PHP - Can't Echo The Count
I am trying to echo the count for blog posts that have not been review yet, but I get this error.
Notice: Undefined index: status but I feel like I have defined my status in my query. Plus my query is working as I have checked it in phpmyadmin Here is my function: Code: [Select] function unapproved_post_counter() { $sql = "SELECT COUNT(`approved`) AS `status` FROM `blog_posts` WHERE `approved` = 0 "; $result = mysql_query($sql); $data = array(); while (($row = mysql_fetch_assoc($result)) !== false) { $data[] = $row; } return $data; } Here is how I am echoing it. Code: [Select] $count = unapproved_post_counter(); <h1>Blog<span class="blogadds f_right"><?php echo $count['status']; ?> New Articles</span> print_r($count) tells me this... Array ( => Array ( [status] => 2 ) ) Similar TutorialsI have a html table displaying data from mysql database. I can count rows properly with mysql_num_rows but is there a way to echo which row number it is. What I want to do is count the rows ordered by cities and echo row number. Thanks for any help. Hi guys, I need your help. I am trying to insert the rows in the mysql database as I input the values in the url bar which it would be like this: Code: [Select] www.mysite.com/testupdate.php?user=tester&pass=test&user1=tester&email=me@shitmail.com&ip=myisp However i have got a error which i don't know how to fix it. Error: Column count doesn't match value count at row 1 <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'mydbusername'); define('DB_PASSWORD', 'mydbpassword'); define('DB_DATABASE', 'mydbname'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $username = clean($_GET['user']); $password = clean($_GET['pass']); $adduser = clean($_GET['user1']); $email = clean($_GET['email']); $IP = clean($_GET['ip']); if($username == '') { $errmsg_arr[] = 'username is missing'; $errflag = true; } if($password == '') { $errmsg_arr[] = 'PASSWORD is missing'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $sql = "INSERT INTO `members` (`username`,`email`,`IP`) VALUES ('$adduser','$email','$IP')"; if (!mysql_query($sql,$link)) { die('Error: ' . mysql_error()); } echo "The information have been updated."; } ?> Here's the name of the columns i have got in my database: Code: [Select] username IP I have input the correct columns names, so I can't correct the problem I am getting. Please can you help? This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=331562.0 Why am I getting this error when there are 3 Fields with 3 values? Column count doesn't match value count at row 1 Code: [Select] $sql5="INSERT INTO participants (participant_name, team_no, sport) VALUES ('".implode("','$_POST[team]'),('",$_POST['participant_name'])."','$_POST[team]','$sport')"; Anyone can help me? <?php $tbl_name="menu"; $kategorije = mysql_query("SELECT * FROM $tbl_name WHERE Left(menu_name, 1) BETWEEN 'A' AND 'M' ORDER BY menu_name ASC"); if (!$kategorije) { die("Database query failed: " . mysql_error()); } while ($row=mysql_fetch_array( $kategorije )) { echo "<div id=lista-header><h4>{$row["menu_name"]}</h4></div>"; $id_sub=$row['id_menu']; $podkategorije = mysql_query("SELECT * FROM submenu WHERE id_menu=$id_sub ORDER BY sub_name ASC", $connection); if (!$podkategorije) { die("Database query failed: " . mysql_error()); } echo "<ul class=\"pod\">"; while ($pod=mysql_fetch_array( $podkategorije )) { echo "<li><a href=index.php?=podkate?".$pod["id_sub"]." class=black>{$pod["sub_name"]}</a><hr size=1 align=left width=100px color=#cccccc></li>"; } echo "</ul>"; } ?> In this way, I get list with categories and hes subcategories. How to count how many subcategories have each categories, and count how many articles have each categories? Example (I wanna get this kind of look): Categories name (3) subcategoriesname (2) subcategoriesname (4) subcategoriesname (7) Categories name (5) subcategoriesname (1) subcategoriesname (14) subcategoriesname (9) subcategoriesname (2) subcategoriesname ( Categories name (2) subcategoriesname (28) subcategoriesname (17) Where the numbers represent how many categories and sub-items have articles OK, have no idea what's going on... I've done this a million times... why wont this output!?? I must have a major brain meltdown and dont know it yet!!! Code: [Select] <?php // this echoes just fine: echo $_POST['testfield']; // but this wont echo: echo if (isset($_POST['testfield'])) { $_POST['testfield'] = $test; } echo $test; /// or even this DOESNT echo either!: $_POST['testfield'] = $test; echo $test; ?> Hi All, I'm trying to echo the response from an SLA query, the query works and returns the data when I test it on an SQL application.. but when I run it on my webpage it won't echo the result. Please help? <?php $mysqli = mysqli_connect("removed", "removed", "removed", "removed"); $sql = "SELECT posts.message FROM posts INNER JOIN threads ON posts.pid=threads.firstpost WHERE threads.firstpost='1'"; $result = mysqli_query($mysqli, $sql); echo {$result['message']}; ?> I have a log system that allows 10 logs on each side(Left and right). I am trying to make it so that the left side has the 10 most recent logs, then the right as the next 10. Any ideas? So I need to echo a row from my database with php, but where i need to echo is already inside an echo. This is my part of my code: $con = mysql_connect("$host","$username","$password"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("main", $con); $result = mysql_query("SELECT * FROM Vendor"); while($row = mysql_fetch_array($result)) { //I need to echo right here .................. but I get a blank page when I try this. Please Help. echo '<option value=$row['vendor_id']>'; echo $row['vendor_id']; echo '</option>'; } mysql_close($con); Result: A Blank page. Thanks in advance! Hi
I try to echo out random lines of a html file and want after submit password to whole content of the same html file. I have two Problems.
1st Problem When I echo out the random lines of the html file I don't get just the text but the code of the html file as well. I don't want that. I just want the text. How to do that?
for($x = 1;$x<=40;$x++) { $lines = file("$filename.html"); echo $lines[rand(0, count($lines)-1)]."<br>"; }I tried instead of "file("$filename.html");" "readfile("$filename.html");" But then I get the random lines plus the whole content. Is there anything else I can use instead of file so that I get the random lines of text without the html code?P.S file_get_contents doesn't work either have tried that one. 2nd Problem: As you could see in my first problem I have a file called $filename.html. After I submit the value of a password I want the whole content. But it is like the program did forget what $filename.html is. How can I make the program remember what $filename.html is? Or with other words how to get the whole content of the html file? My code: if($_POST['submitPasswordIT']){ if ($_POST['passIT']== $password ){ $my_file = file_get_contents("$filename.html"); echo $my_file; } else{ echo "You entered wrong password"; } }If the password isn't correct I get: You entered wrong password. If the password is correct I get nothing. I probably need to create a path to the file "$filename.html", but I don't know exactly how to do that. hey i'm just woundering how i would go abouts on echoing unread messages sent to a user i want to beable to see how many unread messages i have Hey All, I need to count and display the number of rows I have. <?php //declare the SQL statement that will query the database $query = " SELECT COUNT(id) FROM connectvisits WHERE staffid = '$staffid' "; //execute the SQL query and return records $result = mssql_query($query); $thismonth = mssql_num_rows($result); echo "Total Meetings This Month: "; echo $thismonth; echo "<br />"; ?> For some reason I keep getting a result of 1. Everything else I try gives me a "Resource ID" number. Any help would be apprecaited. Maybe I have the completely wrong code. Thanks! returns -30. how can i remove the minus so it just returns 30 Code: [Select] $days = (strtotime(date("Y-m-d")) - strtotime($info['expiredate'])) / (60 * 60 * 24) echo $days Hello, I'm using the following code to count the amount of SQL results and divide it. Works great as is but I want to alter it so it works with categories. $amount = mysql_query('SELECT COUNT(`car_id`) AS num FROM `tbl_cars`') or die(mysql_error()); $amount_row = mysql_fetch_assoc($amount); I tried adding... WHERE car_cat = '".$cat."' which causes an error. Why doesn't this work, and how can I get the amount of results as a variable? Hello,
I am trying to get the report of my sales table. I want to get total number of leads for each month. And then i have to use this data to create graph using google graph.
But i am not getting how eactly i can get this.
Here is my code
<?php if(isset($_POST['submit'])) { $type = $_POST['type']; $sql="select * from leads where lead_customer='".$type."'"; $query=mysql_query($sql); while ($row = mysql_fetch_array($query)) { list($year,$month,$day)=explode("-", $row['last_modified']); $l = $row['last_modified']; $count=mysql_num_rows($month); echo $count; $myurl[] = "['".$month."', ".$count."]"; } print_r($myurl); echo implode(",", $myurl); } ?>But for $count, it doesn't show any values. below is my database. Capture.PNG 18.16KB 0 downloads Please suggest me Hi, I want the user to be able to go to one page and hit a button, once he hits the butten I want a count down to start from 10 minutes. When the user goes to the page within those 10 minutes, I want it to show him a live countdown until those ten minutes run out. Any ideas how I would go about doing this? Hello guys. I have the following code and I would like to count the number of times that the word has been replaced. I tried echo count ($replacement) and echo count ($match) but both only show "1". I want to be able to tell if there has been a replacement and send a warning email and/or add a flag in the db. Code: [Select] $search_for_bad_words = mysql_query("SELECT * FROM badwords WHERE 1"); $ad_title2 = $ad_title; $ad_body2 = $ad_body; //$wordlist = "shit:cr*p|dang:d*ng|shoot:sh**t"; $seperate_text = "|"; $entry_seperate_text = ":"; while($row = mysql_fetch_array($search_for_bad_words)) { $wordlist = $wordlist.$seperate_text.$row[word].$entry_seperate_text.$row[r_word]; } $words = explode('|', $wordlist); foreach ($words as $word) { list($match, $replacement) = explode(':', $word); $ad_title2 = preg_replace("/([^a-z^A-Z]?)($match)([^a-z^A-Z]?)/i", "$1".$replacement."$3", $ad_title2); $ad_body2 = preg_replace("/([^a-z^A-Z]?)($match)([^a-z^A-Z]?)/i", "$1".$replacement."$3", $ad_body2); } Thanks for all your help! $count=0; $numb=50; foreach ($sepkeys as $dbkey) { for ($page=10;$page<=$numb;$page=$page + 10) { // the if block $count=$count+1; } } I am trying to maintain a separate a count for each key number in the above code. Eg: key- 574, it searches from pages 10-50 and increments the count by 1. The problem that I have is the count is continuous. After searching for the first key and moves on to the next key and them I need the count to start from the beginning rather than being continuous. Eg: key-874 : count = 22, in my case the next key 875 : the count is 23 I need to make it 1. I removed the if block and several lines because the code is too long. Can someone please suggest me a way how to do it I am printing question id and corect answers and incorrect one. Example: Questionid =8 Incorrect Incorrect Incorrect Correct Correct Incorrect This will be repeated for other question id. Now I would like to count total answers. In my example it will be 6. I would like to count the correct one so that will be 2. And print 2 as outcome. So far I have Code: [Select] $sqlll=questionid(); while($infoo = mysql_fetch_array( $sqlll)) { echo "<hr><br><strong>{$infoo['Que_ID']}</strong><br />\n"; $_Session1=$infoo['Que_ID']; $que=question($_Session1); while($infooo = mysql_fetch_array( $que)) { $answer1 = $infooo['Que_Answer1'] == $infooo['Ans_Answer1']; $answer2 = $infooo['Que_Answer2'] == $infooo['Ans_Answer2']; $answer3 = $infooo['Que_Answer3'] == $infooo['Ans_Answer3']; $answer4 = $infooo['Que_Answer4'] == $infooo['Ans_Answer4']; if ( $answer1 && $answer2 && $answer3 && $answer4) { echo ('<b><p style="color: green; text-align: left"> Correct </p></b>'); $scoree = $scoree + 1; } else { echo ('<br><b><p style="color: red; text-align: left"> Incorrect </p></b></br>'); $intt= $intt +1; } } |