jotorres1
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Everything posted by jotorres1
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OnClick Button to Display Database Table
jotorres1 replied to javinladish's topic in PHP Coding Help
Glad it worked for ya! -
OnClick Button to Display Database Table
jotorres1 replied to javinladish's topic in PHP Coding Help
Yes. Try change this line: $("#result_table").append(output_string); to this: $("#result_table").html(output_string); The html will rewrite what you have thus making it appear only once. -
OnClick Button to Display Database Table
jotorres1 replied to javinladish's topic in PHP Coding Help
I can probably give you basic help on this. Edit: had to modify your first 2 echo's for json_encode, sorry. I have modified your file just a bit to be able to handle json output. <?php // This file I have named it getdata.php // And you will see why include("connect.php"); //Query of facebook database $facebook = mysql_query("SELECT * FROM facebook") or die(mysql_error()); //Output results if(!$facebook) { mysql_close(); echo json_encode("There was an error running the query: " . mysql_error()); } elseif(!mysql_num_rows($facebook)) { mysql_close(); echo json_encode("No results returned"); } else { $header = false; $output_string = ""; $output_string .= "<table border='1'>\n"; while($row = mysql_fetch_assoc($facebook)) { if(!$header) { $output_string .= "<tr>\n"; foreach($row as $header => $value) { $output_string .= "<th>{$header}</th>\n"; } $output_string .= "</tr>\n"; } $output_string .= "<tr>\n"; foreach($row as $value) { $output_string .= "<th>{$value}</th>\n"; } $output_string .= "</tr>\n"; } $output_string .= "</table>\n"; } mysql_close(); // This echo for jquery echo json_encode($output_string); ?> The html file, (php for me) I have named it display.php <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <script language="JavaScript" type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.6.4/jquery.min.js"></script> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Display Page</title> </head> <body> <button type="button" name="getdata" id="getdata">Get Data.</button> <div id="result_table"> </div> <script type="text/javascript" language="javascript"> $('#getdata').click(function(){ $.ajax({ url: "getdata.php", type:'POST', dataType: 'json', success: function(output_string){ $("#result_table").append(output_string); } // End of success function of ajax form }); // End of ajax call }); </script> </body> </html> Since I have no idea what your page looks like, I just made a simple button, that when clicked, it will look inside the DB, grab data, and return it, placing it inside the div. Now I have not tested this, but this is how it should be looking like. If you get any errors let me know, I will help out.