get value in textbox from database according to seleceted dropdown item in php -

so i've change text box data according option selected in dropdown. want text box data fetched database.

for eg: if choose airmail current frank value of airmail should displayed in text box database.

here code

<div>     <form action="alertfrank.php" method="post" class="boxin" id="dropdown">         <select name="type_post" required id="type_post" >             <option value="select post type" selected>select post type</option>             <option value="airmail" >airmail</option>             <option value="airmail ad">airmail ad</option>             <option value="airmail speed post">airmail speed post</option>             <option value="speed post">speed post</option>             <option value="book post">book post</option>             <option value="ordinary post">ordinary post</option>             <option value="registered post">registered post</option>             <option value="registered ad">registered ad</option>             <option value="registered parcel">registered parcel</option>             <option value="parcel">parcel</option>             <option value="courier">courier</option>         </select>         <br/>         <br/>         current frank value <?php echo '<input type="number" name="value_frank" value="">'; ?> <br/>         <br/>         new frank value         <input type="number" name="new_frank" placeholder="new frank value"/>         <br/>         <br/>         <input type="submit" name="submit" value="change value" id="sub"/>         <br/>         <br/>         <br/>         <br/>     </form> </div> <?php if (null !== (filter_input(input_post, 'submit'))) {     mysql_connect("localhost", "root", "") or die(mysql_error());     mysql_select_db("disp") or die(mysql_error());     $type_post = (filter_input(input_post, 'type_post'));     $_session['type_post'] = $type_post;     $q = mysql_query("select value_frank frank type_post='" . $_session['type_post'] . "'") or die(mysql_error());     $value_frank = mysql_fetch_row($q);     $_session['value_frank'] = $value_frank; } ?> 

i want current frank value displayed in value_frank feild..and before submit.

add onchange() select this,

<select name="type_post" required id="type_post" onchange="get_data(this.value)" >      <option value="select post type" selected>select post type</option>      <option value="airmail" >airmail</option>      <option value="airmail ad">airmail ad</option>      <option value="airmail speed post">airmail speed post</option>      <option value="speed post">speed post</option>      <option value="book post">book post</option>      <option value="ordinary post">ordinary post</option>      <option value="registered post">registered post</option>      <option value="registered ad">registered ad</option>      <option value="registered parcel">registered parcel</option>      <option value="parcel">parcel</option>      <option value="courier">courier</option>  </select> 

then in script write ajax function fetch data server.

this script.

function get_data(value){     $.ajax({         url: "nameof ajax file .php",         type: "post",         datatype: "html",         async: false,         data: {value: value}         success: function(data) {            //here can populate required fields based on value database                         }      });  } 

in ajax file value db based on value selected.and display in text boxes.

you have move database part file,

mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db("disp") or die(mysql_error()); $type_post = (filter_input(input_post, 'type_post'));  $q = mysql_query("select value_frank frank type_post='" . $type_post . "'") or die(mysql_error()); $value_frank = mysql_fetch_row($q); $val_fra = $value_frank;  echo $val_fra;// here no need use sessions @ all.