根据php jQuery中的下拉选项单击按钮时填充文本框
[英]Fill textbox when click a button according to dropdown selection in php jQuery
问题描述
My process is like this: I have a dropdown menu and text box. 
我的过程是这样的：我有一个下拉菜单和文本框。 When I select an id (unique id) from dropdown and then click submit button want to display corresponding name to text box. 当我从下拉列表中选择一个id（唯一ID）然后单击“提交”按钮时，要在文本框中显示相应的名称。
My database fields : 我的数据库字段：
-
id(Auto increment)id（自动增量） -
AgencyName_id(unique id)AgencyName_id（唯一ID） - Name 名称
dispay.html 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 $row['AgencyName_id'];?></option>
<?php
}
?>
</select>
// for input text
<input type="text" id="testid">
// submit button
<input type="submit" name="lookupSubmit">
dataGet.php dataGet.php
<?php
if (isset($_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 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);
$('#testid').val(data.Name);
});
});
});
</script>
When i click submit button I got the database results as array in "dataGet.php".But in textbox did not display the result.Any mistake in my code? 当我点击提交按钮时，我在“dataGet.php”中将数据库结果作为数组。但是在文本框中没有显示结果。我的代码中有任何错误？
2 个解决方案
解决方案1
2 2015-03-21 10:07:45
try like this in dataGet.php, 在dataGet.php中尝试这样，
while ($row = mysqli_fetch_assoc($result)) {
echo $row["Name"];
}
解决方案2
1 已采纳 2015-03-21 10:27:34
here is your answer 这是你的答案
your index.php 你的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 和你的dataGet.php文件如下
<?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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