在php jQuery中根据下拉列表选择点击一个按钮时填写文本框 [英] Fill textbox when click a button according to dropdown selection in php jQuery

问题描述

我的过程是这样的：我有一个下拉菜单和文本框。当我从下拉列表中选择一个id（唯一ID），然后单击提交按钮，以显示相应的名称到文本框。

我的数据库字段：

1. id （自动增量）


2. AgencyName_id （唯一ID）


3. 名称

dispay。 html

 < select name =agencyID_dwnclass =idLookup_dwnid =agencyID_dwn> 
< option selected> ...选择...< / option> 
<？php 
 while（$ row = mysqli_fetch_array（$ result））{
？> 
 
< option value =<？php echo $ row ['AgencyName_id'];？>> 
<？php echo $ row ['AgencyName_id'];？>< / option> 
<？php 
} 
？> 
< / select> 
 //输入文本
< input type =textid =testid> 
 //提交按钮
< input type =submitname =lookupSubmit> 
  

dataGet.php

 <？php 
 if（isset（$ _ POST [lookupSubmit]））{
 
 $ user_id = $ _ POST ['agencyID_dwn']; 
 
 $ query =select * from AgencyHome其中AgencyName_id ='$ user_id'; 
 $ result = mysqli_query（$ db，$ query）; 
 
 $ data = mysqli_fetch_assoc（$ result）; 
 
 echo json_encode（$ data）; 
 exit（）; 
 
} 
 
？> 
  

myjson.js

 < script src =// code.jquery.com/jquery-1.11.2.min.js\"> < /脚本> 
< script src =// code.jquery.com/jquery-migrate-1.2.1.min.js\"> < /脚本> 
< script type =text / javascript> 
 $（document）.ready（function（）{
 
 $（'＃agencyID_dwn'）。change（function（）{
 
 var reg_number = $这个）.val（）; 
 var data_String; 
 data_String ='reg_number ='+ reg_number; 
 $ .post（'dataGet.php'，data_String，function（data）{
 var data = jQuery.parseJSON（data）; 
 
 $（'＃testid'）。val（data.Name）; 
 
 
}）; 
}）; 
}）; 
 
< / script> 
  

当我点击提交按钮
我将数据库结果作为数组在dataGet.php 但是在文本框中没有显示结果。我的代码中有任何错误？

解决方案

这是你的答案

你的index.php

 <？php 
 $ conn = mysqli_connect（ 本地主机， 根， ， TEST_DB）; 
 
？> 
 
<！DOCTYPE> 
< html xmlns =http://www.w3.org/1999/xhtml> 
< head> 
< meta http-equiv =Content-Typecontent =text / html; charset = utf-8/> 
< title> Untitled Document< / title> 
< script src =// code.jquery.com/jquery-1.11.2.min.js\"> < /脚本> 
< script src =// code.jquery.com/jquery-migrate-1.2.1.min.js\"> < /脚本> 
< script type =text / javascript> 
 $（document）.ready（function（）{
 
 $（'＃agencyID_dwn'）。change（function（）{
 
 var reg_number = $这个）.val（）; 
 var data_String; 
 data_String ='reg_number ='+ reg_number; 
 $ .post（'dataGet.php'，data_String，function（data）{
 console.log（data）; 
 var data = jQuery.parseJSON（data）; 
 
 $（'＃testid'）。val（data.Name）; 
 
 
}）; 
}）; 
}）; 
 
< / script> 
  

 <身体GT; 
< form> 
< select name =agencyID_dwnclass =idLookup_dwnid =agencyID_dwn> 
< option selected> ...选择...< / option> 
<？php 
 $ query =从代理商名称选择代理商名称; 
 $ result = mysqli_query（$ conn，$ query）; 
 while（$ row = mysqli_fetch_array（$ result））{
？> 
 
< option value =<？php echo $ row ['AgencyName_id'];？>> 
<？php echo $ row ['AgencyName_id'];？>< / option> 
<？php 
} 
？> 
< / select> 
 //输入文本
< input type =textid =testid> 
 //提交按钮
< input type =submitname =lookupSubmit> 
< / form> 
< / body> 
< / html> 
  

和您的dataGet.php文件如下

 <？php 
 
 $ conn = mysqli_connect（localhost，root，，test_db）; 
 
 $ reg_number = $ _ POST ['reg_number']; 
 
 $ query =select * from AgencyName其中AgencyName_id ='$ reg_number'; 
 $ result = mysqli_query（$ conn，$ query）; 
 
 $ data = mysqli_fetch_assoc（$ result）; 
 // print_r（$ data）; 
 echo json_encode（$ data）; 
 exit（）; 
 
 
？> 
只是检查你的表名，所有的都将工作
  

My process is like this: I have a dropdown menu and text box. When I select an id (unique id) from dropdown and then click submit button want to display corresponding name to text box.

My database fields :

1. id (Auto increment)
2. AgencyName_id (unique id)
3. Name

dispay.html

<select name="agencyID_dwn" class="idLookup_dwn"  id="agencyID_dwn" >
<option selected>...Select...</option>
  <?php
  while($row = mysqli_fetch_array($result)){
  ?>

   <option value="<?php echo $row['AgencyName_id'];?>"> 
   <?php echo $row['AgencyName_id'];?></option>
 <?php
  }
   ?>
    </select>
        // for input text
        <input type="text" id="testid">
        // submit button
         <input type="submit" name="lookupSubmit">

dataGet.php

            <?php
            if (isset($_POST["lookupSubmit"])) {

            $user_id=$_POST['agencyID_dwn'];

             $query = "select * from AgencyHome where AgencyName_id = '$user_id'" ;
        $result=mysqli_query($db, $query);

        $data =  mysqli_fetch_assoc($result);

       echo json_encode($data);
         exit();

       }

        ?>

myjson.js

           <script src="//code.jquery.com/jquery-1.11.2.min.js">                </script>
<script src="//code.jquery.com/jquery-migrate-1.2.1.min.js">  </script>
    <script type="text/javascript">
      $(document).ready(function(){

      $('#agencyID_dwn').change(function(){

    var reg_number = $(this).val();
      var data_String;
     data_String = 'reg_number='+reg_number;
    $.post('dataGet.php',data_String,function(data){
      var data= jQuery.parseJSON(data);

       $('#testid').val(data.Name);


          });
       });
      });

      </script>

When i click submit button I got the database results as array in "dataGet.php".But in textbox did not display the result.Any mistake in my code?

解决方案

here is your answer

your index.php

    <?php
    $conn = mysqli_connect("localhost","root","","test_db");

 ?> 

<!DOCTYPE>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
<script src="//code.jquery.com/jquery-1.11.2.min.js">                     </script>
<script src="//code.jquery.com/jquery-migrate-1.2.1.min.js">  </script>
 <script type="text/javascript">
  $(document).ready(function(){

  $('#agencyID_dwn').change(function(){

var reg_number = $(this).val();
  var data_String;
 data_String = 'reg_number='+reg_number;
$.post('dataGet.php',data_String,function(data){
    console.log(data);
  var data= jQuery.parseJSON(data);

   $('#testid').val(data.Name);


      });
   });
  });

  </script>

<body>
<form>
<select name="agencyID_dwn" class="idLookup_dwn"  id="agencyID_dwn" >
<option selected>...Select...</option>
 <?php
  $query = "select AgencyName_id from AgencyName";
 $result = mysqli_query($conn,$query);
 while($row = mysqli_fetch_array($result)){
  ?>

   <option value="<?php echo $row['AgencyName_id'];?>"> 
   <?php echo $row['AgencyName_id'];?></option>
  <?php
   }
   ?>
    </select>
    // for input text
    <input type="text" id="testid">
    // submit button
       <input type="submit" name="lookupSubmit">
  </form>
 </body>
 </html>

and your dataGet.php file as bellow

     <?php

   $conn = mysqli_connect("localhost","root","","test_db");

        $reg_number=$_POST['reg_number'];

   $query = "select * from AgencyName where AgencyName_id = '$reg_number'" ;
    $result=mysqli_query($conn, $query);

    $data =  mysqli_fetch_assoc($result);
   // print_r($data);
      echo json_encode($data);
     exit();


    ?>
     just ccheck your table name and all will work

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