PHP将DB的返回值放入输入框

Question

美好的一天！，我有一个问题。 我下面的代码从我的数据库返回pos_name和train_name的值。 我的问题是如何将返回的值放在每个文本框都有唯一名称的文本框中。 因为我将要保存到另一个数据库。

我想要这样的东西

 <input type="text" name="<?php echo $row["pos_name"][0] ?>" value="<?php echo $row["train_name"][0] ?>"
 <input type="text" name="<?php echo $row["pos_name"][1] ?>" value="<?php echo $row["train_name"][1] ?>"
// The number of textbox will depend on the number of returned value or the ($i)



//MY PHP CODE =========
<?php 

    $con = mysql_connect("localhost","root","");
        if (!$con){
        die("Can not connect: " . mysql_error());
        }
        mysql_select_db("tms",$con);

 $Query="SELECT id,pos_name,train_name FROM pos_train_db ORDER BY id DESC LIMIT 15";
 $sql = mysql_query($Query, $con);

            $dyn_table = '<table border="1" cellpadding="10">';
           while( $row = mysql_fetch_array($sql)){

                $id = $row["id"];
                $pos_name = $row["pos_name"];
                $train_name = $row["train_name"];

                    if ($i % 3 == 0) { // if $i is divisible by our target number (in this case "3")

            $dyn_table .= '<tr><td>' . $pos_name . '</td>';
        } else {
            $dyn_table .= '<td>' . $pos_name . '</td>';
        }
        $i++;
    }
    $dyn_table .= '</tr></table>';

            ?>    





       <?php 
//SOME TESTING CODE
               echo $dyn_table;

               echo $i;

               echo "<input type='text' value=" . "$pos_name" . ">";


                ?>

请帮忙谢谢！

Answer 1

关于OP的要求还不是很清楚，但是-我想这可能是答案的一部分：

<?php 

    $con = mysql_connect("localhost","root","");
        if (!$con){
        die("Can not connect: " . mysql_error());
        }
        mysql_select_db("tms",$con);

 $Query="SELECT id,pos_name,train_name FROM pos_train_db ORDER BY id DESC LIMIT 15";
 $sql = mysql_query($Query, $con);

            $dyn_table = '<table border="1" cellpadding="10">';
           while( $row = mysql_fetch_array($sql)){

                $id = html_entity_decode($row["id"]);
                $pos_name = html_entity_decode($row["pos_name"]);
                $train_name = html_entity_decode($row["train_name"]);
echo "<input type=\"text\" name=\"".$id."\" value=\"".$train_name."\"></input>";

                    if ($i % 3 == 0) { // if $i is divisible by our target number (in this case "3")

            $dyn_table .= '<tr><td>' . $pos_name . '</td>';
        } else {
            $dyn_table .= '<td>' . $pos_name . '</td>';
        }
        $i++;
    }
    $dyn_table .= '</tr></table>';

            ?>