%s", $row[0]);            echo "
";        }    }    //mysql_select_db('web', $conn);    ?>

輸出的結果是這樣的：

PHP code

資料庫連接成功show databasesResource id #4 information_schemainformation_schemamysqlmysqlperformance_schemaperformance_schematesttestwebweb

????為什麼，每個庫名都輸出了兩遍，我也用print_r($row);執行過，結果如下：

PHP code

資料庫連接成功show databasesResource id #4 Array ( [0] => information_schema [Database] => information_schema ) Array ( [0] => mysql [Database] => mysql ) Array ( [0] => performance_schema [Database] => performance_schema ) Array ( [0] => test [Database] => test ) Array ( [0] => web [Database] => web )

------解決方案--------------------
mysql_fetch_array()返回的既有數字數組，又有關聯陣列。你可以換用 mysql_fetch_row()
------解決方案--------------------
http://cn2.php.net/manual/zh/function.mysql-fetch-array.php

你自己看看你這程式寫的:

PHP code

        for($i = 0; $i != count($row); ++$i)        {            printf("%s", $row[0]);            echo "
";        }
