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PHP - Displaying More Than One Found Result In A Query
Hello!
I am busy developing a script that returns all users that have the cell "Available" set to 1 but I am unsure how to do this, I am familiar with basic SQL and I have made this: $sql = "SELECT * FROM `clans` WHERE `available` =1 LIMIT 0 , 30"; if ($result = mysql_query($sql)) { if (mysql_num_rows($result)) { $row = mysql_fetch_assoc($result); echo ($row['name']); but it only displays information from the first row found, and I want it to display all the rows found, how can I do this? Thanks - GreenFanta Similar TutorialsHello - Suppose I had a MySQL result set that was something like this: ITEM: COLOR: ball red ball blue book red book green book black If I were to use the code: Code: [Select] do { echo $row_myQuery['item'] . " - " . echo $row_myQuery['color'] . "<br />"; } while ($row_myQuery= mysql_fetch_assoc($myQuery)); Then I end up with this: Code: [Select] ball - red ball - blue book - red book - green book - black MY QUESTION: What I'd like to know is, how could I re-write that php code so that I could get an output like this?: Code: [Select] ball - red, blue book - red, green, black Thanks! Hello, First time posting here, my MYSQL version is 5.1.52 and I have a created a search function to find a user record and if the name is typed correctly the name and details are displayed on the search page, but if the name is typed incorrectly nothing is appearing. Basically I would like to be able to say if no record found and redirect back to the previous page. Code: Code: [Select] <?php add = $_POST['add']; // Database details $host=""; // Host name $username=""; // Mysql username $password=""; // Mysql password $db_name=""; // Database name $tbl_name="Rewards"; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot SELECT DB"); //Add query $sql = mysql_query("UPDATE Rewards SET Rewards = (Rewards + 10) WHERE Name '%$add%'"); // Database details $host=""; // Host name $username=""; // Mysql username $password=""; // Mysql password $db_name=""; // Database name $tbl_name="UserList"; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot SELECT DB"); //Update Total Points query $sql = mysql_query("UPDATE UserList SET TotalPoints = (TotalPoints + 10) WHERE Name '%$add%'"); // Database details $host=""; // Host name $username=""; // Mysql username $password=""; // Mysql password $db_name=""; // Database name $tbl_name="UserList"; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot SELECT DB"); //Search Query $sql = mysql_query("SELECT * FROM UserList WHERE Name '%$add%'"); while ($row = mysql_fetch_array($sql)){ echo '<br/><img src="images/line1.jpg" alt="" width="200" height="2" />'; echo '<br/> Username: '.$row['Username']; echo '<br/>'; echo '<br/><img src="images/line2.jpg" alt="" width="200" height="2" />'; echo '<br/> Name: '.$row['Name']; echo '<br/> Year Group: '.$row['YearGroup']; echo '<br/> Form Class: '.$row['FormClass']; echo '<br/>'; echo '<br/><img src="images/line3.jpg" alt="" width="200" height="2" />'; } // Database details $host=""; // Host name $username=""; // Mysql username $password=""; // Mysql password $db_name=""; // Database name $tbl_name="Rewards"; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot SELECT DB"); $sql = mysql_query("SELECT * FROM Rewards WHERE Name '%$add%'"); while ($row = mysql_fetch_array($sql)){ echo '<br/> Rewards: '.$row['Rewards']; } // Database details $host=""; // Host name $username=""; // Mysql username $password=""; // Mysql password $db_name=""; // Database name $tbl_name="Sanctions"; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot SELECT DB"); $sql = mysql_query("SELECT * FROM Sanctions WHERE Name '%$add%'"); while ($row = mysql_fetch_array($sql)){ echo '<br/> Sanctions: '.$row['Sanctions']; } // Database details $host=""; // Host name $username=""; // Mysql username $password=""; // Mysql password $db_name=""; // Database name $tbl_name="UserList"; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot SELECT DB"); $sql = mysql_query("SELECT * FROM UserList WHERE Name '%$add%'"); while ($row = mysql_fetch_array($sql)){ echo '<br/> Total Points: '.$row['TotalPoints']; echo '<br/>'; echo '<br/><img src="images/line4.jpg" alt="" width="200" height="2" />'; } Any help would be greatly appreciated. Many Thanks P Quinn MOD EDIT: [code] . . . [/code] tags added. i have it set up so the user picks the year and session and then the database shows all the games in that specific year and session and then displays them with a link to their photo galleria and also shows the record(wins - losses - ties) but the way i have it know its not displaying the first result here is the code Code: [Select] <?php //declares $y as year played and $s as session $y = ' '; $s = ' '; //handles submition of form for picking year and session if (isset($_POST['submit'])) { $data = mysql_real_escape_string($_POST['session']); $exploeded = explode(",", $data); $y = $exploeded[0]; $s = $exploeded[1]; } //Query for the selection menu $squery = "SELECT DISTINCT(Year_Played), Sessions FROM pinkpanther_games"; $sresults = mysql_query($squery) or die("squery failed ($squery) - " . mysql_error()); //checks $y and $s to see if their is a vaule if ($y == ' '){$y = '20112012';} if ($s == ' '){$s = '2';} //Query for getting record and for getting all the appropriate info for displaying the games $query = " SELECT (SELECT COUNT(id) FROM pinkpanther_games WHERE Year_Played = $y AND Sessions = $s AND Win_Loss = 'Tie' ) AS 'Tie', (SELECT COUNT(id) FROM pinkpanther_games WHERE Year_Played = $y AND Sessions = $s AND Win_Loss = 'Win') AS 'Win', (SELECT COUNT(id) FROM pinkpanther_games WHERE Year_Played = $y AND Sessions = $s AND Win_Loss = 'Loss') AS 'Lose', Opponent AS Opponent, Score AS Score, Gallery_no AS Gal, Win_Loss AS Record FROM pinkpanther_games WHERE Year_Played = $y AND Sessions = $s"; $results = mysql_query($query) or die("Query failed ($query) - " . mysql_error()); $row_a = mysql_fetch_assoc($results); //Set record variables $wins = $row_a['Win']; $loss = $row_a['Lose']; $tie = $row_a['Tie']; //creates from and selection menu for year and session echo "<tr><td colspan='2'><form method='post' action=''><select name='session' class='txtbox2'>"; while ($srow = mysql_fetch_assoc($sresults)){ echo "<option value='" . $srow['Year_Played'] . "," . $srow['Sessions'] . "'>" . $srow['Year_Played'] . " Session " . $srow['Sessions'] . "</option>"; } echo "</select><input type='submit' name='submit' value='Go' class='txtbox2' /></form></td></tr>"; //displays record echo "<tr><td colspan='2'>" .$wins . " - " . $loss . " - " . $tie . "</td></tr>"; echo "<tr></tr>"; //displays games and scores in appropriate year and session while ($row = mysql_fetch_assoc($results)){ $opp = $row['Opponent']; $score = $row['Score']; $gal = $row['Gal']; $wlt = $row['Record']; //styles games acording to if win lose or tie if ($wlt == 'Win'){ echo "<tr><td class='win'>"; echo "<a href='galleries.php?g=$gal'" . $gal . " >" . $opp . "</a>"; echo "</td><td class='win'>"; echo $score . ""; echo "</td></tr>"; } if ($wlt == 'Loss'){ echo "<tr><td class='loss'>"; echo "<a href='galleries.php?g=$gal'" . $gal . " >" . $opp . "</a>"; echo "</td><td class='loss'>"; echo $score . ""; echo "</td></tr>"; } if ($wlt == 'Tie'){ echo "<tr><td class='tie'>"; echo "<a href='galleries.php?g=$gal'" . $gal . " >" . $opp . "</a>"; echo "</td><td class='tie'>"; echo $score . ""; echo "</td></tr>"; } } ?> i cannot figure out why my first result is not displaying properly it successfully shows all results but me first myurl.com/other.php?id=2 now showing any thing kindly help to solved this problem! it seems like problem is in variable $http$ids but dunno how to combine them to get result $Ids always b numeric number <?php ini_set ("display_errors", "1"); error_reporting(E_ALL); $http = "myurl.com/other.php?id=" ; $conn=odbc_connect('greeting','',''); if (!$conn) {exit("Connection Failed: " . $conn);} $sql="SELECT * FROM mytable"; $rs=odbc_exec($conn,$sql); if (!$rs) {exit("Error in SQL");} echo "<table border= 1><tr>"; echo "<th>ID</th>"; echo "<th>Like</th>"; echo "<th>title</th></tr>"; while (odbc_fetch_row($rs)) { $ids=odbc_result($rs,"ID"); $title=odbc_result($rs,"title"); echo "<tr><td>$ids</td>"; echo "<td><fb:like href=\"$http$ids\" layout=\"button_count\" show_faces=\"false\" width=\"100\" font=\"tahoma\" colorscheme=\"dark\"></fb:like> </td>"; echo "<td>$title</td></tr>"; } echo "</table>"; odbc_close($conn); ?> Hi! I want the image of my site to be clickable so that it shows the next image contained in the result set of the sql query. Code: [Select] <?php require_once('includes/connection.inc.php'); $conn = dbConnect(); $sql = 'SELECT filename FROM images ORDER BY image_id DESC LIMIT 1'; $msg = ''; if(!$sql) { echo "Something went wrong with the query. Please try again."; } $result = $conn->query($sql) or die(mysqli_error()); if(mysqli_num_rows($result) == 0) { header('Location: empty_blog.php'); $conn->close(); } else { $row = $result->fetch_row(); $image = $row[0]; } ?> <?php include('header.php'); ?> <div class="content main"> <img id="image" src="images/<?php echo $image; ?>"/> </div> <?php include('includes/footer.inc.php'); ?> All the filenames of the images are contained correctly in the $result variable, all I need to do now is to fetch the next image in the set. How would I go about this? Any help is greatly appreciated! Hi, I am new to php but not to programming. I created a script on a windows platform which connects to the mysql database and returns the results of a table. A very basic script which I wrote to simply test my connection worked. The script works fine on my windows machine but not on my new mac. On the mac it simply does not display any records at all. I know that the database connection has been established because there is no error but I can not see why the result set is not being displayed on screen, as I said it worked fine on my windows machine. The Mac has mysql (with data) and apache running for php. Please could someone help as I have no idea what to do now? Script below: Code: [Select] $dbhost = 'localhost'; $dbuser = 'root'; $dbpass = 'root'; $conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); $dbname = 'test'; mysql_select_db($dbname); mysql_select_db("test", $conn); $result = mysql_query("SELECT * FROM new_table"); while($row = mysql_fetch_array($result)) { echo $row['test1'] . " " . $row['test2'] . " " . $row['test3']; echo "<br />"; } mysql_close($con); Hi I am building a product search engine using MySql and PHP (Jquery / AJAX). As i am getting the search results i wondered is it possible to count the instances of a category? For example: If i search for "Dell Laptops" i get 400+ results back because it has lots of laptops and accessories. What i am aiming for it to do is while looping through look at the categories and if its the first instanc eof it i want it to begin a count. until it has finished looping and then i can display it. Results: 436 [ Laptops(127) - Accesories (244) - Components(65) ] Im not even sure where to start here so any advice will be helpful Here is the code im using to generate the search: $id = str_replace(" ", "%", "$id"); $q = "SELECT * FROM products WHERE prod_name LIKE '%$id%' order by fee asc LIMIT 20"; This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=330221.0 Hello all,
Am trying to calculate some data and display the result of the current month in a chart form. i just don't know how to get the current month. this is not working..
<?php
include('mysql_connect.php');
$date = 'MONTH(CURRENT_DATE())';
$status = 'paid';
//mysql_select_db("hyprops", $con);
$query = mysql_query("SELECT sum(amount) 'amount', department FROM requisition WHERE status = '$status' AND date='$date' ");
$category = array();
$category['name'] = 'department';
//$series1 = array();
//$series1['name'] = 'actual';
$series2 = array();
$series2['name'] = 'amount';
//$series3 = array();
//$series3['name'] = 'Highcharts';
while($r = mysql_fetch_array($query)) {
$category['data'][] = $r['department'];
// $series1['data'][] = $r['actual'];
$series2['data'][] = $r['amount'];
// $series3['data'][] = $r['highcharts'];
}
$result = array();
array_push($result,$category);
//array_push($result,$series1);
array_push($result,$series2);
//array_push($result,$series3);
print json_encode($result, JSON_NUMERIC_CHECK);
mysql_close($con);
?>
Please how can i solve this issue?
Thanks in advance
Hello all. I am using a premade login script which has some very powerful function built in, the one I am looking at using is as follows:
/**
* query - Performs the given query on the database and
* returns the result, which may be false, true or a
* resource identifier.
*/
function query($query){
return mysqli_query( $this->connection, $query);
}
The issue is, displaying the results. I call is via the following code:
$database->query('SELECT LastName, Firstname FROM Patrons WHERE LastName = "Delude"');
But I have no idea how to display it. I have attempted to echo it out, but I get the following error in the logs:
PHP Catchable fatal error: Object of class mysqli_result could not be converted to string
Hi there
I am new to this forum and not much of a php expert. I hope you can assist me.
My problem is as follows:
I have a column consisting of several values (in this case: sceience, art, humanistics and music)
I would like to use the following mysqli query in order to count the appearances of each value in this column:
$query = "SELECT SUM (IF(department = 'science', 1, 0)) AS science, SUM(IF(department = 'art', 1, 0))-> AS art FROM new2"
I have tried lots of ways to loop the results but couldn't get the it working.
Can anyone suggest any idea?
Thanks
Hanan
hello I'm using this code: Code: [Select] $query="SELECT * FROM `second_content` WHERE CHANGED =0 limit 0,1";to query first row data when I want see it and echo it I recived : Code: [Select] Resource id #2can anyone help me ? thank you Hi everyone. I have this code that suppose to divide the query result 3 columns in a row. Like this : Shoes Hats Jeans Shorts Shirts Belts But for some reason this one doesn't work. It gave me everything in a single row. Like this: Shoes Hats jeans Shorts Shirts Belts. I noticed that before I put this two 2 lines of code, it's working fine. It didn't give me any error messages or anythings. Can someone point me out what did I did wrong? This is two lines I put in: Code: [Select] $vuong_query = tep_db_query("select categories_id from " . TABLE_GENDER_CATEGORIES . " where parent_id = 1"); while ($product_check = tep_db_fetch_array($vuong_query)) { This is the whole piece of code: Code: [Select] <tr> <td><table width = "100%" border="1" bordercolor="maroon"> <tr> <?php $vuong_query = tep_db_query("select categories_id from " . TABLE_GENDER_CATEGORIES . " where parent_id = 1"); while ($product_check = tep_db_fetch_array($vuong_query)) { $categories_querys = tep_db_query("select c.categories_id, cd.categories_name, c.categories_image from " . TABLE_CATEGORIES . " c, " . TABLE_CATEGORIES_DESCRIPTION . " cd where c.categories_id = '" . $product_check['categories_id'] . "' and c.categories_id = cd.categories_id and cd.language_id = '" . (int)$languages_id . "' order by sort_order, cd.categories_name"); $number_of_categories = tep_db_num_rows($categories_querys); $rows = 0; while ($categories = tep_db_fetch_array($categories_querys)) { $rows++; $cPath_new = tep_get_path($categories['categories_id']); $width = (int)(100 / MAX_DISPLAY_CATEGORIES_PER_ROW) . '%'; echo ' <td align="center" class="smallText" width="' . $width . '" valign="top"><a href="' . tep_href_link(FILENAME_DEFAULT, $cPath_new) . '">' . tep_image(DIR_WS_IMAGES . $categories['categories_image'], $categories['categories_name'], SUBCATEGORY_IMAGE_WIDTH, SUBCATEGORY_IMAGE_HEIGHT) . '<br>' . $categories['categories_name'] . '</a></td>' . "\n"; if ((($rows / MAX_DISPLAY_CATEGORIES_PER_ROW) == floor($rows / MAX_DISPLAY_CATEGORIES_PER_ROW)) && ($rows != $number_of_categories)) { echo ' </tr>' . "\n"; echo ' <tr>' . "\n"; } } } ?> </tr> </table> </td> </tr> Thanks anyone for helping me out. Vanvoquan Hi all !,
I am stuck on the following piece of code which does not give an error nor does it give a result. ( i.e. it gives 0 num_rows which should be > 1).
If, however, I execute the query in phpmyadmin by simply substituting the values of $pp,$ll and $room_no in the query it gives the correct result.
Please can someone tell me what I may be doing wrong here. Thanks !
$fcon = mysqli_connect($db_host,$db_user,$db_pass,$db_database) or die('Unable to establish a DB connection');
$pp = "(ms.level = 'Beginner' || ms.level = 'Intermediate')";
$ll = 'ms.diff <= 7';
$room_no = 4;
$query = "SELECT md.Member_reg_id, md.fname, md.lname, md.email, md.cell,
ms.level, ms.diff, ms.score,
r.ID_Status
FROM register as r
JOIN member_detail as md ON r.ID = md.Member_reg_id
JOIN memstatus as ms On r.ID = ms.ID
WHERE r.CENTERCODE = ? AND r.ID_Status ='A' AND ? AND ?
ORDER by level, diff, score DESC";
$stmt=$fcon->prepare($query);
$stmt->bind_param('iss',$room_no,$pp,$ll);
if(!$stmt->execute()) die('Failed to execute the query'.$fcon->error);
else
{
echo "Executed";
$stmt->bind_result($Member_reg_id,$fname,$lname,$email,$cell,$level,$diff ,$score,$ID_Status);
$numrows = $stmt->num_rows;
$stmt->store_result();
// echo $numrows;
while($stmt->fetch())
{
echo "<br>".$fname.' '.$lname;
echo "<br>".$level;
echo "<br>".$diff;
echo "<br>".$score;
echo "<br>".$cell;
echo "<br>".$email;
}
}
Edited by ajoo, 03 January 2015 - 08:00 AM. Hello, So I have a content management site set up, but I'm stuck on one part. I have a query. Code: [Select] Select `email` FROM author WHERE author.id IN(SELECT authorid FROM job WHERE id = '$_POST[id]') I'm a beginner in PHP and SQL. Basically I need the query to run, it will come out with one result and I need that to be where the email is sent to. This is the existing code I have for sending out the email $subject = " Job Application Confirmation"; $message = "Hello $_POST[name] , \r\rYou, or someone using your email address, has applied for the job: $text Using the Resource Locator at Prahan.com. Keep this as future reference. \r\r Full Name: $_POST[name] \r Location: $_POST[location] \r Email: $_POST[email] \r Additional Information: $_POST[info] \r Job ID: $id \r Job Name: $text \r \r Expect a response shortly. \r\r Regards, Prahan.com Team"; $headers = 'From: Prahan.com Team <noreply@prahan.com>'; 'Reply-To: noreply@prahan.com' . "\r\n" . 'X-Mailer: PHP/' . phpversion(); $to1 = 'pratik@prahan.com'; mail($to1, $result, $message, $headers); Thanks in Advance! I have this query to show the current class I'm in: Code: [Select] $GetTimeTable = "SELECT * FROM timetable WHERE username='$username' AND Subject_date='$today_date'"; Then I have this to display current class: Code: [Select] if($TimeNow >= $Start_time && $TimeNow <= $End_time){ echo "Current Class: "; echo $Subject;} How would I find next class that day? Hi, I have attached sql script for table structure and sample data. When I run following sql query ,I get result set. Code: [Select] select * from tablename order by col10 desc I want to create array using result set. Following is the require structure of array. Code: [Select] <?php $resultSet[]=array ( ('0') =>array ( ('col10value1') => array ( ('0') => array ( ('col1') => 'col1', ('col2') => 'col2', ('col3') => 'col3', ('col4') => 'col4', ('col5') => 'col5', ('col6') => 'col6', ('col7') => 'col7', ('col8') => 'col8', ), ('1') => array ( ('col1') => 'col1', ('col2') => 'col2', ('col3') => 'col3', ('col4') => 'col4', ('col5') => 'col5', ('col6') => 'col6', ('col7') => 'col7', ('col8') => 'col8', ), ), ), ('1') => array ( ('col10value2') => array ( ('0') => array ( ('col1') => 'col1', ('col2') => 'col2', ('col3') => 'col3', ('col4') => 'col4', ('col5') => 'col5', ('col6') => 'col6', ('col7') => 'col7', ('col8') => 'col8', ), ), ), ); ?> Hi all, Just a pretty basic question that I can't find the asnwer to: I have a MySQL query of the following form: Code: [Select] $result = mysql_query (" SELECT $cols FROM tablename WHERE......."); Where $cols is as follows: Code: [Select] $cols = firstname, lastname, $price1, $price2, etc. ($price1 and $price2 are variable column names that contains prices) How do I find the lowest value for $price1 and/or $price2 out of the results set of this query? Thanks! Hello, I am new to the board and PHP. I created a simple store locator app with PHP. It pulls the info from a database. I want to take the website field from the table, once it is returned from the query, and have it print out a a hyperlink. Please help. Thanks, I have posted most of the relevant code. <?php if (isset ($stores)) { if (!empty ($stores)) { echo '<p><strong>' . count ($stores) . ' results were found.</strong></p>'; foreach ($stores as $value) { echo '<p><strong>' . $value['name'] . '</strong><br />'; echo $value['address'] . '<br />'; echo $value['town'] . ', ' . $value['state'] . ' ' . $value['postal']; echo ' <a target="_blank" href="http://maps.google.com/maps?q=', $value['address'], ' ', $value['town'], ', ', $value['state'], ' ', $value['postal'],'">Map this location</a><br />'; echo 'Phone: ' . $value['phone'] . '<br />'; echo 'Website: ' . $value['website'] . '<br />'; echo 'Email: ' . $value['email'] . '<br />'; echo 'Hours: ' . $value['hours'] ; echo '</p>'; } } else { echo '<p><strong>No results found</strong></p>'; } } ?> Hello. This one is driving me up the wall. It used to work, I have a sports league database (MsSql) and one of the tables contains team Captain's first name, last name, etc.. To create a unique but simple login, the query searches for the captain's first initial of their first name + lastname + league. However since there is the remote possibility of two similar usernames (i.e. David Smith) if the last name is found, the query adds an integer to the end of the last name. So the result would look like this: DSmithM4A At the moment, my database is completely blank. However what is happening when I run the query below is I get: DSmith1M4A Can someone figure out why this code is not working correctly? If there is not more than one "Smith" no integer is supposed to be returned. There is something in the "Captain Count" that I think is not working correctly. Code: [Select] /* * Gather captian information */ $strCaptainFirstInitial = substr($arrPost['cpt_first'],0,1); $strCaptainLastName = preg_replace("/[^a-zA-Z]/","",$arrPost['cpt_last']); $strCaptainName = strtolower($strCaptainFirstInitial.$strCaptainLastName); $sqlCaptainSearch = sprintf ( $strCaptainSearch, $strCaptainName ); $resCaptainSearch = mssql_query($sqlCaptainSearch,$conDB); /* * Create unique captain login and Password */ $strCaptainCount = (mssql_num_rows($resCaptainSearch) > 0) ? mssql_num_rows($resCaptainSearch) : "" ; $strLeagueTypeInitial = substr($arrLeagueSearch['type'],0,1); $strCaptainLogin = $strCaptainName . $strCaptainCount . $strLeagueTypeInitial . $arrLeagueSearch['size'] . $arrLeagueSearch['division']; |
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