mac007 Posted October 28, 2008 Share Posted October 28, 2008 Hi, I am trying to work out this SELECT/UPDATE script here below and I almost got it to work, but no matter what I do, can't get the script to display back the "Updated" record correctly; unless I like refresh my browser twice, then it displays it right... Appreciate any feedback! <CODE> <?php // connection info $host = 'housername'; $password = 'password'; $database = 'database_name'; $con = mysql_pconnect($host,$user,$password) or die('No Connection Found'); mysql_select_db($database,$con) or die('No Database Found'); $submit = $_POST['submit']; $id = $_GET['id']; $id_update = $_POST['recordID']; $title = $_POST['title']; $body = $_POST['body']; $result = mysql_query("SELECT * from blog WHERE blog_id = $id"); if(isset($id)) { $row = mysql_fetch_array($result) or die('No record found'); } if (isset($submit)){ mysql_query("UPDATE blog SET blog_title = '$title', blog_body = '$body' WHERE blog_id = '$id_update'"); } ?> <form id="form1" name="form1" method="POST" action=""> <p>post title: <input name="title" type="text" id="title" value="<?php echo $row['blog_title'];?>"/> </p> <p>post body: <input name="body" type="text" id="body" value="<?php echo $row['blog_body'];?>" /> </p> <p> <input name="recordID" type="hidden" id="recordID" value="<?php echo $row['blog_id'];?>" /> </p> <p> <input type="submit" name="submit" value="submit" /> </p> </form> <?php mysql_close($con); ?> </CODE> Link to comment https://forums.phpfreaks.com/topic/130474-solved-how-come-i-cant-get-updated-record-to-display-back-unles-i-refresh-page-twice/ Share on other sites More sharing options...
wildteen88 Posted October 28, 2008 Share Posted October 28, 2008 Move if (isset($submit)){ mysql_query("UPDATE blog SET blog_title = '$title', blog_body = '$body' WHERE blog_id = '$id_update'"); } before $result = mysql_query("SELECT * from blog WHERE blog_id = $id"); if(isset($id)) { $row = mysql_fetch_array($result) or die('No record found'); } Change if (isset($submit)) to if (isset($_POST['submit'])) Link to comment https://forums.phpfreaks.com/topic/130474-solved-how-come-i-cant-get-updated-record-to-display-back-unles-i-refresh-page-twice/#findComment-676838 Share on other sites More sharing options...
mac007 Posted October 28, 2008 Author Share Posted October 28, 2008 Thanks Wildteen!!! right on the dot... worked right off I guess I do need to really watch the sequence of executions; I messed with it so much, but the one thing I hadnt done was to place the UPDATE line before the $result!!! makes sense...( first update, then echo back...) was so close!! Thanks a lot! hey, one more quick question: I noticed one can use the mysql_connect, or also the mysql_pconnect, which one is best? and is it correct that one doenst need to close the connection with the latter? Link to comment https://forums.phpfreaks.com/topic/130474-solved-how-come-i-cant-get-updated-record-to-display-back-unles-i-refresh-page-twice/#findComment-676892 Share on other sites More sharing options...
wildteen88 Posted October 28, 2008 Share Posted October 28, 2008 mysql_pconnect use a persistent connections, which means the connection to the mysql server is never closed until you use mysql_close. mysql_connect does not use a persistent connection and the connection is terminated once the script has finished being parsed. Link to comment https://forums.phpfreaks.com/topic/130474-solved-how-come-i-cant-get-updated-record-to-display-back-unles-i-refresh-page-twice/#findComment-676988 Share on other sites More sharing options...
mac007 Posted October 29, 2008 Author Share Posted October 29, 2008 thanks a lot Wildteen... Link to comment https://forums.phpfreaks.com/topic/130474-solved-how-come-i-cant-get-updated-record-to-display-back-unles-i-refresh-page-twice/#findComment-677132 Share on other sites More sharing options...
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