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Found 5 results

  1. Hello guys, I have a page named compare.php with two dropdown lists wich are populated with same values from a specific table. Now what i want to do is when a user is selecting one value from the first dropdown and another value from the secound dropdown to be redirected to a page lets say compare-items.php?id=1&id=2 and here in two different cols to get the values for the items, in the first col the values for the first item, and in the secound col the values from the sec item. How can i achive that? Thank you very much!
  2. Below is my code that is suppose to insert and show records from database using ajax. Retrieving the records from the database works. However it won't insert a record. I do not get any errors. The page simply refreshes when I hit submit button. Can you see what might be wrong in my code? index.php <script> $(document).ready(function(){ function showComment(){ $.ajax({ type:"post", url:"process.php", data:"action=showcomment", success:function(data){ $(".wrapper").html(data); } }); } showComment(); $("#submit").click(function(){ var name=$("#name").val(); var message=$("#details").val(); $.ajax({ type:"post", url:"process.php", data:"name="+name+"&details="+message+"&action=addcomment", success:function(data){ showComment(); } }); }); }); </script> $id = $_GET['id']; $title = $_GET['title']; $_SESSION['id'] = $id; $_SESSION['title'] = $title; <div class="wrapper"></div> <form action="" method="post" enctype="multipart/form-data"> <div class="newfield"> <label for="title">Name <span class="highlight">*</span></label> <input id="name" type="text" name="name"> </div> <div class="newfield"> <label for="details">Details <span class="highlight">*</span></label> <textarea id="details" name="details""></textarea> </div> <input type="submit" name="submit" id="submit" value="Submit Tale"> </form> process.php <?php require_once '/core/init.php'; $id = $_SESSION['id']; $action = $_POST['action']; if($action == 'showcomment') { try { $get = $db->prepare("SELECT * FROM sub_posts WHERE id = :id"); $get->bindParam('id', $id); $get->execute(); $getStmt = $get->fetchAll(PDO::FETCH_ASSOC); if(count($getStmt) > 0) { foreach($getStmt as $row) { $new_name = $row['name']; $new_details = $row['details']; ?> <ul> <li> <?php echo $new_name; ?> </li> <li> <?php echo $new_details; ?> </li> </ul> <?php } } else { // no records found. } } catch(Exception $e) { die($e->getMessage()); } } else if($action == 'addcomment') { $name = $_GET['name']; $details = $_GET['details']; try { $insert = $db->prepare("INSERT INTO sub_posts(id, name, details) VALUES(:id, :name, :details)"); $insert->bindParam('id', $id); $insert->bindParam('name', $name); $insert->bindParam('details', $details); $insert->execute(); if($insert == false){ echo 'record could not be inserted.'; } else { echo 'record has been inserted.'; } } catch(Exception $e) { die($e->getMessage()); } } else { echo 'no results.'; }
  3. Say I have a records table with multiple records. There could be multiple records with the same name but different amounts. For Eg. Looking at the table below, the results should retrieve newest record-1 and record-2 because their amount is equals to or greater than 5.00. Record-3 is not selected because it falls below 3.00. record_id record_name record_amount 1 record-1 4.00 2 record-1 3.00 3 record-2 2.00 4 record-1 5.00 5 record-2 6.00 6 record-3 3.00 $get_records = $db->prepare("SELECT record_id, record_name FROM records WHERE record_amount >= :record_amount ORDER BY record_id DESC"); $get_records->bindValue(':record_amount', 5.00); $get_records->execute(); $result_records = $get_records->fetchAll(PDO::FETCH_ASSOC); if(count($result_records) > 0) { foreach($result_records as row) { $record_id = $row['record_id']; $record_name = $row['record_name']; echo $record_name; } } Currently the query above outputs ALL the rows that matches the >=. What I want to do is select only ONE row of each unique record name that matches the criteria. And that row is typically the last row that was inserted. So the output would be like only these two. How do I do that? 4 record-1 5.00 5 record-2 6.00
  4. So I've followed this code, corrected about 12 error, talked to my hosting and I am so done. I have tried everything but the error won't go away. The code is pasted below. As it's really late any help would be appreciated that would make my day the next day... <?PHP //connect to server $connect = mysql_connect("localhost","cello10_import","brigite27"); //connection to database mysql_select_db("cello10_import"); //query the database $query = mysql_query("SELECT * FROM users WHERE cover_image = 'http://d1w7fb2mkkr3kw.cloudfront.net/assets/images/book/small/9781/2500/9781250038821.jpg' "); //fetch the results / convert into array WHILE($rows = mysql_fetch_array(query)): $cover_image = $rows['cover_image']; $title = $rows['title']; $author = $rows['author']; echo "$cover_image<br>$title<br>$author<br><br><br>"; endwhile; ?> The error experienced is: Warning: mysql_fetch_array() expects parameter 1 to be resource, string given in /home/cello10/public_html/display.php on line 10 Thanks for your help
  5. So I have a simple script that adds and returns mysql data without refreshing the page. Everything works as it should. The only issue that I've noticed is that if I login as a user and insert a record(text) the FIRST time, it won't return that record in the row below. Not until I refresh the page. Now if I delete that record and insert a new record, it will then automatically return the record without the page refresh, as it should. Can you tell me why this is happening? Here is my js code. <script> $(document).ready(function(){ var recordId = $("#record-id").val(); function showRecord(){ $.ajax({ type: "POST", url: "bid-actions.php", data: "recordId="+recordId+"&action=showRecord", success:function(data){ $(".show-records").html(data); } }); } showRecord(); $(document).on('click','#record-submit',function() { var message = $("#message").val(); var recordId = $("#record-id").val(); $.ajax({ type: "POST", url: "bid-actions.php", data: "message="+message+"&recordId="+recordId+"&action=insertRecord", success:function(data){ showRecord(); $("#message").val(''); } }); }); }); </script> Html code. <div id="record-submit-form"> <input type="hidden" id="record-id" value="<?php echo $record_id; ?>"> <textarea id="message"></textarea> <input type="button" id="record-submit" value="Submit"> </div> <div class="show-records"> </div>
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