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Error help '[object HTMLFormElement]'


Darkmatter5

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Here are the three snippets of code that I'm using.  I have a text box with a button that launches a function in the javascript and displays a result at a div in the search.php file.  The problem is when I click the button the result I get is "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '[object HTMLFormElement]' at line 3".  Can someone please help me diagnose this error?

 

search.php

<form><div align="center" class="style7"><span class="style6">
    <label>Search by LastName:
    <input name="LastName" type="text" id="LastName" /></label></span>
    <input name="searchclients" type="button" id="searchclients" onclick="showClientTXT(this.form)" value="Search" />
    </div></form>

 

select_TXTsearchclients.js

// JavaScript Document
var xmlHttp

function showClientTXT(str)
{
xmlHttp=GetXmlHttpObject()
if (xmlHttp==null)
{
	alert ("Browser does not support HTTP Request")
	return
}
var url="http://byrndb01/byrndb/library/get_TXTsearchclients.php"
url=url+"?r1="+str
url=url+"&sid="+Math.random()
xmlHttp.onreadystatechange=stateChanged
xmlHttp.open("GET",url,true)
xmlHttp.send(null)
}

function stateChanged()
{
if (xmlHttp.readyState==4 || xmlHttp.readyState=="complete")
{
	document.getElementById("results").innerHTML=xmlHttp.responseText
}
}

function GetXmlHttpObject()
{
var xmlHttp=null;
try
{
	// Firefox, Opera 8.0+, Safari
	xmlHttp=new XMLHttpRequest();
}
catch (e)
{
	//Internet Explorer
	try
	{
		xmlHttp=new ActiveXObject("Msxml12.XMLHTTP");
	}
	catch (e)
	{
		xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
	}
}
return xmlHttp;
}

 

get_TXTsearchclients.php

<?php
$lastname=$_GET["r1"];

include 'dbconfig.php';
include 'opendb.php';

$query="SELECT ClientID, FirstName, LastName, CompanyName, HomePhone, WorkPhone
	FROM byrnjobdb.clients
	WHERE LastName=$lastname";
$result=mysql_query($query) OR die (mysql_error());

echo "<table border='1'>
  <tr>
  <th>ClientID</th>
  <th>First Name</th>
  <th>Last Name</th>
  <th>Company Name</th>
  <th>Home Phone</th>
  <th>Work Phone</th>
  </tr>";

while($row=mysql_fetch_array($result))
{
echo "<tr>";
echo "<td><a href='addclient.php' target='mainFrame'>" . $row['ClientID'] . "</a></td>";
echo "<td>" . $row['FirstName'] . "</td>";
echo "<td>" . $row['LastName'] . "</td>";
echo "<td>" . $row['CompanyName'] . "</td>";
echo "<td>" . $row['HomePhone'] . "</td>";
echo "<td>" . $row['WorkPhone'] . "</td>";
echo "</tr>";
}
echo "</table>";

include 'closedb.php';
?>

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SELECT ClientID, FirstName, LastName, CompanyName, HomePhone, WorkPhone

FROM byrnjobdb.clients

WHERE LastName=$lastname

 

Kindly run this in your query interface i think you have error somewhere here.

 

special in "FROM byrnjobdb.clients" i.e if you are keeping table aliases then byrnjobdb as clients i.e tablename as alias

 

hope it help to solve your query.

 

Gr8 day!!!

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