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  • Can't Insert Data Into Tables Containing Auto Increment Primary Key Using PHP Prepared Statements

    - by Drew
    I know I have that my connection to the database works, and a test I did using no auto-increment id worked fine for me. The code below refuses to work and I can't find a a way around it. My table has 3 columns, ID (auto increment), name and value. What do I need to change in order to get this to work? Thanks in advance //create placeholder query $query = "INSERT INTO prepared_test (name, value) VALUES (?,?)"; //prepare the query $stmt = mysqli_prepare($connection, $query); $name = 'steve'; $value = 45; mysqli_bind_param($stmt, array(MYSQLI_BIND_STRING, MYSQLI_BIND_INT), $name, $value); mysqli_execute($stmt); if(mysqli_stmt_affected_rows($stmt) != 1) die("issues"); mysqli_stmt_close($stmt); $connection-close();

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  • mysqli_stmt_bind_param SQL Injection

    - by profitphp
    Is there still an injection risk when using prepared statements and mysqli_stmt_bind_param? For example: $malicious_input = 'bob"; drop table users'; mysqli_stmt_bind_param($stmt, 's', $malicious_input); Behind the scenes does mysqli_stmt_bind_param pass this query string to mysql: SET @username = "bob"; drop table users"; Or does it perform the SET command through the API, or use some type of protection to keep this from happening?

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  • PHP, MySQL prepared statements - can you use results of execute more than once by calling data_seek(

    - by Carvell Fenton
    Hello, I have a case where I want to use the results of a prepared statement more than once in a nested loop. The outer loop processes the results of another query, and the inner loop is the results of the prepared statement query. So the code would be something like this (just "pseudoish" to demonstrate the concept): // not showing the outer query, it is just a basic SELECT, not prepared statement // we'll call it $outer_query $obj_array = array(); // going to save objects in this $ids = array(18,19,20); // just example id numbers $query = "SELECT field1, field2 FROM table1 WHERE id=?"; $stmt = $db->prepare($query); foreach ($ids as $id) { $stmt->bind_param("i", $id); $stmt->execute(); $stmt->bind_result($var1, $var2); $stmt->store_result(); // I think I need this for data_seek while ($q1 = $outer_query->fetch_object()) { while ($stmt->fetch()) { if ($q1->field1 == $var1) { // looking for a match $obj = new stdClass(); $obj->var1 = $var1; $obj->var2 = $var2; $obj_array[] = $obj; $stmt->data_seek(0); // reset for outer loop break; // found match, so leave inner } } } } The problem I seem to be experiencing is that the values are not getting bound in the variables as I would expect after the first time I use fetch in the inner loop. Specifically, in one example I ran with 3 ids for the foreach, the first id was processed correctly, the second was processed incorrectly (matches were not found in the inner loop even though they existed), and then the third was processed correctly. Is there something wrong with the prepared statment function calls in the sequence I am doing above, or is this an invalid way to use the results of the prepared statement? Thanks.

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  • Syntax for "RETURNING" clause in Mysql PDO

    - by dmontain
    I'm trying to add a record, and at the same time return the id of that record added. I read it's possible to do it with a RETURNING clause. $stmt->prepare("INSERT INTO tablename (field1, field2) VALUES (:value1, :value2) RETURNING id"); but the insertion fails when I add RETURNING. There is an auto-incremented field called id in the table being added to. Can someone see anything wrong with my syntax? or maybe PDO does not support RETURNING?

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  • PHP not talking to MySQL

    - by KodeSeeker
    Im trying to set up a AMP installation on my windows 8 laptop. However, Im not being able to get PHP to talk to MySQL. When I start apache server, the following error log is generated PHP Warning: PHP Startup: Unable to load dynamic library 'ext\\php_mysql.dll' - %1 is not a valid Win32 application.\r\n in Unknown on line 0 PHP Warning: PHP Startup: Unable to load dynamic library 'ext\\php_mysqli.dll' - %1 is not a valid Win32 application.\r\n in Unknown on line .. Any idea why this could be happening?

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  • MySQL break out group clause from subquery

    - by Anton Gildebrand
    Here is my query SELECT COALESCE(js.name,'Lead saknas'), count(j.id) FROM jobs j LEFT JOIN job_sources js ON j.job_source=js.id LEFT JOIN (SELECT * FROM quotes GROUP BY job_id) q ON j.id=q.job_id GROUP BY j.job_source The problem is that it's allowed for each job to have more than one quote. Because of that i group the quotes by job_id. Now sure, this works. But i don't like the solution with a subquery. How can i break out the group clause from the subquery to the main query? I have tried to add q.job_id to the main group clause, both before and after the existing one but don't get the same results.

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  • Strange sql query result from Mysql and from PHP mysqli_query!

    - by qinHaiXiang
    this is the query command echo from php web page: SELECT DISTINCT FT.file_type_name AS type,FT.file_type_en AS tp,FT.file_type_id AS fti, MATCH(keywords) AGAINST ('words <2' IN BOOLEAN MODE ) AS score FROM movie AS M,file_type AS FT WHERE MATCH (keywords) AGAINST ('words <2' IN BOOLEAN MODE ) AND M.type_cn = FT.file_type_id HAVING score >=1 ORDER BY FT.file_type_order; I am running above query in MySQL tools HeidiSQL and got only tow row records which score are 1.66666 and 2. If I remove the HAVING clause I would got three row records with one's score less than 1. But the same query I get from PHP mysqli_query() were all the three records and the one which score less than 1 became 1. What is the problem. Any tips will be pleasure. Thank you very much!!

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  • Having a problem inserting into database

    - by neo skosana
    I have a stored procedure: CREATE PROCEDURE busi_reg(IN instruc VARCHAR(10), IN tble VARCHAR(20), IN busName VARCHAR(50), IN busCateg VARCHAR(100), IN busType VARCHAR(50), IN phne VARCHAR(20), IN addrs VARCHAR(200), IN cty VARCHAR(50), IN prvnce VARCHAR(50), IN pstCde VARCHAR(10), IN nm VARCHAR(20), IN surname VARCHAR(20), IN eml VARCHAR(50), IN pswd VARCHAR(20), IN srce VARCHAR(50), IN refr VARCHAR(50), IN sess VARCHAR(50)) BEGIN INSERT INTO b_register SET business_name = busName, business_category = busCateg, business_type = busType, phone = phne, address = addrs, city = cty, province = prvnce, postal_code = pstCde, firstname = nm, lastname = surname, email = eml, password = pswd, source = srce, ref_no = refr; END; This is my php script: $busName = $_POST['bus_name']; $busCateg = $_POST['bus_cat']; $busType = $_POST['bus_type']; $phne = $_POST['phone']; $addrs = $_POST['address']; $cty = $_POST['city']; $prvnce = $_POST['province']; $pstCde = $_POST['postCode']; $nm = $_POST['name']; $surname = $_POST['lname']; $eml = $_POST['email']; $srce = $_POST['source']; $ref = $_POST['ref_no']; $result2 = $db->query("CALL busi_reg('$instruc', '$tble', '$busName', '$busCateg', '$busType', '$phne', '$addrs', '$cty', '$prvnce', '$pstCde', '$nm', '$surname', '$eml', '$pswd', '$srce', '$refr', '')"); if($result) { echo "Data has been saved"; } else { printf("Error: %s\n",$db->error); } Now the error that I get: Commands out of sync; you can't run this command now

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  • How should I handle the case in which a username is already in use?

    - by idealmachine
    I'm a JavaScript programmer and new to PHP and MySQL (want to get into server-side coding). Because I'm trying to learn PHP by building a simple online game (more specifically, correspondence chess), I'm starting by implementing a simple user accounts system. Of course, user registration comes first. What are the best practices for: How I should handle the (likely) possibility that when a user tries to register, the username he has chosen is already in use, particularly when it comes to function return values?($result === true is rather ugly, and I'm not sure whether checking the MySQL error code is the best way to do it either) How to cleanly handle varying page titles?($gPageTitle = '...'; require_once 'bgsheader.php'; is also rather ugly) Anything else I'm doing wrong? In some ways, PHP is rather different from JavaScript... Here is a (rather large) excerpt of the code I have written so far. Note that this is a work in progress and is missing security checks that I will add as my next step. function addUser( $username, $password ) { global $gDB, $gPasswordSalt; $stmt = $gDB->prepare( 'INSERT INTO user(user_name, user_password, user_registration) VALUES(?, ?, NOW())' ); $stmt || trigger_error( 'Failed to prepare statement: ' . htmlspecialchars( $gDB->error ) ); $hashedPassword = hash_hmac( 'sha256', $password, $gPasswordSalt, true ); $stmt->bind_param( 'ss', $username, $hashedPassword ); if( $stmt->execute() ) { return true; } elseif( $stmt->errno == 1062) { return 'exists'; } else { trigger_error( 'Failed to execute statement: ' . htmlspecialchars( $stmt->error ) ); } } $username = $_REQUEST['username']; $password = $_REQUEST['password']; $result = addUser( $username, $password ); if( $result === true ) { $gPageTitle = 'Registration successful'; require_once 'bgsheader.php'; echo '<p>You have successfully registered as ' . htmlspecialchars( $username ) . ' on this site.</p>'; } elseif( $result == 'exists' ) { $gPageTitle = 'Username already taken'; require_once 'bgsheader.php'; echo '<p>Someone is already using the username you have chosen. Please try using another one instead.'; } else { trigger_error('This should never happen'); } require_once 'bgsfooter.php';

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  • Using array instead of lots of db queries in PHP

    - by Tural Teyyuboglu
    My function looks like that. It works but does lots of work (recursively calls itself and does lots of db queries.). There must be another way to do same thing but with array (with one query). I can't figure out how to modify this function to get it work with array. Please help. function genMenu($parent, $level, $menu, $utype) { global $db; $stmt=$db->prepare("select id, name FROM navigation WHERE parent = ? AND menu=? AND user_type=?") or die($db->error); $stmt->bind_param("iii", $parent, $menu, $utype) or die($stmt->error); $stmt->execute() or die($stmt->error); $stmt->store_result(); /* bind variables to prepared statement */ $stmt->bind_result($id, $name) or die($stmt->error); if ($level > 0 && $stmt->num_rows > 0) { echo "\n<ul>\n"; } while ($stmt->fetch()) { echo "<li>"; echo '<a href="?page=' . $id . '">' . $name . '</a>'; //display this level's children genMenu($id, $level+1, $menu, $utype); echo "</li>\n\n"; } if ($level > 0 && $stmt->num_rows > 0) { echo "</ul>\n"; } $stmt->close(); }

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  • How to SET tiggers 0 in MySQL?

    - by Grijesh Chauhan
    In my MySQL database I have some Triggers ON DELETE and ON INSERT. Some time I need to switched-off my some Triggers, And I have to DROP e.g. DROP TRIGGER IF EXISTS hostgroup_before_insert // and reinstall. Is there any shortcut to SET triggers hostgroup_before_insert = 0 like we have for foreign keys mysql> SELECT version(); +-------------------------+ | version() | +-------------------------+ | 5.1.61-0ubuntu0.10.10.1 | +-------------------------+ 1 row in set (0.00 sec) I am new learner and I could not find regarding this on web.

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  • Why does this php/ajax query fail?

    - by Ashley Brown
    I'm ajaxing over to this php file. $a = 'old'; $b = 'new'; if ($_POST['info-type'] == $a || $b) { $info = $_POST['info-type']; $query = "SELECT * FROM `tld` WHERE type = '".$var."'"; } $query = "SELECT * FROM `tld` "; $result = mysqli_query($link,$query); while($row = mysqli_fetch_assoc($result)) { echo '<div>'.$row['something'].'</div>'; } The data posted is either 'all' 'new' or 'old'. If i send the data as either new or old, the script works and outputs as expected. If the posted data is neither new or old but all instead, it fails and don't show any errors or respond anything back. (I've monitored via dev tools aswell) So, I tried this if ($_POST['info-type'] == $a || $b) { $info = $_POST['info-type']; $var = "SELECT * FROM `tld` WHERE type = '".$var."'"; } elseif ($_POST['info-type'] == 'all'){ $query = "SELECT * FROM `tld` "; } But the script still fails. If i fully remove the IF statements and use the query without the WHERE clause like it is after the elseif, it works?

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  • checking is username exists on two tables PHP PDO?

    - by PHPLOVER
    Me again. I have a users table and a users_banlist table. On my registration form i want to check all in one query whether the username someone entered on form exists in the users table and see if it also exists on the users_banlist table. I can do them on there own in individual queries but would rather do it all in one. Here is what i got, but even thou i enter a username that is taken it does not tell me its already taken. $stmt = $dbh->prepare(" SELECT users.user_login, users_banlist.user_banlist FROM users , users_banlist WHERE users.user_login = ? OR users_banlist.user_banlist = ?"); // checker if username exists in users table or users_banlist table $stmt->execute(array($username, $username)); if ( $stmt->rowCount() > 0 ) { $error[] = 'Username already taken'; } Basically i think it is something to do with the execute or rowCount(), could anyone tell me where i am going wrong ? being new to pdo im finding it a little confusing at the moment until i get my pdo book delivered to learn pdo. Thank you as always, phplover

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  • Someone please can see why the following prepared statment returns nothing?

    - by jartaud
    $stmt = mysqli_prepare($link,"SELECT *FROM ads INNER JOIN dept ON dept.id_dept = ads.in_dpt INNER JOIN members ON members.idMem = ads.from_Mem INNER JOIN sub_cat_ad ON id_sub_cat = ads.ads_in_Cat INNER JOIN cat_ad ON idCat_ad = sub_cat_ad.from_cat_ad WHERE ads_in_Cat = ? "); if(isset($_GET['fromSCat'])){ $fromSCat = mysqli_real_escape_string($link,$_GET['fromSCat']);} mysqli_stmt_bind_param($stmt,'i',$fromSCat); mysqli_stmt_execute($stmt); mysqli_stmt_fetch($stmt); $tot=mysqli_stmt_num_rows($stmt); //Ouput: 0

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  • saving mySql row checkpoint in table ?

    - by Keet
    hello, I am having a wee problem, and I am sure there is a more convenient/simpler way to achieve the solution, but all searches are throw in up a blanks at the moment ! I have a mysql db that is regularly updated by php page [ via a cron job ] this adds or deletes entries as appropriate. My issue is that I also need to check if any details [ie the phone number or similar] for the entry have changed, but doing this at every call is not possible [ not only does is seem to me to be overkill, but I am restricted by a 3rd party api call limit] Plus this is not critical info. So I was thinking it might be best to just check one entry per page call, and iterate through the rows/entires with each successive page call. What would be the best way of doing this, ie keeping track of which entry/row in the table that the should be checked next? I have 2 ideas of how to implement this: 1 ) The id of current row could be save to a file on the server [ surely not the best way] 2) an extra boolean field [check] is add to the table, set to True on the first entry and false to all other. Then on each page call it; finds 'where check = TRUE' runs the update check on this row, 'set check = FALSE' 'set [the next row] check = TRUE' Si this the best way to do this, or does anyone have any better sugestion ? thanks in advance ! .k PS sorry about the title

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  • Need help fetching an array using prepared statments

    - by eldan221
    I have wrote the following code to fetch a string. But for some reason its only returning 1. I have doubled checked everything and it seems like its correct. I am not sure why its only returning 1? Any help here would be really appreciated! //Class Defined here function MenuCat($id){ $query = "SELECT menu_category_description FROM menu_categories WHERE id = ?"; $stmt = $this->db->prepare($query); $stmt->bind_param("i", $id); $stmt->execute(); $stmt->bind_result($menu_category_description); $row = $stmt->fetch(); return $row; } $display_category = $cat_des->MenuCat($id); echo $display_category

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  • laravel multiple where clauses within a loop

    - by user1424508
    Pretty much I want the query to select all records of users that are 25 years old AND are either between 150-170cm OR 190-200cm. I have this query written down below. However the problem is it keeps getting 25 year olds OR people who are 190-200cm instead of 25 year olds that are 150-170 OR 25 year olds that 190-200cm tall. How can I fix this? thanks $heightarray=array(array(150,170),array(190,200)); $user->where('age',25); for($i=0;$i<count($heightarray);i++){ if($i==0){ $user->whereBetween('height',$heightarray[$i]) }else{ $user->orWhereBetween('height',$heightarray[$i]) } } $user->get(); Edit: I tried advanced wheres (http://laravel.com/docs/queries#advanced-wheres) and it doesn't work for me as I cannot pass the $heightarray parameter into the closure. from laravel documentation DB::table('users') ->where('name', '=', 'John') ->orWhere(function($query) { $query->where('votes', '>', 100) ->where('title', '<>', 'Admin'); }) ->get();

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  • PHP, MySQL - would results-array shuffle be quicker than "select... order by rand()"?

    - by sombe
    I've been reading a lot about the disadvantages of using "order by rand" so I don't need update on that. I was thinking, since I only need a limited amount of rows retrieved from the db to be randomized, maybe I should do: $r = $db->query("select * from table limit 500"); for($i;$i<500;$i++) $arr[$i]=mysqli_fetch_assoc($r); shuffle($arr); (i know this only randomizes the 500 first rows, be it). would that be faster than $r = $db->("select * from table order by rand() limit 500"); let me just mention, say the db tables were packed with more than...10,000 rows. why don't you do it yourself?!? - well, i have, but i'm looking for your experienced opinion. thanks!

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  • Warning: date() expects parameter 2 to be long, string given in

    - by Simon
    its the $birthDay = date("d", $alder); $birthYear = date("Y", $alder); i dont know what it is here is my code //Dag $maxDays = 31; $birthDay = date("d", $alder); echo '<select name="day">'; echo '<option value="">Dag</option>'; for($i=1; $i<=$maxDays; $i++) { echo '<option '; if($birthDay == $i){ echo 'selected="selected"'; } echo ' value="'.$i.'">'.$i.'</option>'; } echo '</select>'; //Måned echo '<select name="month">'; $birthMonth = date("m", $alder); $aManeder = 12; echo '<option value="">Måned</option>'; for($i = 1; $i <= $aManeder; $i++) { echo '<option '; if($birthMonth == $i) { echo 'selected="selected"'; } echo ' value="'.$i.'">'.$ManderArray[$i].'</option>'; } echo '</select>'; //År $startYear = date("Y"); $endYear = $startYear - 30; $birthYear = date("Y", $alder); echo '<select name="year">'; echo '<option value="">år</option>'; while($endYear <= $startYear) { echo '<option '; if($birthYear == $endYear) { echo 'selected="selected"'; } echo ' value="'.$endYear.'">'.$endYear.'</option>'; $endYear++; } echo '</select>';

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  • PHP function to handle most database queries has a problem with results. I am getting the right numb

    - by asdasds
    Here is my little function. It does not handle the results correctly. I do get all the rows that I want, but all the rows of the $results array contain the exact same values. So i make 2 arrays, a temporary array to hold the values after each fetch, and another array to hold all the temporary arrays. First i take the temp array and map its keys to the column names. Then i give it to bind_result, and call fetch() and use it like I would any other result value. Could this be because I re-use the $results array? numresults is the number of values you are taking from each row. if 0, you are not getting any results back. function db_query($db, $query, $params = NULL, $numresults = 0) { if($stmt = $db -> prepare($query)) { if($params != NULL) { call_user_func_array(array($stmt, 'bind_param'), $params); } if(!$stmt -> execute()) { //echo 'exec error:',$db->error; return false; } if($numresults > 0) { $results = array(); $tmpresult = array(); $meta = $stmt->result_metadata(); while ($columnName = $meta->fetch_field()) $tmpresult[] = &$results[$columnName->name]; call_user_func_array(array($stmt, 'bind_result'), $tmpresult); $meta->close(); $results = array(); while($stmt -> fetch()) $results[] = $tmpresult; } $stmt -> close(); } else { //echo 'prepare error: ',$db->error; return false; } if($numresults == 0) return true; return $results; }

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  • Query to bring count from comma seperated Value

    - by Mugil
    I have Two Tables One for Storing Products and Other for Storing Orders List. CREATE TABLE ProductsList(ProductId INT NOT NULL PRIMARY KEY, ProductName VARCHAR(50)) INSERT INTO ProductsList(ProductId, ProductName) VALUES(1,'Product A'), (2,'Product B'), (3,'Product C'), (4,'Product D'), (5,'Product E'), (6,'Product F'), (7,'Product G'), (8,'Product H'), (9,'Product I'), (10,'Product J'); CREATE TABLE OrderList(OrderId INT NOT NULL PRIMARY KEY AUTO_INCREMENT, EmailId VARCHAR(50), CSVProductIds VARCHAR(50)) SELECT * FROM OrderList INSERT INTO OrderList(EmailId, CSVProductIds) VALUES('[email protected]', '2,4,1,5,7'), ('[email protected]', '5,7,4'), ('[email protected]', '2'), ('[email protected]', '8,9'), ('[email protected]', '4,5,9'), ('[email protected]', '1,2,3'), ('[email protected]', '9,10'), ('[email protected]', '1,5'); Output ItemName NoOfOrders Product A 4 Product B 3 Product C 1 Product D 3 Product E 4 Product F 0 Product G 2 Product H 1 Product I 2 Product J 1 The Order List Stores the ItemsId as Comma separated value for every customer who places order.Like this i am having more than 40k Records in my dB table Now I am assigned with a task of creating report in which I should display Items and No of People ordered Items as Shown Below I Used Query as below in my PHP to bring the Orders One By One and storing in array. SELECT COUNT(PL.EmailId) FROM OrderList PL WHERE CSVProductIds LIKE '2' OR CSVProductIds LIKE '%,2,%' OR CSVProductIds LIKE '%,2' OR CSVProductIds LIKE '2,%'; 1.Is it possible to get the same out put by using Single Query 2.Does using a like in mysql query slows down the dB when the table has more no of records i.e 40k rows

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  • MySQL error problem

    - by comma
    I keep getting the error listed below but it only says line 1 what does this mean and how do I fix it? Here is the error I keep getting? You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1 here is the code. if (isset($_POST['info_submitted'])) { $user_id = '5'; $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT learned_skills.*, users_skills.* FROM learned_skills INNER JOIN users_skills ON learned_skills.id = users_skills.skill_id WHERE user_id='$user_id'"); $skill = $_POST['skill']; $experience = $_POST['experience']; $year = $_POST['year']; if (mysqli_num_rows($dbc) == 0) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $query1 = mysqli_query($mysqli,"INSERT INTO learned_skills (skill, experience, year) VALUES ('" . $skill . "', '" . $experience . "', '" . $year . "')"); } if (!mysqli_query($mysqli, $query1)) { print mysqli_error($mysqli); return; } $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT id FROM learned_skills WHERE id='" . $skill . "' AND experience='" . $experience . "' AND year='" . $year . "'"); if (!$dbc) { print mysqli_error($mysqli); } else { while($row = mysqli_fetch_array($dbc)){ $id = $row["id"]; } } $query2 = "INSERT INTO users_skills (skill_id, user_id, date_created) VALUES ('$id', '$user_id', NOW())"; if (!mysqli_query($mysqli, $query2)) { print mysqli_error($mysqli); return; } if ($dbc == TRUE) { $dbc = mysqli_query($mysqli,"UPDATE learned_skills JOIN users_skills ON (users_skills.skill_id = learned_skills.id) SET skill = '$skill', experience = '$experience', year = '$year'"); echo '<p class="changes-saved">Your changes have been saved!</p>'; } if (!$dbc) { print mysqli_error($mysqli); return; } }

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  • PHP 5.3 Not Logging

    - by BHare
    I have set error_log = "/var/log/apache2/php_errors.log" and made sure errors were being logged. I have set the file to be owned by the www-data owner and group and even set the permissions to 777. I have confirmed with phpinfo() that the error_log is correctly set, however The logging still only happens in my vhost's apache error log. The following is my php.ini for 5.3.3-7 on Debian Squeeze Apache 2: The top is populated with comments on what I have been interested, or have changed. I have deleted all comments to save space. Full versions here: http://pastebin.com/AhWLiQBR [PHP] ;short_open_tag = On ;allow_call_time_pass_reference = On ;error_reporting = E_ALL & ~E_NOTICE & ~E_DEPRECATED ;display_errors = On ;display_startup_errors = Off ;log_errors = On ;html_errors = On error_log = "/var/log/apache2/php_errors.log" engine = On short_open_tag = On asp_tags = Off precision = 14 y2k_compliance = On output_buffering = 4096 zlib.output_compression = Off implicit_flush = Off unserialize_callback_func = serialize_precision = 100 allow_call_time_pass_reference = On safe_mode = Off safe_mode_gid = Off safe_mode_include_dir = safe_mode_exec_dir = safe_mode_allowed_env_vars = PHP_ safe_mode_protected_env_vars = LD_LIBRARY_PATH disable_functions = disable_classes = expose_php = On max_execution_time = 30 max_input_time = 60 memory_limit = 128M error_reporting = E_ALL & ~E_NOTICE & ~E_DEPRECATED display_errors = On display_startup_errors = Off log_errors = On log_errors_max_len = 1024 ignore_repeated_errors = Off ignore_repeated_source = Off report_memleaks = On track_errors = Off html_errors = On variables_order = "GPCS" request_order = "GPC" register_globals = Off register_long_arrays = Off register_argc_argv = Off auto_globals_jit = On post_max_size = 100M magic_quotes_gpc = Off magic_quotes_runtime = Off magic_quotes_sybase = Off auto_prepend_file = auto_append_file = default_mimetype = "text/html" doc_root = user_dir = enable_dl = Off file_uploads = On upload_tmp_dir = /tmp upload_max_filesize = 100M max_file_uploads = 20 allow_url_fopen = On allow_url_include = Off default_socket_timeout = 60 [Date] [filter] [iconv] [intl] [sqlite] [sqlite3] [Pcre] [Pdo] [Pdo_mysql] pdo_mysql.cache_size = 2000 pdo_mysql.default_socket= [Phar] [Syslog] define_syslog_variables = Off [mail function] SMTP = localhost smtp_port = 25 mail.add_x_header = On [SQL] sql.safe_mode = Off [ODBC] odbc.allow_persistent = On odbc.check_persistent = On odbc.max_persistent = -1 odbc.max_links = -1 odbc.defaultlrl = 4096 odbc.defaultbinmode = 1 [Interbase] ibase.allow_persistent = 1 ibase.max_persistent = -1 ibase.max_links = -1 ibase.timestampformat = "%Y-%m-%d %H:%M:%S" ibase.dateformat = "%Y-%m-%d" ibase.timeformat = "%H:%M:%S" [MySQL] mysql.allow_local_infile = On mysql.allow_persistent = On mysql.cache_size = 2000 mysql.max_persistent = -1 mysql.max_links = -1 mysql.default_port = mysql.default_socket = mysql.default_host = mysql.default_user = mysql.default_password = mysql.connect_timeout = 60 mysql.trace_mode = Off [MySQLi] mysqli.max_persistent = -1 mysqli.allow_persistent = On mysqli.max_links = -1 mysqli.cache_size = 2000 mysqli.default_port = 3306 mysqli.default_socket = mysqli.default_host = mysqli.default_user = mysqli.default_pw = mysqli.reconnect = Off [mysqlnd] mysqlnd.collect_statistics = On mysqlnd.collect_memory_statistics = Off [OCI8] [PostgresSQL] pgsql.allow_persistent = On pgsql.auto_reset_persistent = Off pgsql.max_persistent = -1 pgsql.max_links = -1 pgsql.ignore_notice = 0 pgsql.log_notice = 0 [Sybase-CT] sybct.allow_persistent = On sybct.max_persistent = -1 sybct.max_links = -1 sybct.min_server_severity = 10 sybct.min_client_severity = 10 [bcmath] bcmath.scale = 0 [browscap] [Session] session.save_handler = files session.use_cookies = 1 session.use_only_cookies = 1 session.name = PHPSESSID session.auto_start = 0 session.cookie_lifetime = 0 session.cookie_path = / session.cookie_domain = session.cookie_httponly = session.serialize_handler = php session.gc_probability = 0 session.gc_divisor = 1000 session.gc_maxlifetime = 1440 session.bug_compat_42 = Off session.bug_compat_warn = Off session.referer_check = session.entropy_length = 0 session.cache_limiter = nocache session.cache_expire = 180 session.use_trans_sid = 0 session.hash_function = 0 session.hash_bits_per_character = 5 url_rewriter.tags = "a=href,area=href,frame=src,input=src,form=fakeentry" [MSSQL] mssql.allow_persistent = On mssql.max_persistent = -1 mssql.max_links = -1 mssql.min_error_severity = 10 mssql.min_message_severity = 10 mssql.compatability_mode = Off mssql.secure_connection = Off [Assertion] [COM] [mbstring] [gd] [exif] [Tidy] tidy.clean_output = Off [soap] soap.wsdl_cache_enabled=1 soap.wsdl_cache_dir="/tmp" soap.wsdl_cache_ttl=86400 soap.wsdl_cache_limit = 5 [sysvshm] [ldap] ldap.max_links = -1 [mcrypt] [dba]

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  • PHP and MySQL problem?

    - by TaG
    When my code is stored and saved and out putted I keep getting these slashes \\\\\ in my output text. how can I get rid of these slashes? Here is the PHP code. if (isset($_POST['submitted'])) { // Handle the form. require_once '../../htmlpurifier/library/HTMLPurifier.auto.php'; $config = HTMLPurifier_Config::createDefault(); $config->set('Core.Encoding', 'UTF-8'); // replace with your encoding $config->set('HTML.Doctype', 'XHTML 1.0 Strict'); // replace with your doctype $purifier = new HTMLPurifier($config); $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT users.*, profile.* FROM users INNER JOIN contact_info ON contact_info.user_id = users.user_id WHERE users.user_id=3"); $about_me = mysqli_real_escape_string($mysqli, $purifier->purify($_POST['about_me'])); $interests = mysqli_real_escape_string($mysqli, $purifier->purify($_POST['interests'])); if (mysqli_num_rows($dbc) == 0) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"INSERT INTO profile (user_id, about_me, interests) VALUES ('$user_id', '$about_me', '$interests')"); } if ($dbc == TRUE) { $dbc = mysqli_query($mysqli,"UPDATE profile SET about_me = '$about_me', interests = '$interests' WHERE user_id = '$user_id'"); echo '<p class="changes-saved">Your changes have been saved!</p>'; } if (!$dbc) { // There was an error...do something about it here... print mysqli_error($mysqli); return; } } Here is the XHTML code. <form method="post" action="index.php"> <fieldset> <ul> <li><label for="about_me">About Me: </label> <textarea rows="8" cols="60" name="about_me" id="about_me"><?php if (isset($_POST['about_me'])) { echo mysqli_real_escape_string($mysqli, $_POST['about_me']); } else if(!empty($about_me)) { echo mysqli_real_escape_string($mysqli, $about_me); } ?></textarea></li> <li><label for="my-interests">My Interests: </label> <textarea rows="8" cols="60" name="interests" id="interests"><?php if (isset($_POST['interests'])) { echo mysqli_real_escape_string($mysqli, $_POST['interests']); } else if(!empty($interests)) { echo mysqli_real_escape_string($mysqli, $interests); } ?></textarea></li> <li><input type="submit" name="submit" value="Save Changes" class="save-button" /> <input type="hidden" name="submitted" value="true" /> <input type="submit" name="submit" value="Preview Changes" class="preview-changes-button" /></li> </ul> </fieldset> </form>

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