PHP基础有关问题

PHP基础有关问题

PHP基础问题

PHP code

";
    $sql = "show databases";  //查找某个库是否存在
    print("$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);
    
?>

输出的结果是这样的：

PHP code

数据库连接成功
show databases
Resource id #4 
information_schema
information_schema
mysql
mysql
performance_schema
performance_schema
test
test
web
web

????为什么，每个库名都输出了两遍，我也用print_r($row);执行过，结果如下：

PHP code

数据库连接成功
show databases
Resource 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 )

------解决方案--------------------
mysql_fetch_array()返回的既有数字数组，又有关联数组。你可以换用 mysql_fetch_row()
------解决方案--------------------
http://cn2.php.net/manual/zh/function.mysql-fetch-array.php

你自己看看你这程序写的:

PHP code

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

/phprm/55531.htmlwww.phpzy.comtrue/phprm/55531.htmlTechArticlePHP基础有关问题 PHP基础问题 PHP code "; $sql = "show databases"; //查找某个库是否存在 print("$sql "); $sql_return = mysql_query($sql); echo "$sql_return "; while($row = mysql_fetch_array($sql_return)) { for($i = 0; $i !=...