mysql - php mysqli with insert and update queries -
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- php mysqli insert , update queries 3 answers
i creating feedback form users can write feedback , store in database using php mysqli without refreshing whole page . got success message without entered data can me ? asked yesterday same question php mysqli insert , update queries
feedback_form.php
<?php session_start(); $login = ($_session['login']); $userid = ($_session['user_id']); $login_user = ($_session['username']); $fname = ($_session['first_name']); $lname = ($_session['last_name']); $sessionaddres =($_session['address']); ?> <!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"> <head> <meta http-equiv="content-type" content="text/html; charset=utf-8" /> <title>feedback page</title> <script type = "text/javascript" src = "http://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script> <link href="style/stylesheet.css"rel="stylesheet" type="text/css"/> <script type = "text/javascript"> $(function(){ $('#submit').click(function(){ $('#container').append('<img src = "images/loading.gif" alt="currently loading" id = "loading" />'); var comments = $('#comments').val(); $.ajax({ url: 'feedback_process.php', type: 'post', data: {"comments": comments}, success: function(result){ $('#response').remove(); $('#container').append('<p id = "response">' + result + '</p>'); $('#loading').fadeout(500, function(){ $(this).remove(); }); } }); return false; }); }); </script> </head> <?php require_once('header.php'); ?> <body> <form action = "feedback_form.php" method = "post"> <div id = "container"> <h2><?php echo $login_user ?></h2> <label = "comments">comments</label> <textarea rows = "5"cols = "35" name = "comments" id = "comments"></textarea> <br /> </div> </form> <input type = "submit" name = "submit" id = "submit" value = "send feedback" /> </body> </html> feedback_process.php
<?php session_start(); $login = ($_session['login']); $userid = ($_session['user_id']); $login_user = ($_session['username']); $fname = ($_session['first_name']); $lname = ($_session['last_name']); $sessionaddres =($_session['address']); $conn = new mysqli('localhost', 'root', 'root', 'lam_el_chamel_db'); echo"<pre>"; print_r($_post); echo"</pre>"; if(isset($_post['comments'])){ $comments = $_post['comments']; $query = "insert feedback (feedback_text, user_name,) values(?,?)"; $stmt = $conn->stmt_init(); if($stmt->prepare($query)) { $stmt->bind_param('ss', $comments, $login_user); $stmt->execute(); } $query2 = "update feedback set feedback_text = ?, user_name = ? user_name = ? "; $stmt = $conn->stmt_init(); if($stmt->prepare($query2)) { $stmt->bind_param('sss', $comments, $login_user, $login_user); $stmt->execute(); } if($stmt){ echo "thank .we in touch <br />"; } else{ echo "there error. try again later."; } } else echo"it big error"; ?>
first: don't ask questions 2 times. won't better or quicker answers because answer them twice...
the success message tells you've accessed file success (and not else). based on this, try run feedback_process.php alone (without involvement of feedback_form.php) "dummy" comment , "dummy" login_user. added output of "failed" when insert query did not work... (your code implemented success or not the update feedback-query (last one))
i hope code down below you...
<?php session_start(); $login = ($_session['login']); $userid = ($_session['user_id']); $login_user = ($_session['username']); $fname = ($_session['first_name']); $lname = ($_session['last_name']); $sessionaddres =($_session['address']); $conn = new mysqli('localhost', 'root', 'root', 'lam_el_chamel_db'); echo"<pre>"; print_r($_post); echo"</pre>"; //some dummys debugging $comments = 'this comments'; $login_user = 'foo'; $query = "insert feedback (feedback_text, user_name) values(?,?)"; $stmt = $conn->stmt_init(); if($stmt->prepare($query)) { $stmt->bind_param('ss', $comments, $login_user); $stmt->execute(); } else { echo 'failed!'; } $query2 = "update feedback set feedback_text = ?, user_name = ? user_name = ? "; $stmt = $conn->stmt_init(); if($stmt->prepare($query2)) { $stmt->bind_param('sss', $comments, $login_user, $login_user); $stmt->execute(); } else { echo 'failed!'; } if($stmt){ echo "thank .we in touch <br />"; } else { echo "there error. try again later."; } ?> 
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