文本框由下拉选定的值填充

这是我迄今为止尝试过的。我打算根据下拉列表中的选择来填充文本框。随着下拉列表的更改，我希望文本框的值也会发生更改。文本框由下拉选定的值填充

<form class="form-horizontal" role="form" action="edit_task.php" method="post">                   <div class="form-group"> 
                <div class="col-lg-6"> 

                <input class="form-control" type="hidden" id="id" name="id" /> 
                </div> 
                </div> 


                <div class="form-group"> 
                 <label class="col-lg-2 control-label">Plate Number</label> 
                 <div class="col-lg-6"> 
                 <?php $query = mysql_query("select * from vehicle"); // Run your query 

                 echo '<select name="plateno" id="plateno">'; // Open your drop down box 
                  // Loop through the query results, outputing the options one by one 
                  while ($row = mysql_fetch_array($query)) { 
                  echo '<option value="'.$row['plateno'].'">'.$row['plateno'].'</option>'; 
                   } 

                 echo '</select>'; ?> 
                 </div> 
                </div> 


                  <input type="text" class="form-control" id="cartype" name="cartype" placeholder=" " required> 


                  <input type="text" class="form-control" id="make" name="make" placeholder=" " required> 
                 </div> 
                </div> 


                  <input type="text" class="form-control" id="transmission" name="transmission" placeholder=" " required> 

                  <input type="text" class="form-control" id="caryear" name="caryear" placeholder=" " required> 

                  <input type="text" class="form-control" id="condition" name="condition" placeholder=" " required> 

                  <input type="text" class="form-control" id="task" name="task" placeholder=" " required> 

                  <input type="text" class="form-control" id="item_used" name="item_used" placeholder=" " required> 

                  <input type="text" class="form-control" id="remark" name="remark" placeholder=" " required> 

                  <button type="submit" class="btn btn-primary">Save</button> 
                  <button type="button" class="btn btn-danger">Cancel</button> 
                 </div> 
                </div> 
<script type="text/javascript"> 
     $(document).ready(function(){ 
    $('#plateno').change(function(){ 
    var plateno = $(this).val(); 
    var data_String; 
    data_String = 'plateno='+plateno; 
    $.post('ajax_file.php',data_String,function(data){ 
      var data= jQuery.parseJSON(data); 
      $('#cartype').val(data.cartype) 
      $('#make').val(data.make) 
      $('#transmission').val(data.transmission) 
      $('#caryear').val(data.caryear) 

     }); 
    }); 
}); 
</script> 
               </form>

这是连接到JSON文件 Ajax_file.php

<?php 
     $plateno =$_POST['plateno']; 

    echo json_encode($data); 
    exit();` 

?>

来源

2016-11-02 ojulari

这里是你上面的代码工作示例：

<!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 http-equiv="Content-Type" content="text/html; charset=utf-8" /> 
    <title>Document</title> 
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.4/jquery.min.js"></script> 
    <script type="text/javascript"> 
     $(document).ready(function() { 
      $('#plateno').change(function() { 
       var plateno = $(this).val(); 
       var data_String = 'plateno=' + plateno; 
       $.get('json.php', data_String, function(result) { 

        $.each(result, function(){ 
         $('#cartype').val(this.cartype); 
         $('#make').val(this.make); 
         $('#transmission').val(this.transmission); 
         $('#caryear').val(this.caryear); 
        }); 


       }); 
      }); 
     }); 
    </script> 
    </head> 
    <body> 
    <?php  
    $con=mysqli_connect("localhost","root","","my_db"); 
    // Check connection 
    if (mysqli_connect_errno()){ 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    } 
?> 
<form class="form-horizontal" role="form" action="edit_task.php" method="post">                      
    <div class="form-group"> 
     <div class="col-lg-6"> 
      <input class="form-control" type="hidden" id="id" name="id" /> 
     </div> 
    </div> 
    <div class="form-group"> 
     <label class="col-lg-2 control-label">Plate Number</label> 
     <div class="col-lg-6"> 
      <?php 
      $query = mysqli_query($con,"select * from vehicle"); // Run your query 
      echo '<select name="plateno" id="plateno">'; // Open your drop down box 
      // Loop through the query results, outputing the options one by one 
      while ($row = mysqli_fetch_array($query, MYSQL_BOTH)) { 
       echo '<option value="' . $row['plateno'] . '">' . $row['plateno'] . '</option>'; 
      } 
      echo '</select>'; 
      ?> 
     </div> 
    </div> 
    <input type="text" class="form-control" id="cartype" name="cartype" placeholder=" " required> 
    <input type="text" class="form-control" id="make" name="make" placeholder=" " required> 
    <input type="text" class="form-control" id="transmission" name="transmission" placeholder=" " required> 
    <input type="text" class="form-control" id="caryear" name="caryear" placeholder=" " required> 
    <input type="text" class="form-control" id="condition" name="condition" placeholder=" " required> 
    <input type="text" class="form-control" id="task" name="task" placeholder=" " required> 
    <input type="text" class="form-control" id="item_used" name="item_used" placeholder=" " required> 
    <input type="text" class="form-control" id="remark" name="remark" placeholder=" " required> 
    <button type="submit" class="btn btn-primary">Save</button> 
    <button type="button" class="btn btn-danger">Cancel</button> 
    </form> 

    </body> 
</html>

这是您的ajax_file.php代码：

<?php 
header('Content-Type: application/json'); 
$response = array(); 
if (isset($_GET['plateno'])){ 
//vul hier je database gebuikersnaam en ww in 
$con=mysqli_connect("localhost","root","","my_db"); 
// Check connection 
if (mysqli_connect_errno()){ 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 

$qry = "SELECT * FROM vehicle WHERE plateno = '".$_GET['plateno']."'"; 

$result = mysqli_query($con, $qry); //mysql_query($qry); 

while ($row = mysqli_fetch_array($result, MYSQL_BOTH)) { 
    array_push($response, $row); 
} 

echo json_encode($response); 
} 
?>

来源

2016-11-03 07:22:06

嗨，请问您在哪里看到json.php？ – ojulari

我已将json.php更改为Ajax_file.php – ojulari

$数据，该数据在编码Ajax_file.php没有值。你没有给它分配任何数据。您可能希望使用$ plateno并从数据库获取记录并将其分配给$ data。

来源

2016-11-02 12:13:36

可以请你给的示例 – ojulari

<？php header（'Content-Type：application/json'）; $ response = array（）;如果（isset（$ _GET ['plateno']））{ // vulh je数据库gebuikersnaam en ww $ con = mysqli_connect（“localhost”，“root”，“”，“my_db”）; //检查连接 if（mysqli_connect_errno（））{ echo“无法连接到MySQL：”。 mysqli_connect_error（）; } $ qry =“SELECT * FROM vehicle WHERE plateno ='”。$ _ GET ['plateno']。“'”; $ result = mysqli_query（$ con，$ qry）; //的mysql_query（$ QRY）; （$ row = mysqli_fetch_array（$ result，MYSQL_BOTH））{ array_push（$ response，$ row）; } echo json_encode（$ response）; } ？> –

上述代码错误添加。 –

替换Ajax_file.php代码：

<?php 
     $plateno =$_POST['plateno']; 

    echo json_encode($plateno); 
    die(); 

?>

除了上面的代码，更改您发送Ajax请求的代码：

$.post('ajax_file.php',data_String,function(data){ 
      var data= jQuery.parseJSON(data); 
      $('#cartype').val(data.cartype) 
      $('#make').val(data.make) 
      $('#transmission').val(data.transmission) 
      $('#caryear').val(data.caryear) 

     });

将这个来代替：

$.ajax({ 
    url: 'ajax_file.php', 
    type: 'post', 
    dataType:'json', 
    data: { 
     plateno: plateno 
    } 
}).done(function(data){ 
    $('#cartype').val(data.cartype) 
    $('#make').val(data.make) 
    $('#transmission').val(data.transmission) 
    $('#caryear').val(data.caryear) 
});

可能的问题可能是您在发送数据时未指定数据类型。 我希望它能帮助

来源

2016-11-02 12:25:43 Learner

不工作。如何获取记录并将其分配给$ data – ojulari

$plateno = $_POST['plateno']; 
if (isset($plateno)){ 
    $con=mysqli_connect("localhost","root","","dn_name"); 

if (mysqli_connect_errno()){ 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 

$qry = "SELECT * FROM vehicle WHERE plateno = '".$plateno."'"; 

$result = mysqli_query($con, $qry); 

while ($row = mysqli_fetch_array($result, MYSQL_BOTH)) { 
    array_push($response, $row); 
} 

echo json_encode($data); 
}

来源

2016-11-03 08:51:31

文本框由下拉选定的值填充

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