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  • PHP MYSQL query result "RANKING"

    - by fkessler
    Hi, I need to get a list of users Ranking by points and from my command line (MySQL) is was able to generate the necessary code: SET @rank=0; SELECT rank, iduser, pontos FROM ( SELECT @rank:=@rank+1 AS rank, SUM(points.points) AS pontos, points.iduser, users.name, users.idade FROM points INNER JOIN users ON (points.iduser = users.id) WHERE (users.idade >= %s) AND (users.idade <= %s) GROUP BY points.iduser ORDER BY pontos DESC) AS totals WHERE iduser = %s The problem is that I need this to run on AMFPHP and I´ve tested it in a test PHP file and seems that I can´t use the SET and SELECT in the same "mysql_query". I´ve looked and some used to mysql_query to do this (I´ve tested it and it works), but can I trust this to be effective and error free? Does it work like in MySQL transactions or setting the @rank in a seperated query may cause unexpected results?

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  • PHP Ajax not working

    - by Kostis
    I have 3 buttons on my page and depending on which one the user is clickingi want to run through ajax call a delete query in my database. When the user clicks on a button the javascript function seems to work but it doesn't run the query in php script. The html page: <?php session_start(); ?> <!DOCTYPE HTML> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-7"> <script> function myFunction(name) { var r=confirm("Are you sure? This action cannot be undone!"); if (r==true) { alert(name); // check if is getting in if statement and confirm the parameter's value var xmlhttp; if (str.length==0) { document.getElementById("clearMessage").innerHTML=""; return; } if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); } else {// code for IE6, IE5 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById("clearMessage").innerHTML= responseText; } } xmlhttp.open("GET","clearDatabase.php?q="+name,true); xmlhttp.send(); } else alert('pff'); } </script> </head> <body> <div id="wrapper"> <div id="header"></div> <div id="main"> <?php if (session_is_registered("username")){ ?> <!--<a href="#">???a????s? pa?a??? µ???µ?t??</a><br /> <a href="#">???a????s? pa?a??? s??ed????</a><br /> <a href="#">???a????s? push notifications</a><br />--> <input type="button" value="???a????s? pa?a??? µ???µ?t??" onclick="myFunction('messages')" /> <input type="button" value="???a????s? pa?a??? s??ed????" onclick="myFunction('conferences')" /> <input type="button" value="???a????s? push notifications" onclick="myFunction('notifications')" /> <div id="clearMessage"></div> <?php } else echo "Login first."; ?> </div> <div id="footer"></div> </div> </body> </html> and the php script: <?php if (isset($_GET["q"])) $q=$_GET["q"]; $host = "localhost"; $database = "dbname"; $user = "dbuser"; $pass = "dbpass"; $con = mysql_connect($host,$user,$pass) or die(mysql_error()); mysql_select_db($database,$con) or die(mysql_error()); if ($q=="messages") $query = "DELETE FROM push_message WHERE time_sent IS NOT NULL"; else if ($q=="conferences") $query = "DELETE FROM push_message WHERE time_sent IS NOT NULL"; else if ($q=="notifications") { $query = "DELETE FROM push_friend WHERE time_sent IS NOT NULL"; } $res = mysql_query($query,$con) or die(mysql_error()); if ($res) echo "success"; else echo "failed"; mysql_close($con); ?>

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  • output all data from db to a php page

    - by Sarit
    Hi, I'm a real beginner with PHP & mysql. Just for studying and for a simple example at school I would like to work this simple query and possibly output all the rows (or maybe even one) to a very basic output on a php page: <?php $user= "root"; $host="localhost"; $password=""; $database = "PetCatalog"; $cxn = mysqli_connect($host,$user,$password,$database) or die ("couldn’t connect to server"); $query = "SELECT * FROM Pet"; $result = mysql_query($cxn,$query) or die ("Couldn’t execute query."); $row = mysqli_fetch_assoc($result); echo "$result"; ?> this is the error message i've been getting: Warning: mysql_query() expects parameter 1 to be string, object given in C:\xampplite\htdocs\myblog\query.php on line 18 Couldn’t execute query. What should I do? Thanks!

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  • Default datetime format in mySQL

    - by Davuz
    I'm coding to search data by compare datetime. In web form, user input value '2012-06-04' store in $_REQUEST['dateSearch']. I create datetime object from that: $dateObj = date_create_from_format("Y-m-d",$_REQUEST['dateSearch']); $dateParam = $dateObj->format("Y-m"); Then I use $dateParam in sql query to find data $sql = " SELECT * FROM `temp` as `t` WHERE `t`.`date` LIKE '$dateParam%' "; Everything is ok, but I'm not sure mySQL alway use only one datetime format "Y-m-d H:i:s". At here, $dateParam = $dateObj->format("Y-m"); I set default format is "Y-m". Is the default format never change? I don't want hard set format "Y-m-d H:i:s" in code, instead of, I think get format string from system to use is better. How do I get default datetime format string from mySQL???

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  • User welcome message in php

    - by user225269
    How do I create a user welcome message in php. So that the user who has been logged on will be able to see his username. I have this code, but it doesn't seem to work. <?php $con = mysql_connect("localhost","root","nitoryolai123$%^"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("school", $con); $result = mysql_query("SELECT * FROM users WHERE Username='$username'"); while($row = mysql_fetch_array($result)) { echo $row['Username']; echo "<br />"; } ?> I'm trying to make use of the data that is inputted in this login form: <form name="form1" method="post" action="verifylogin.php"> <td> <table border="0" cellpadding="3" cellspacing="1" bgcolor=""> <tr> <td colspan="16" height="25" style="background:#5C915C; color:white; border:white 1px solid; text-align: left"><strong><font size="2">Login User</strong></td> </tr> <tr> <td width="30" height="35"><font size="2">Username:</td> <td width="30"><input name="myusername" type="text" id="idnum" maxlength="5"></td> </tr> <tr> <td width="30" height="35" ><font size="2">Password:</td> <td width="30"><input name="mypassword" type="password" id="lname" maxlength="15"></td> </tr> <td align="right" width="30"><td align="right" width="30"><input type="submit" name="Submit" value="Submit" /></td> <td align="right" width="30"><input type="reset" name="Reset" value="Reset"></td></td> </tr> </form> But this, verifylogin.php, seems to be in the way. <?php $host="localhost"; $username="root"; $password="nitoryolai123$%^"; $db_name="school"; $tbl_name="users"; mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $myusername=$_POST['myusername']; $mypassword=$_POST['mypassword']; $myusername = stripslashes($myusername); $mypassword = stripslashes($mypassword); $myusername = mysql_real_escape_string($myusername); $mypassword = mysql_real_escape_string($mypassword); $sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$mypassword'"; $result=mysql_query($sql); $count=mysql_num_rows($result); if($count==1){ session_register("myusername"); session_register("mypassword"); header("location:userpage.php"); } else { echo "Wrong Username or Password"; } ?> How do I do it? I always get this error when I run it: Notice: Undefined variable: username in C:\wamp\www\exp\userpage.php on line 53 Can you recommend of an easier on how I can achieve the same thing?

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  • MySQL how to display data from two tables.

    - by gew
    I'm trying to display the username of the person who has submitted the most articles but I don't know how to do it using MySQL & PHP, can someone help me? Here is the MySQL code. CREATE TABLE users ( user_id INT UNSIGNED NOT NULL AUTO_INCREMENT, username VARCHAR(255) DEFAULT NULL, pass CHAR(40) NOT NULL, PRIMARY KEY (user_id) ); CREATE TABLE users_articles ( id INT UNSIGNED NOT NULL AUTO_INCREMENT, user_id INT UNSIGNED mNOT NULL, title TEXT NOT NULL, acontent LONGTEXT NOT NULL, PRIMARY KEY (id) ); Here is the code I have so far. $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT COUNT(*) as coun, user_id FROM users_articles GROUP BY user_id ORDER BY coun DESC LIMIT 1");

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  • PHP: Simulate multiple MySQL connections to same database

    - by Varun
    An insert query is constantly getting logged in my MySQL slow query log. I want to see how much time the INSERT query is taking with 100 simultaneous insert operations(to the same table). So I need to simulate the follwoing. 500 different simultaneous connections from PHP to the same database on a mysql server, all of which are inserting a row(simultaneously) to the same table. DO I need to use any load testing tool? Or Can I simply write a PHP script to do this? Any thoughts?

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  • MySQL indexes - what are the best practises?

    - by Haroldo
    I've been using indexes on my mysql databases for a while now but never properly learnt about them. Generally I put an index on any fields that i will be searching or selecting using a WHERE clause but sometimes it doesn't seem so black and white. What are the best practises for mysql indexes? example situations/dilemas: If a table has six columns and all of them are searchable, should i index all of them or none of them? . What are the negetive performance impacts of indexing? . If i have a VARCHAR 2500 column which is searchable from parts of my site, should i index it?

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  • undefined offset error in php

    - by user225269
    I don't know why but the code below is working when I have a different query: $result = mysql_query("SELECT * FROM student WHERE IDNO='".$_GET['id']."'") ?> <?php while ( $row = mysql_fetch_array($result) ) { ?> <?php list($year,$month,$day)=explode("-", $row['BIRTHDAY']); ?> <tr> <td width="30" height="35"><font size="2">Month:</td> <td width="30"><input name="mm" type="text" id="mm" onkeypress="return handleEnter(this, event)" value="<?php echo $month;?>"> <td width="30" height="35"><font size="2">Day:</td> <td width="30"><input name="dd" type="text" id="dd" maxlength="25" onkeypress="return handleEnter(this, event)" value="<?php echo $day;?>"> <td width="30" height="35"><font size="2">Year:</td> <td width="30"><input name="yyyy" type="text" id="yyyy" maxlength="25" onkeypress="return handleEnter(this, event)" value="<?php echo $year;?>"> And it works when this is my query: $idnum = mysql_real_escape_string($_POST['idnum']); mysql_select_db("school", $con); $result = mysql_query("SELECT * FROM student WHERE IDNO='$idnum'"); Please help, why do I get the undefined offset error when I use this query: $result = mysql_query("SELECT * FROM student WHERE IDNO='".$_GET['id']."'") I assume that the query is the problem because its the only thing that's different between the two.

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  • 3 small PHP errors I cannot decipher

    - by Dave
    *Notice: Use of undefined constant _ - assumed '_' in /.../uploader.php on line 45* Line 45 $newname = str_replace(array(' ', '&'), array('_', 'and'), trim( strip_tags( $_POST['name'] ) ) ) . _ . $formKey->generateKey() . '_' . time() . '.jpg'; Notice: Undefined index: approve in /.../uploader.php on line 81 Line 81 - the second last line here $query = sprintf("INSERT INTO `$db_name`.`the_table` (`id` , `name` , `photo` , `email` , `date` , `code` , `subscribe` , `approve` , `created` ) VALUES ( NULL , '%s', '%s', '%s', '%s', '%s', '%s', '1', CURRENT_TIMESTAMP );", mysql_real_escape_string($_POST['name']), mysql_real_escape_string($newname), mysql_real_escape_string($_POST['email']), mysql_real_escape_string($date), mysql_real_escape_string($_POST['code']), mysql_real_escape_string($subscribe), mysql_real_escape_string($_POST['approve']) ); Warning: Cannot modify header information - headers already sent by (output started at /.../uploader.php:45) in/.../uploader.php on line 102 Line 45 $newname = str_replace(array(' ', '&'), array('_', 'and'), trim( strip_tags( $_POST['name'] ) ) ) . _ . $formKey->generateKey() . '_' . time() . '.jpg'; Line 102 - the third line here if ($success == 'Done') { $page = 'uploader'; header('Location: ./thanks.php'); } else { echo "error"; }

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  • PHP and MySQL SELECT problem.

    - by R.I.P.coalMINERS
    Trying to check if a name is already stored in the database from the login user. The name is a set of dynamic arrays entered by the user threw a set of dynamic form fields added by the user. Can some show me how to check and see if the name is already entered by the login user? I know my code can't be right. Thanks! MySQL code. SELECT * FROM names WHERE name = '" . $_POST['name'] . "' AND userID = '$userID' Here is the MySQL table. CREATE TABLE names ( id INT UNSIGNED NOT NULL AUTO_INCREMENT, userID INT NOT NULL, name VARCHAR(255) NOT NULL, meaning VARCHAR(255) NOT NULL, PRIMARY KEY (id) );

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  • How to insert form data into MySQL database table

    - by Richard
    So I have this registration script: The HTML: <form action="register.php" method="POST"> <label>Username:</label> <input type="text" name="username" /><br /> <label>Password:</label> <input type="text" name="password" /><br /> <label>Gender:</label> <select name="gender"> <optgroup label="genderset"> <option value="Male">Male</option> <option value="Female">Female</option> <option value="Hermaphrodite">Hermaphrodite</option> <option value="Not Sure!!!">Not Sure!!!</option> </optgroup> </select><br /> <input type="submit" value="Register" /> </form> The PHP/SQL: <?php $username = $_POST['username']; $password = $_POST['password']; $gender = $_POST['gender']; mysql_query("INSERT INTO registration_info (username, password, gender) VALUES ('$username', '$password', '$gender') ") ?> The problem is, the username and password gets inserted into the "registration_info" table just fine. But the Gender input from the select drop down menu doesn't. Can some one tell me how to fix this, thanks.

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  • Connection to mysql works through php but not through a form submission?

    - by Legend
    I have a download.php file that connects to a remote mysql database. If I run it using php download.php it works fine. But if I create another php file form.php and then submit the form to this download.php, it complains the following: Can't connect to MySQL server on '<IP_ADDRESS>' (13) Does anyone know why this might be happening? I can't see a reason why this works directly but fails to work upon form submission...

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  • Having trouble with php and ajax search function

    - by Andy
    I am still quite new to php/ajax/mysql. In anycase, I'm creating a search function which is returning properly the data I'm looking for. In brief, I have a mysql database set up. A php website that has a search function. I'm now trying to add a link to a mysql database search rather than just showing the results. In my search.php, the echo line is working fine but the $string .= is not returning anything. I'm just trying to get the same as the echo but with the link to the mysql php record. Am I missing something simple? //echo $query; $result = mysqli_query($link, $query); $string = ''; if($result) { if(mysqli_affected_rows($link)!=0) { while($row = mysqli_fetch_array($result,MYSQLI_ASSOC)) { echo '<p> <b>'.$row['title'].'</b> '.$row['post_ID'].'</p>'; $string .= "<p><a href='set-detail.php?recordID=".$row->post_ID."'>".$row->title."</a></p>"; } } else { echo 'No Results for :"'.$_GET['keyword'].'"'; }

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  • What is corret way to connect to MySQL?

    - by sky
    what is the correct way to connect to MySQL database without the mysql_fetch_assoc() error? Getting [Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource] with mysql_connect('localhost', 'name', 'pass'); mysql_select_db('dbname'); getting mysql_fetch_assoc() error without mysql_select_db any suggest? CODE are: $result= mysql_query('SELECT DISTINCT username FROM users'); $somethings= array(); while ($row= mysql_fetch_assoc($results)) { $somethings[]= $row['something']; } ? var somethings= ;

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  • Insert date and time into Mysql

    - by Jerry
    Hi..guys I am trying to insert date and time into mysql datetime field. When a user select a date and time, it will generate two POST variables. I have searched internet but still not sure how to do it. My code. //date value is 05/25/2010 //time value is 10:00 $date=$_POST['date']; $time=$_POST['time']; $datetime=$date.$time If I insert $datetime into mysql, the date appears to be 0000-00-00:00:00:00 I appreciate it if anyone could help me about this. Thanks.

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  • How to self-update PHP+MySQL CMS?

    - by SaltLake
    I'm writing a CMS on PHP+MySQL. I want it to be self-updatable (throw one click in admin panel). What are the best practices? How to compare current version of cms and a version of the update (application itself and database). Should it just download zip archive, upzip it and overwrite files? (but what to do with files that are no longer used). How to check if an update is downloaded correctly? Also it supports modules and I want this modules to be downloadable from the admin panel of cms. And how should I update MySQL tables?

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  • PHP & MySQL Pagination Update Help

    - by TaG
    Its been a while since I updated my pagination on my web page and I'm trying to add First and Last Links to my pagination as well as the ... when the search results are to long. For example I'm trying to achieve the following in the example below. Can some one help me fix my code so I can update my site. Thanks Previous First 1 2 3 4 5 6 7 ... 199 200 Last Next I currently have the following displayed using my code. Previous 1 2 3 4 5 6 7 Next Here is the part of my pagination code that displays the links. if ($pages > 1) { echo '<br /><p>'; $current_page = ($start/$display) + 1; if ($current_page != 1) { echo '<a href="index.php?s=' . ($start - $display) . '&p=' . $pages . '">Previous</a> '; } for ($i = 1; $i <= $pages; $i++) { if ($i != $current_page) { echo '<a href="index.php?s=' . (($display * ($i - 1))) . '&p=' . $pages . '">' . $i . '</a> '; } else { echo '<span>' . $i . '</span> '; } } if ($current_page != $pages) { echo '<a href="index.php?s=' . ($start + $display) . '&p=' . $pages . '">Next</a>'; } echo '</p>'; }

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  • Pass PHP variables without being seen when working with a database generated list

    - by Wilcoholic
    Looking for any help regarding the problem. Here's the deal: I have a database that has a teams table and it contains team_id. On one of my pages, I generate a list of links that contain the team_id of the creator in their get URL. I need the team_id on the next page but can't figure out how to pass it through any other way. Using a form and POST isn't an option because this method would only pass through the last links data on the list. Storing in a session isn't an option either because there is no way to discretely pass the the variables I need to a function to set the session variables. I have tried and it can pretty easily be viewed from viewing the source code. So here's some sample code to see exactly what I'm dealing with. <? if(mysql_num_rows($result2)>0){ ?> <a class="fltrt btn btn-danger btn-small" onclick="test()" href="acceptmatch-exec.php?match_id=<?php echo $match_id; ?>&team_id=<?php echo $team_id;?>&action=cancel">Cancel Match</a> <?}else{?> <a class="fltrt btn btn-success btn-small" href="acceptmatch-exec.php?match_id=<?php echo $match_id; ?>&team_id=<?php echo $team_id;?>&action=accept">Accept Match</a> <?} ?> The code above is generated multiple times on a page via a while loop that was excluded. I want to pass the match_id and team_id variables without being seen anywhere. If I made this a form, it wouldn't pass the correct variables unless there is only one result at the time (not likely). I'm sure there has to be an easy method that is eluding me, so please share thoughts on how to solve this. I feel as though I am not explaining it well enough, but it's not really easy to explain. I basically want something that works like GET but acts like POST and can be hidden from people viewing source code or link locations. Thanks

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  • PHP Error Form - Leave Contents of Form on Redirect

    - by user1371500
    I have a simple login form in which if an error occurs such as wrong password, I need it to be able to remember the username which was entered. Would I Go about doing this PHP or Javascript as I am not allowed to use JQuery. My current PHP - (Not Including the HTML Form) <?php //MySQl Connection mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db("clubresults") or die(mysql_error()); //Initiates New Session - Cookie session_start(); // Start a new session // Get the data passed from the form $username = $_POST['username']; $password = md5($_POST['pass']); // Do some basic sanitizing $username = mysql_real_escape_string($username); $password = mysql_real_escape_string($password); //Performs SQL Query to retrieve Login Details from DB $sql = "select * from admin_passwords where username = '$username' and password = '$password'"; $result = mysql_query($sql) or die ( mysql_error() ); //Assigns a Variable Count to 0 $count = 0; //Exectues a loop to increment on Successful Login while ($line = mysql_fetch_assoc($result)) { $count++; } //If count is equal to 1 Redirect user to the Members Page and Set Cookie if ($count == 1) { $_SESSION['loggedIn'] = "true"; header("Location: members.php"); // This is wherever you want to redirect the user to } else { //Else Echo that login was a failure. die('Login Failed. <a href=login.php>Click Here to Try Again</a>'); } ?> Any help would be appreciated. Cheers

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  • Am I underestimating MySQL ?

    - by user281434
    Hi I'm about to implement a feature on my website that recommends content to users based on the content they already have in their library (a la Last.fm). A single table holds all the records of the content they have added, so a row might look something like: -------------------- | userid | content | -------------------- | 28 | a | -------------------- When I want to recommend some content for a user, I use a query to get all the user id's that have content a added in their library. Then, out of those user id's, I make another query that finds the next most common content among those users (fx. 'b'), and show that to the user. My problem is when I'm thinking about the big picture here. Say that eventually my site will hold something like 500.000 rows in the table, will this make the MySQL response very slow or am I underestimating MySQL here?

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  • MySQL comparisons between multiple rows

    - by Hurpe
    I have a MySQL table with the following columns: id(int), date (timestamp), starttime(varchar), endtime(varchar), ... I need to find time slots that are occupied by two or more rows. Here is an example table id| date |starttime|endtime | __|_____________________|_________|________| 1 | 2010-02-16 17:37:36 |14:35:00 |17:37:00| 2 | 2010-02-17 12:24:22 |12:13:00 |14:32:00| 3 | 2010-02-16 12:24:22 |15:00:00 |18:00:00| Rows 1 and 3 collide, and need to be corrected by the user. I need a query to identify such colliding rows - something that would give me the ID of all rows in the collision. When inserting data in the database I find collisions with this query: SELECT ID FROM LEDGER WHERE DATE(DATE) = DATE('$timestamp') AND ( STR_TO_DATE('$starttime','%H:%i:%s') BETWEEN STR_TO_DATE(STARTTIME,'%H:%i:%s') AND STR_TO_DATE(ENDTIME,'%H:%i:%s') OR STR_TO_DATE('$endtime','%H:%i:%s') BETWEEN STR_TO_DATE(STARTTIME,'%H:%i:%s') AND STR_TO_DATE(ENDTIME,'%H:%i:%s') ) AND FNAME = '$fname'"; Is there any way to accomplish this strictly using MySQL or do I have to use PHP to find the collisions?

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