passing php variables

Question

kgizo 0 Light Poster

13 Years Ago

Hi

what i want to do is pass a php variable in a script i've already got a drop down list
now what i want to do is as soon as you click on any province in the drop down menu the cities appear in the text area. what is happening now iz i have to put the id in the writeclass function which iz kinda not what i want.

heres the code of the funtion

<form name="myform">

        <select name = "province" onChange="onChangeDropBox();"> 
        <? populateDropBox(); ?>     
        <textarea name ="textArea" readonly="true"></textarea>     
        </form>      
<script type="text/javascript">
function onChangeDropBox()

    {
        var selected =0;
        selected = document.myform.province.value;                  

        var t = "<? writeCities(9);?>";
        document.myform.textArea.value = t;

        }
</script>

thanxx in advance!!!!

php

Edited 11 Years Ago by mike_2000_17 because: Fixed formatting

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7 Replies
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Latest Post 13 Years Ago Latest Post by diafol

diafol

13 Years Ago

This usually requires Ajax.

Either that or you create javascript arrays of data via php before the page displays.

<script type="text/javascript">
//place data into js from php 
var myCities=new Array(<?php echo $arr_city;?>);
</script>

diafol

13 Years Ago

CODE tags please. Can't read it.

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kgizo 0 Light Poster · Answer 1 · 2011-03-06T16:40:33+00:00

thanxx ardav i'll try it out, but i dont want to use ajax, thanxx alot for your interest!!!

kgizo 0 Light Poster · Answer 2 · 2011-03-07T19:01:52+00:00

ardav i've changed my mind now i think i'll go with ajax ive already created a submit button for the drop down i have never used ajax before but i saw a similar code at w3 schools but i need guidane on how i'm going to about implementing this code.

here's my code again

function writeCities($id)
{       
    $con = mysql_connect("localhost","root","");
    if (!$con) die('Could not connect: ' . mysql_error());
    mysql_select_db("msansi", $con);
    $query  = "SELECT cities FROM provinces WHERE id =";
    $query .= $id;
    $result = mysql_query($query);          

    $row = mysql_fetch_array($result);
    echo $row[0];       
}



function populateDropBox()
{
    $con = mysql_connect("localhost","root","");
    if (!$con) die('Could not connect: ' . mysql_error());
    mysql_select_db("msansi", $con);
    $result = mysql_query("SELECT id,title,cities FROM provinces");

    while($row = mysql_fetch_array($result))
    {   
        echo "<option value=$row[0]>" . $row['title']."</option>";
    }
}
?>

<form name="myform">

    <select name = "province" onChange="onChangeDropBox();"/> 
    <? populateDropBox(); ?>     
    <input type="submit" value ="submit";>     
    </form>

the ajax code is this

function showCustomer(str)
{
var xmlhttp;
if (str=="")
  {
  document.getElementById("txtHint").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("txtHint").innerHTML=xmlhttp.responseText;
    }
  }
xmlhttp.open("GET","getcustomer.asp?q="+str,true);
xmlhttp.send();
}

and its output is this when a custumer is selected from the drop menu and its in a table

CustomerID  ALFKI 
CompanyName Alfreds Futterkiste 
ContactName Maria Anders 
Address         Obere Str. 57 
City            Berlin 
PostalCode  12209 
Country         Germany

your help will be highly appreciated!!!

kgizo 0 Light Poster · Answer 3 · 2011-03-07T20:09:04+00:00

ok heres the code for my drop down and my submit button

<head><title><?php echo $title;?></title>

<h1><?php echo $heading;?></h1>





</head>

<body>


<?
	function writeCities($id)
	{		
		$con = mysql_connect("localhost","root","");
		if (!$con) die('Could not connect: ' . mysql_error());
	  	mysql_select_db("msansi", $con);
		$query  = "SELECT cities FROM provinces WHERE id =";
		$query .= $id;
		$result = mysql_query($query);			
				
		$row = mysql_fetch_array($result);
		echo $row[0];		
	}
	


	function populateDropBox()
	{
		$con = mysql_connect("localhost","root","");
		if (!$con) die('Could not connect: ' . mysql_error());
	  	mysql_select_db("msansi", $con);
		$result = mysql_query("SELECT id,title,cities FROM provinces");
		
		while($row = mysql_fetch_array($result))
		{	
		    echo "<option value=$row[0]>" . $row['title']."</option>";
		}
	}
?>
 
	<form name="myform">
	
		<select name = "province" onChange="onChangeDropBox();"/> 
		<? populateDropBox(); ?> 	
		<input type ="submit" value="submit";/>   
		</form>	  	


	  
	  
	  
</body>

///////////////////////////////////////////////////////////////////////////////////
and the code from w3 schools is this
which is Ajax and outputs the table in the previous message when a customer is chosen from a drop down menu

I hope i did enough to let you see!!!

Evilfairy -4 Newbie Poster · Answer 4 · 2011-03-07T21:30:21+00:00

You need to type [code=language] for code tags, kgizo ;-)

Additionally, you might want to try asking in the JavaScript / DHTML / AJAX Forum. You're more likely to get an answer there than in the PHP forum

diafol · Answer 5 · 2011-03-07T23:16:47+00:00

[ CODE ] tags!!!!