PHP 将DB中的return值放入输入框

Question

美好的一天！，我有一个问题。我有下面的代码，return 来自我的数据库的 pos_name 和 train_name 的值。我的问题是如何将 returned 值放入每个文本框都有唯一名称的文本框内。因为我要把它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

对于 OP 的要求不是很清楚，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>';

            ?>

PHP 将DB中的return值放入输入框

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