Question

大家好，任何人都可以帮助我，我已经解决了我的问题，但我无法将值从div传递到输入字段，需要请 page1.php中

<!doctype html>
<?php
$conn = mysql_connect('localhost','root','') or die (mysql_error);
$db   = mysql_select_db('employee', $conn) or die (mysql_error);
$select = "SELECT * FROM employee ";
$result = mysql_query($select);
$option = '';
while($row = mysql_fetch_assoc($result)){ 
$option .= '<option       value="'.$row['id'].'">'.$row['employee_name'].'</option>';
}
?>
<html>
<head>
    <title>Retrieve data from database using Ajax</title>
    <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
    <script type="text/javascript">
        function getData(q, a){
            $.ajax({
                url: 'loademployeedata.php?empid='+q, //call storeemdata.php to store form data
                success: function(html) {
                    var ajaxDisplay = document.getElementById(a);
                    ajaxDisplay.innerHTML = html;
                }
            });
        }
         $(document).ready(function(){
    $("#tyt").live("change", function() {
        $("#ab  ").val($(this).find("option:selected").attr("value"));

    });
});
    </script>
</head>
<body>
    <form method="post">
        <select id="tyt" onchange="getData(this.value, 'dis')" >
          <?php
            echo $option;
          ?>
        </select>
        <div id="dis" >
        </div><input name="" type="text"  id="ab" >
    </form>
</body>

第二个连接页面是

<?php
$empid = $_GET['empid'];
$connection = mysql_connect("localhost", "root", ""); // Establishing    Connection with Server..
$db = mysql_select_db("employee", $connection); // Selecting Database
if (isset($empid)) {
$query = " SELECT * FROM employee WHERE id = '$empid'";
$result = mysql_query($query);

while($row = mysql_fetch_assoc($result)){
    $a=$row['employee_salary'];?>
    <?php echo $a; ?>
 <?php }

  }
  mysql_close($connection); // Connection Closed
  ?>

我只需要将值而不是div均值放入输入字段

Answer 1

尝试这样的事情（未经测试）：

{{1}}

如何将下拉结果放在php中的textbox字段中

1 个答案: