Question

这是一个非常常见的问题，但是尝试了一些没有用的东西。我试图根据从下拉列表中选择的值来填充数据库中的文本框值。我的index.php

<?php
 # Connect 
 mysql_connect('localhost', 'user', 'pass') or die('Could not connect: ' . mysql_error());

# Choose a database
mysql_select_db('db_name') or die('Could not select database');

# Perform database query
$query = "SELECT * from customer_reg";
$result = mysql_query($query) or die('Query failed: ' . mysql_error());
?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" 
  "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
   <html xmlns="http://www.w3.org/1999/xhtml">
 <head>
    <meta charset="utf-8">
  <title>whatever</title>
 </head>

<body>
  <label for="ddcust_nm">Customer Name: </label>
  <select id="customer_name">
   <option value="0">Please select</option>
   <?php
    while ($row = mysql_fetch_assoc($result)) {
        echo '<option value="' . $row['id'] . '">' . $row['customer_name'] . '</option>';
    }
  ?>
</select>

<div>
  <label for="address">Street Address</label>
  <input id="address" name="address" class="element text large" type="text">
</div>

<div>
  <span class="floatLeft">
    <label for="phoneno">phone No:</label>
    <input id="phoneno" name="phoneno" class="element text medium" style="width:14em" type="text">
  </span>


</div>

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script>
  function insertResults(json){
    $("#address").val(json["StreetAddress"]);
    $("#phone").val(json["City"]);

  }



  function makeAjaxRequest(placeId){
    $.ajax({
      type: "POST",
      data: { placeId: placeId },
      dataType: "json", 
      url: "search.php",
      success: function(json) {
        insertResults(json);
      }
    });
  }

  $("#customer_name").on("change", function(){
    var id = $(this).val();
    if (id === "0"){
      clearForm();
    } else {
      makeAjaxRequest(id);
    }
  });
  </script>
   </body>
  </html>

Search.php

<?php
  mysql_connect('localhost', 'user', 'pass') or die('Could not connect: ' . 
mysql_error());

mysql_select_db('db_name') or die('Could not select database');

 $placeId = $_POST['placeId'];

  $query = "SELECT * from customer_reg";
  $result = mysql_query($query) or die('Query failed: ' . mysql_error());
  while ($row = mysql_fetch_assoc($result)) {
    if ($placeId == $row['id']){
     echo json_encode($row);
    }
  }
?>

我可以从下拉列表中获取价值，但是，在下拉列表中选择任何值之后，什么也不会发生。我在每个功能中都尝试过警报，但它什么也没显示。我尝试单击（我知道我不应该这样做），但没有触发我从HERE

获得了这段代码

请帮助

如何从数据库填充文本框中的下拉所选值

0 个答案: