cleary1981 Posted July 1, 2008 Share Posted July 1, 2008 Hi if a was querying a database to find a single result how would I output the result? <?php require "config.php"; $model = $_REQUEST['model']); $sql = "SELECT mod_desc FROM module WHERE type='$model'"; $result = mysql_query($sql) or trigger_error(mysql_error()); $row = mysql_fetch_object($result); $description=$row->description; echo $description; ?> The result should only be one item. How do I get this to output from the script? Quote Link to comment Share on other sites More sharing options...
kenrbnsn Posted July 1, 2008 Share Posted July 1, 2008 You almost have it, but you're trying to get the item by the wrong name: <?php <?php require "config.php"; $model = $_REQUEST['model']); $sql = "SELECT mod_desc FROM module WHERE type='$model'"; $result = mysql_query($sql) or trigger_error(mysql_error()); $row = mysql_fetch_object($result); $description=$row->mod_desc; //this has to match what you have in the select statement. echo $description; ?> Ken Quote Link to comment Share on other sites More sharing options...
trq Posted July 1, 2008 Share Posted July 1, 2008 Your selecting mod_desc then trying to display description. Your code ought be.... <?php require "config.php"; if (isset($_REQUEST['model'])) { $model = mysql_real_escape_string($_REQUEST['model']); $sql = "SELECT mod_desc FROM module WHERE type='$model'"; if ($result = mysql_query($sql)) { if (mysql_num_rows($result)) { $row = mysql_fetch_object($result); echo $row->mod_desc; } } } ?> Quote Link to comment Share on other sites More sharing options...
Wolphie Posted July 1, 2008 Share Posted July 1, 2008 From what I understand, you want to return only a single result even if there are more results of the same criteria. I'm not really sure how you want to organize it; but there are plenty of SQL statements that can organize it by ID etc.. <?php $model = $_REQUEST['model']; if(isset($model)) { $query = sprintf("SELECT `mod_desc` FROM `module` WHERE `type` = '%s' LIMIT 1", mysql_real_escape_string($model)); $result = mysql_query($query) or trigger_error(mysql_error()); if(mysql_num_rows($query) > 0) { while($obj = mysql_fetch_object($query)) { print $obj->mod_desc; // You were calling it by the wrong name, dammit those 2 got there before me >.< } } else { print 'No results!'; } ?> I'm not entirely sure what you are looking for, if this is not acceptable. Please try to make yourself clearer. Quote Link to comment Share on other sites More sharing options...
cleary1981 Posted July 1, 2008 Author Share Posted July 1, 2008 kenrbnsn your answer works best for me but i still have one problem. when the result passes back into my javascript I get "undefined". Any ideas why? Heres my code <!doctype html public "-//w3c//dtd html 3.2//en"> <html> <head> <title>TES - Pricing System</title> <link rel="stylesheet" type="text/css" href="style.css" /> </script> <script type="text/javascript" src="prototype.js"></script> <script type="text/javascript" src="scriptaculous.js"></script> <script language="javascript" src="list.php"></script> <script type="text/javascript" src="getdimensions.js"> </script> <script type="text/javascript" src="text-utils.js"> </script> <script type="text/javascript"> var request = null; try { request = new XMLHttpRequest(); } catch (trymicrosoft) { try { request = new ActiveXObject("Msxml2.XMLHTTP"); } catch (othermicrosoft) { try { request = new ActiveXObject("Microsoft.XMLHTTP"); } catch (failed) { request = null; } } } if (request == null) alert("Error creating XMLHttp Request!"); window.onload = function() { new Draggable('object', {snap:25} ); fillCategory(); } function get_description() { var model = document.getElementById("model").value; var url = "lookupdescription.php?model=" + escape(model); request.open("GET", url, true); request.onreadystatechange = updatePage; request.send(null); } function updatePage() { if (request.readyState ==4) { var modeldescription = request.reponseText; var x = document.getElementById("description"); replaceText(x, modeldescription); } } </script> </head> <body bgcolor="#ffffff" text="#000000" link="#0000ff" vlink="#800080" alink="#ff0000"> <div id="container"> <div id="top"> <h1>TES - Pricing System (Design Mode)</h1> </div> <div id="leftSide"> <fieldset> <legend>Module</legend> <FORM name="drop_list"> <label for="type">Type</label> <div class="div_texbox"> <SELECT NAME="Category" onChange="SelectSubCat();" > <Option value="">Select Type</option> </SELECT> </div> <label for="mod_named">Model</label> <div class="div_texbox"> <SELECT id="model" NAME="SubCat" onChange="get_description();"> <Option value="">Select Model</option> </SELECT> </div> <fieldset> <legend>Description</legend> <span id="description">description goes here</span> </fieldset> <label for="mod_name">Name</label> <div class="div_texbox"> <input name="username" type="text" class="username" id="username" value="" /> </div> <div class="button_div"> <input name="Submit" type="button" value="Generate" class="buttons" onClick="generate_module()"/> </div> </form> </fieldset> <fieldset> <fieldset> <legend>Delete</legend> </fieldset> <fieldset> <legend>Generated Module</legend> <img id="object"> </fiedset> </fieldset> </div> <div id="content"> <div id="drop_area"> </div> </div> <div class="clear"></div> </div> </body> </html> Quote Link to comment Share on other sites More sharing options...
trq Posted July 1, 2008 Share Posted July 1, 2008 Your question is no longer php related, Id'e suggest asking your new question in either the Javascript or Ajax boards. Quote Link to comment Share on other sites More sharing options...
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