按钮单击时，数据未从MySql加载到文本框中

i have a little problem here. What i am trying to achieve is User types in unique id into textbox then presses load button, script is looking for this parameter in MySql database. When found match data should be loaded into textbox below.

UPDATED VERSION

<?php

mysql_connect ("localhost", "root","84946dff6e1")  or die (mysql_error());
mysql_select_db ("employees");


if(isset($_POST["loadbtn"]))
{
    $load = $_POST["loadbtn"];

    $sql = mysql_query("SELECT * FROM titles WHERE emp_no = '$load' ");
    $details = mysql_fetch_array($sql);

    $savedTitle = $details["title"];     
}
?>
<form method="post" action="changeTitleView.php">
<table width="400" border="0" cellspacing="1" cellpadding="2" align="center">
<tr>
<td width="150">Employee number</td>
<td><input type="text" name="load" /></td>
<td><input type="submit" name="loadbtn" value="Load" /></td>
</tr>
</table>
</form>
<br />
<br />
<form method="get" action="changeTitleView.php">
<table width="400" border="0" cellspacing="1" cellpadding="2" align="center">
<td width="150">Employee Title</td>
<td><input name="salary" type="text" value="<?php echo $savedTitle; ?>"></td>
</tr>
<td width="150"> </td>
<td>
<input name="add" type="submit" id="add" value="Update">
</td>
</tr>
</table>
</form>

thank you for looking and help :)

Try this

<?php
error_reporting(-1);// show all errors when debugging

// don't use these database commands
//mysql_connect ("localhost", "root","84946dff6e1")  or die (mysql_error());
//mysql_select_db ("employees");

// do it this way and don't show us your database credentials.
$conn = new mysqli('localhost', "root", "84946dff6e1",  'employees');

//declare your variables so if POST isn't true you don't have errors later

$load = "some Id";
$savedTitle = "no value yet!"; // use something interesting when testing

//you want the value of the textbox which name is load
if(isset($_POST["load"]))
{
    //never trust the user directly
     //$load = $_POST["load"]; 
     // do this
     $load = $conn->real_escape_string($_POST["load"]);

    $result = $conn->query("SELECT * FROM titles WHERE emp_no = '$load' ");

    $details = $result->fetch_assoc();

    $savedTitle = $details["title"];     
    // show me error when testing to see if something is wrong with query
     echo $conn->error;
}
?>
<form method="post" action="changeTitleView.php">
<table width="400" border="0" cellspacing="1" cellpadding="2" align="center">
<tr>
<td width="150">Employee number</td>
<td><input type="text" name="load" value="<?php echo $load; ?>" /></td>
<td><input type="submit" name="loadbtn" value="Load" /></td>
</tr>
</table>
</form>
<br />
<br />
<form method="get" action="changeTitleView.php">
<table width="400" border="0" cellspacing="1" cellpadding="2" align="center">
<td width="150">Employee Title</td>
<td><input name="salary" type="text" value="<?php echo $savedTitle; ?>"></td>
</tr>
<td width="150"> </td>
<td>
<input name="add" type="submit" id="add" value="Update">
</td>
</tr>
</table>
</form>

You will want to add another if branch to handle the second form which updates the title And remember that form is GET instead of POST like the first form.

<td><input name="salary" type="text" value="<? echo $empTitle; ?>"></td>

try change that line to this:

<td><input name="salary" type="text" value="<?php echo $empTitle; ?>"></td>