Question

下面给出了用于动态添加html表的

php代码，但它无法在数据库中插入所有值

    <?php
    session_start();
    require_once("include/config.php");
    $num=$_POST['n'];
    $i=0;
    if( isset($_POST['submit']) )
    {
    $num=$_POST['n'];
    echo "ji".$num;
    for($i=0;$i<=$num; $i++)
    {
       $j=addslashes($_POST['t1'][$i]);
       $m=addslashes($_POST['t2'][$i]);
       $sql= "INSERT INTO ba (a,b)values('$j','$m');"; 
       $res =mysqli_query($con,$sql) ;
       $i++;
       if($res)
       {
        echo "<script>alert('Family details submitted successfully');document.location.href='b.php';</script>"; 
        }
        else
        {
        echo "<script>alert('Error. Please Retry');document.location.href='b.php';</script>";
        }
       }
    }

?>

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

    <html xmlns="http://www.w3.org/1999/xhtml">

    <head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <title>Untitled Document</title>

    </head>

    <body>
    <form name="sdfs" method="post">
    <table border="3">
    <?php

    while($i<$num)
    {

        echo "<tr><td><input type='text' id='t1[$i]' name='t1[$i]'></td><td>
        <input type='text' id='t2[$i]' name='t2[$i]'></td>
        ";
            $i++;
    }
    ?>
    <input type="submit" name="submit" id="submit" value="submit" />
    </tr>

    </form>
    </body>
    </html>

num的值是从另一页获得的.... 替换值插入db ...但我想将所有值插入到通过表单获取的数据库中请找到解决这个问题的方法........................

如何动态添加html表和在php中插入数据库的值

0 个答案: