dynamic select list AJAX and insertion in database table through posting form

**nadeem14375** · Jul 30, 2012 22:39

I am wondering to find a solution for my dynamic select list of cities, now I am successfully populated the cities but now unable to post the selected city to database.

here is php file:

Code:

<html>  
<head>  
<script type="text/javascript" src="show_cities.js"> 
</script>   
</head>  
<body>  
<?php   
session_start;  
include("conn.php");  
?>
  <form enctype='multipart/form-data' action='posting_process.php' method='post'>");  
<!-- **************************** Country select list ********************** -->  
Country:   <select name="country" onChange="showCities(this.value)">  
<?php
 $sql = 'SELECT country_no, country_name FROM country '.'ORDER BY country_no';      
$rs = mysql_query($sql);      
echo "<option value='0'>"."Select a Country"."</option>\n  ";
      while($row = mysql_fetch_array($rs))      
{  echo "<option value=\"".$row['country_no']."\">".$row['country_name']."</option>\n  ";      }  
?>  
</select>  City: 
<div id="txtCity" class="city_no"> </div> 
<input type=submit name=action value=Post>
</form>  
</body>  
</html>

here is javascript: show_cities.js

Code:

// JavaScript Document  
/* <script type="text/javascript"> */  
function showCities(str)  {  if (str=="")  {  document.getElementById("txtCity").innerHTML="";  return;  }   
if (window.XMLHttpRequest)  {
// code for IE7+, Firefox, Chrome, Opera, Safari  
xmlhttp=new XMLHttpRequest();  }  
else  {
// code for IE6, IE5  
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");  
}  
xmlhttp.onreadystatechange=function()  {  
if (xmlhttp.readyState==4 && xmlhttp.status==200)  
{  document.getElementById("txtCity").innerHTML=xmlhttp.responseText;  }  
}  
xmlhttp.open("GET","get_cities.php?q="+str,true);  xmlhttp.send();  }  
/* </script> */

here is php file: get_cities.php

Code:

<?php  
session_start;  
//library     
include("conn.php");  
$q=$_GET["q"];    
$sql="SELECT * FROM city WHERE country_no = ".$q;  
$result = mysql_query($sql);    
echo "<select name='city'>";    
while($row = mysql_fetch_array($result)) 
 {    
echo "<option value=\"".$row['city_no']."\">".$row['city_name'] . "</option>\n";  
}  
echo "</select>";    
?>

don't know how to mention the txtcity in my insert statement to insert in database, all the other columns are ok, but city is blank.

regards:

**MarPlo** · Jul 31, 2012 05:18

Hi,
From what I saw in your code, the cities are added in <select name='city'>, so, in your php script you can get the selected option with $_POST['city'] .

**nadeem14375** · Jul 31, 2012 09:32

Dear Please help me, I am wondering for many weeks to solve this problem, but could not.

here is the insert script:

[CODE]
session_start();

include("conn.php");
$today = date("Y-m-d");

$sql_ad_no = mysql_query("select max(ad_no) from ad_info") or die(mysql_error());
$info_ad_no = mysql_fetch_row( $sql_ad_no );
$ad_no = $info_ad_no[0];
$ad_no = $ad_no+1;

$result = mysql_query("INSERT INTO ad_info (
ad_no,
ad_date,
city,
country,
CATEGORY_NO,
sub_category_no,
title,
price,
description,
email,
website,
user_type,
phone_no,
address
) VALUES (
$ad_no,
'$today',
'$_POST[city]',
'$_POST[country]',
'$_POST[category]',
'$_POST[sub_category_no]',
'$_POST[title]',
'$_POST[price]',
'$_POST[description]',
'$_POST[email]',
'$_POST[website]',
'$_POST[user_type]',
'$_POST[phone_no]',
'$_POST[address]'
)");

[CODE]

**cpradio** · Jul 31, 2012 09:46

Originally Posted by nadeem14375

Dear Please help me, I am wondering for many weeks to solve this problem, but could not.

That is because you are using GET as your Ajax Submission, not POST. So if you change $_POST to $_GET, it will work. However, you should really adjust this further, as you have opened a SQL Injection.

You should instead change your code to as follows:

PHP Code:


session_start();

include("conn.php");
$today = date("Y-m-d"); 

$sql_ad_no = mysql_query("select max(ad_no) from ad_info") or die(mysql_error());
$info_ad_no = mysql_fetch_row( $sql_ad_no );
$ad_no = $info_ad_no[0]; 
$ad_no = $ad_no+1;

$city = mysql_real_escape_string($_GET['city']);
$country = mysql_real_escape_string($_GET['country']);
$category = mysql_real_escape_string($_GET['category']);
$sub_category_no = mysql_real_escape_string($_GET['sub_category_no']);
$title = mysql_real_escape_string($_GET['title']);
$price = mysql_real_escape_string($_GET['price']);
$description = mysql_real_escape_string($_GET['description']);
$email = mysql_real_escape_string($_GET['email']);
$website = mysql_real_escape_string($_GET['website']);
$user_type = mysql_real_escape_string($_GET['user_type']);
$phone_no = mysql_real_escape_string($_GET['phone_no']);
$address = mysql_real_escape_string($_GET['address']);

$result = mysql_query("INSERT INTO ad_info (
ad_no,
ad_date,
city,
country,
CATEGORY_NO,
sub_category_no,
title,
price,
description,
email,
website,
user_type,
phone_no,
address
) VALUES (
$ad_no,
'$today',
'$city',
'$country',
'$category',
'$sub_category_no',    
'$title',
'$price',
'$description',
'$email',
'$website',    
'$user_type',
'$phone_no',
'$address'
)");

**nadeem14375** · Jul 31, 2012 10:01

Dear cpradio,

in my <form > I used the method=post so why I use the get?

this also not work for me. $city = mysql_real_escape_string($_GET['city']);

the other values are ok, and can be inserted.

why this city have problem?

regards:

**cpradio** · Jul 31, 2012 10:49

Sorry, I misread your original post and the corresponding code. You are correct, your form does use $_POST.

Can you add the following to your posting_process.php file?

PHP Code:


var_dump($_POST);

I'd like to see what fields are being sent to it.

**nadeem14375** · Jul 31, 2012 20:58

here is the out put and there is no city again.

array(11) { ["country"]=> string(1) "1" ["title"]=> string(3) "pak" ["price"]=> string(2) "23" ["description"]=> string(2) "23" ["email"]=> string(14) "emai@email.com" ["website"]=> string(10) "www.abc.om" ["user_type"]=> string(1) "1" ["phone_no"]=> string(2) "33" ["address"]=> string(2) "22" ["MAX_FILE_SIZE"]=> string(7) "1000000" ["action"]=> string(4) "Post" }

**codeatar** · Jul 31, 2012 21:42

I believe this has to do with your use of innerHTML to dynamically create your city dropdown. This person had an experience similar to yours:

http://www.webdeveloper.com/forum/sh...d.php?t=218311

Try doing some Googling for stuff about innerHTML not passing form fields and you'll find a few more discussions with possible solutions for you. Hope that helps!

**nadeem14375** · Jul 31, 2012 22:01

thanks for your reply,

can you guide me to some other option to gain the result, because this a very lengthy discussion / research and I am beginner in web development.

Please guide me to some other solution.

**nadeem14375** · Aug 1, 2012 01:20

the following tutorial help me solve the problem.

http://www.9lessons.info/2010/08/dyn...box-using.html

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