PHP - Count Rows In Phpmyadmin

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i need help .. with calculating how many days are between example:  january = 1  ; day =23 to feb =2; day 23
in php ... how can i do that

Hi Everyone,

I am working on implementing a blog comment system using the query below to display comments. My question is what is the best way to do a row count for the id column with this query or do I need to do a second db query to accomplish this? The purpose of this is to echo out the number of comments for that post, before displaying them.

Any help is appreciated.

Thanks in advance,

kaiman


<?php
// get url variables
$post_id = mysql_real_escape_string($_GET['id']);

include ("../../scripts/includes/nl2p.inc.php");

// connects to server and selects database
include ("../../scripts/includes/dbconnect.inc.php");

// table name
$tbl_name3="blog_comments";

// select info from comments database
$result3 = mysql_query ("SELECT count(*) FROM $table_name3 WHERE id='$post_id' ORDER BY id DESC LIMIT 1") 
or trigger_error("A mysql error has occurred!");
if (mysql_num_rows($result3) > 0 ) {
while($row = mysql_fetch_array($result3)) {
extract($row);

// display number of comments


// display date
$row_date = strtotime($row['date']);
putenv("TZ=America/Denver");
echo "<p class=\post\">On ".date('F, jS, Y', $row_date)." ";

// display commenter name
if($row['url'] == "") { 
echo $row['name']." wrote:</p>\n"; 
} 
     else { 
echo "<a href=\"".$row['url']."\" target=\"_blank\">".$row['name']."</a> wrote:</p>\n"; 
} 

// display content
$comments = $row['comment'];
echo nl2p($comments);
}
}
else {
echo "<p class=\"large_spacer\">No Comments</p>";
}
?>

 $result = mysql_query("SELECT *  FROM `friend_system` WHERE `friend_system_accepter` = '$myusername' ");
echo mysql_num_rows($result);

This displays the total rows in the table.

$result = mysql_query("SELECT COUNT(friend_system_accepter)  FROM `friend_system` WHERE `friend_system_accepter` = '$myusername' ");
echo mysql_num_rows($result);

This displays '1', which is incorrect.   I want to echo out the number of rows where 'friend_system_accepter' = $myusername   Thanks

Folks,

Having trouble here. Forgot how you use the COUNT. Not interested in the num_rows due to it being slow compared to COUNT.

I have users table like this:

id|username|sponsor_username

0|barand|requinix

1|phpsane|alison

2|mickey_mouse|requinix

Now, I want to check if a sponsor username exists or not. Such as does the entry "requinix" exists or not in one of the rows in the "sponsor_username" column.

In our example it does twice and so the variable $sponsor_username_count should hold the value "2". Else hold "0".

I get error:

Fatal error: Uncaught mysqli_sql_exception: You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near '?' at line 1 

 

My code:

	$stmt_1 = mysqli_prepare($conn,"SELECT COUNT(sponsor_username) from users = ?"); 
            mysqli_stmt_bind_param($stmt_1,'s',$sponsor_username_count); 
            mysqli_stmt_execute($stmt_1); 
            
            //Show error if 'users' tbl was not successfully queried". 
            if (!$stmt_1) 
            { 
                echo "ERROR 1: Sorry! Our system is currently experiencing a problem loading this page!"; 
                exit(); 
            } 
            else 
            { 
                mysqli_stmt_bind_result($stmt_1,$sponsor_username_count);     
                mysqli_stmt_fetch($stmt_1); 
                mysqli_stmt_close($stmt_1); 
                
                //Check if Sponsor Username in Url ($_GET) is already registered or not. If not then show "Invalid Url" or Link message. 
                if ($sponsor_username_count < 1) 
                { 
                    echo "<b>Invalid Url or Link!<br> Invalid Sponsor: \"$sponsor_username\"!</b><?php "; 
                    exit(); 
                } 
            } 
	

The following works, but I don't want to have to write 50 query statements for 50 states.  How can I just write one query (using a variable I presume) and then echo the variable count each time for each state?  I'm just trying to list the total number of rows for each state, that exist in the db. 

<?         
$result = mysql_query("SELECT * FROM mytable WHERE source = 'alabama'") 
or die(mysql_error());  
$num_rows = mysql_num_rows($result);

?>
 
href="/state/alabama/">alabama</a> (<?php echo("$num_rows");?>) <br>
 
 <?
$result = mysql_query("SELECT * FROM mytable WHERE source = 'alaska'") 
or die(mysql_error());  
$num_rows = mysql_num_rows($result);
?>
href="/state/alaska/">alaska</a>  (<?php echo("$num_rows");?>) <br>   

 
The above would give output such as:
alabama (122)
alaska (212)

I 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.

I insert data in mysql table row using multiple method : 1#2#3 . 1 is ID of country, 2 is Id of state, 3 is ID of town . now i have this table for real estate listings. for each list(home) i have country/state/town (1#2#3). in country table i have list of country - in country table i have list of state - in country table i have list of town. i need to The number of houses in country / state / town . my mean is :
Code: [Select]
USA [ 13 ] <!-- This Is equal of alabama+alaska+arizona -->
----Alabama [8] <!-- This Is equal of Adamsville+Addison+Akron -->
-------Adamsville [2]
-------Addison[5]
-------Akron[1]
......(list of other City)
----Alaska [ 3 ]
-------Avondale[3]
......(list of other City)
----Arizona [ 2 ]
-------College[2]
......(list of other City)
Lisintg Table :

Code: [Select]
ID -- NAME -- LOCATION -- DATEJOIN -- ACTIVE
 1 -- TEST -- 1#2#3    -- 20110101 -- 1
 2 -- TEST1 -- 1#2#3    -- 20110101 -- 1
 3 -- TEST2 -- 1#3#5    -- 20110101 -- 1
 4 -- TEST3 -- 1#7#6    -- 20110101 -- 1
Country Table :

Code: [Select]
id       -- name 
 1        -- USA 
stats Table :

Code: [Select]
id        -- countryid -- name 
 1        -- 1             -- albama
 2        -- 1             -- alaska
 3        -- 1             -- akron
town Table :
Code: [Select]
id        -- countryid -- statsid -- name 
 1        -- 1             -- 1         -- adamsville
 2        -- 1             -- 1         -- addison
 3        -- 1             -- 1         -- akron
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
email
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

what Im basically trying to do is just like a phpmyadmin function... you select rows you want to update with a checkbox and then it takes you to a page where the rows that are clicked are shown in forms so that you can view and edit info in them... and then have 1 submit button to update them all at once.

i have been working with a website but only locally. i am now trying to put it online which is ok. when i was working with it locally i have been able to create multiple users which have different privileges to each page. for example i said one of my users can only read and the other can read write.
the problem is when i have went to create these users online i can't seem to find the create user option in phpmyadmin.

i am missing the obvious or will i have to code them in and if so can you direct me to a tutorial which shows me how.

thanks in advance     

Hi guys

I have two questions:

Question 1: phpmyadmin
If my host does not provide phpmyadmin and I want to install it myself, can I just download phpmyadmin and copy it to my public_html folder or is there another way to install it on my site??

Question 2: Email Script
I have a script that sends an email to a system that sends out multiple sms's. The system works like this, I send an email to an address with an attachment called cellnumbers.txt with a list of all the cellnumbers. I must add a specific subject. When I send the email to my own email address, it comes correctly through. But when it is sent to the address of the server, no sms are received. I attached my script. Please take a look. Your help will be greatly appretiated.

Thank you