Darkmatter5 Posted April 29, 2008 Share Posted April 29, 2008 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>" . $row['WorkPhone'] . "</td>"; echo "</tr>"; } echo "</table>"; include 'closedb.php'; ?> Link to comment https://forums.phpfreaks.com/topic/103489-help-with-phpajax/ Share on other sites More sharing options...
mrdamien Posted April 30, 2008 Share Posted April 30, 2008 1. When you say showClientTXT(this.form) your passing the form object. Thats what [object HTMLFormElement] is. You need to use this.form['LastName'].value to send the value in your LastName textbox. 2. "SELECT ClientID, FirstName, LastName, CompanyName, HomePhone, WorkPhone FROM byrnjobdb.clients WHERE LastName=$lastname"; If you don't put quotes around string values ($lastname), MySQL thinks its something else. "SELECT ClientID, FirstName, LastName, CompanyName, HomePhone, WorkPhone FROM byrnjobdb.clients WHERE LastName='$lastname'"; Link to comment https://forums.phpfreaks.com/topic/103489-help-with-phpajax/#findComment-530063 Share on other sites More sharing options...
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