PHP Put return value from DB to input box

Question

good Day!, i have a question. I have the code below which return the value of pos_name and train_name from my DB. my question is how can i put the returned value inside the textbox in which each textbox has a unique name. cause i will going to save it to another db.

i want something like this

 <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" . ">";


                ?>

Please Help Thank you!

Is it the $row["pos_name"] at the top of your example or the $pos_name at the bottom that you are referring to? I mean in terms of which <input> you want modified? — Rasclatt
– Rasclatt, Commented Jan 9, 2015 at 6:20

mike510a · Accepted Answer · 2015-01-09 06:20:28Z

1

Wasn't very clear as to what the OP was asking but -- I suppose this could be part of the answer:

<?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>';

            ?>

answered Jan 9, 2015 at 6:20

mike510a

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1 Comment

Nixxx Over a year ago

Wow thank you! yes this is what i needed $id = html_entity_decode($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>"; im new you help me alot! thank you!

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