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PHP - Problem With Php+mysqli
hi
What is the correct way to do a function like: Code: [Select] public function check_($db, $skills) { $arr_tags = array('16', '17', '36', '546'); $z = implode(', ', array_fill(0, count($arr_tags), '?')); $str = implode('', array_fill(0, count($arr_tags),'s')); $par = "'" . implode("','", $arr_tags) . "'"; $c_arr_tags = count($arr_tag); $sql = $db -> prepare(" SELECT offer_id_offer FROM offer_has_tags WHERE tags_id_tags IN ($z) GROUP BY offer_id_offer HAVING COUNT(*) = ? "); $sql -> bind_param("$srt.'i'", $par, $c_arr_tags); $sql -> execute(); $sql -> bind_result($id_offer); return $id_offer; } At the moment i got: Number of elements in type definition string doesn't match number of bind variables Similar TutorialsHi I'm trying to insert unique info retrieved to my database but seems like I'm doing something wrong with my quary my current setup is as follow
mxit.php
<?php
$con=mysqli_connect("*****","*******","*******","******");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_close($con);
?>
<?
define('TIMEZONE', 'Africa/Harare');
date_default_timezone_set(TIMEZONE);
$ip = $_SERVER["REMOTE_ADDR"];
$post_time = date("U");
$mxitua = $_SERVER["HTTP_X_DEVICE_USER_AGENT"];
$mxitcont = $_SERVER["HTTP_X_MXIT_CONTACT"];
$mxituid = $_SERVER["HTTP_X_MXIT_USERID_R"];
$mxitid = $_SERVER["HTTP_X_MXIT_ID_R"];
$mxitlogin = $_SERVER["HTTP_X_MXIT_LOGIN"];
$mxitnick = $_SERVER["HTTP_X_MXIT_NICK"];
$mxitloc = $_SERVER["HTTP_X_MXIT_LOCATION"];
$mxitprof = $_SERVER["HTTP_X_MXIT_PROFILE"];
if(!isset($mxitid))
{
$mxitid = "DEFAULT";
}
mysqli_query($con,"INSERT INTO mxit (ip,time,user_agent,contact,userid,id,login,nick,location,profile) VALUES ($ip,$post_time,$mxitua,$mxitcont,$mxituid,$mxitid,$mxitlogin,$mxitnick,$mxitloc,$mxitprof)");
mysqli_close($con);
?>
Hello guys, i'm currently building my own cms, a personal project, and now im stucked on an error "Call to a member function query() on a non-object in.. please help
after creating this function.. I know the db connection and everything else worked out because i have a similar function that works just without the switch or the numrow if statement.
protected function _pageStatus($option, $id){ //check if page exists, if it does return the status, or return 404
switch($option){
case 'alpha' : $sql = "SELECT status FROM pages WHERE nick = '$id'"; break;
case 'num' : $sql = "SELECT status FROM pages WHERE id = '$id'"; break;
}
if($result = $this->_db->query($sql)){ //<--- THE ERROR WAS ON THIS LINE.
if($result->num_rows > 0){
while ($status = $result->fetch_object()) {
return $status;
}
return $status;
$result->close();
} else {
return 404;
}
}
}
Hi I have an infuriating problem which is stalling me with two large projects for a well known NGO right now. I am using mysqli and bound variables where the number of variables to bind is dependent on user input. I have a version of the code below working in php 5.2 but as of php 5.3 this method is no longer valid, specifically due to a change in the behavior of call_user_func_array with bound variables as arrays. I have read about this problem eslewhere but cannot get any of the workarounds to work with my example. Any help would be greatly appreciated. Code: [Select] # $parts is an array with variable number of values # $type is an array with variable number of values # $params is an array with variable number of values $query = 'SELECT SQL_CALC_FOUND_ROWS DISTINCT taxon.TaxonID, FROM taxon WHERE . join('', $parts) . " ORDER BY taxon.TaxonID"; # Prepare stmt if ($stmt = $mysqli->prepare($query)) { call_user_func_array (array($stmt, 'bind_param'),array_merge(array(join('', $type)), $params)); # execute $stmt->execute(); # bug info echo $stmt->errno, ':', $stmt->error; #store result $stmt->store_result(); # bind results $stmt->bind_result($ID); # fetch values while ($stmt->fetch()) { # results code goes here! } # free memory $stmt->free_result(); # close statement $stmt->close(); } I've been starting to play around the mysqli class and I've been having trouble using it due to various error it gives me from simple queries like this one. I'm not sure what the error is really, I've been following the php manual. Any help would be greatly appreciated. Code: [Select] <?php $mysqli = new MySQLi('localhost', 'root', 'root', 'jaipai'); if ($mysqli->connect_errno) { echo "There was a connection error: ". $mysqli->connecterrno; } class testClass { private $db; function __construct($mysqli) { $this->db = $mysqli; } public function pageInfo() { $query = "SELECT * FROM users WHERE username = jaipai"; $results = $this->db->query($query); $result = $this->db->fetch_assoc($results); return $result['username']; } } $testClass = new testClass($mysqli); echo $testClass->pageInfo(); ?> This gives me this error: Code: [Select] Fatal error: Call to undefined method mysqli::fetch_assoc() in /Users/JPFoster/Sites/Research & Development/Programs/Object Sandbox/DatabaseConnection.php on line 30 Just to be a little more informative I've also tried this method Code: [Select] $results = $this->db->query($query); $result = $results->fetch_assoc(); return $result['username']; This gives me an error: Code: [Select] Fatal error: Call to a member function fetch_assoc() on a non-object in Sites/Research & Development/Programs/Object Sandbox/DatabaseConnection.php on line 30 I'm not sure which is on the best path to go. Any help would be greatly appreciated.
<?php
if (isset($_POST['reset-submit'])) {
$selector = $_POST['selector'];
$validator = $_POST['validator'];
$password = $_POST['password'];
$password2 = $_POST['password2'];
// probably better to check this earlier
if (empty($password) || empty($password2)) {
header("Location: ../create-new-password.php?newpassword=empty&selector=$selector&validator=$validator");
} elseif ($password !== $password2) {
header("Location: ../create-new-password.php?newpassword=passwordsnotmatch");
}
$currentDate = date("U");
require "dbh.inc.php";
$sql = "SELECT * FROM reset_password WHERE selector=? AND expires >= $currentDate";
$stmt = mysqli_stmt_init($conn);
if (!mysqli_stmt_prepare($stmt, $sql)) {
echo "SQL error 1";
exit();
} else {
mysqli_stmt_bind_param($stmt, 'ss', $selector, $currentDate);
mysqli_stmt_execute($stmt);
$result = mysqli_stmt_get_result($stmt);
if (!$row = mysqli_fetch_assoc($result)) {
echo 'You need to re-submit your reset request.';
exit();
} else {
$tokenBin = hex2bin($validator);
$tokenCheck = password_verify($tokenBin, $row['token']);
if (!$tokenCheck) {
echo 'You need to re-submit your reset request.';
exit();
} else {
$email = $row['email'];
$sql = "SELECT * FROM users WHERE email = $email";
$stmt = mysqli_stmt_init($conn);
if (!mysqli_stmt_prepare($stmt, $sql)) {
echo "SQL error 2";
exit();
} else {
mysqli_stmt_bind_param($stmt, 's', $email);
mysqli_stmt_execute($stmt);
$result = mysqli_stmt_get_result($stmt);
if (!$row = mysqli_fetch_assoc($result)) {
echo "SQL error 3";
exit();
} else {
$sql = "UPDATE users SET password=? WHERE email=?";
$stmt = mysqli_stmt_init($conn);
if (!mysqli_stmt_prepare($stmt, $sql)) {
echo "SQL error4 ";
exit();
} else {
$hashed_password = password_hash($password, PASSWORD_DEFAULT);
mysqli_stmt_bind_param($stmt, 'ss', $hashed_password, $email);
mysqli_stmt_execute($stmt);
$sql = 'DELETE FROM reset_password WHERE email=?';
$stmt = mysqli_stmt_init($conn);
if (!mysqli_stmt_prepare($stmt, $sql)) {
echo 'SQL error5';
exit();
} else {
mysqli_stmt_bind_param($stmt, 's', $email);
mysqli_stmt_execute($stmt);
header("Location: ../signup.php?newpassword=updated");
}
}
}
}
}
}
}
mysqli_stmt_close($stmt);
mysqli_close($conn);
header('Location: ../reset-password.php?reset=success');
} else {
header('Location: ../index.php');
}
I always get this errors:
Warning: mysqli_stmt_bind_param(): Number of variables doesn't match number of parameters in prepared statement in C:\xampp\htdocs\php_login_system-master\includes\reset-password.inc.php on line 26
But i dont find the mistake in the Code. Can someone help me please hi guys I have two functions and they work well, but now i have a problem in the last if the email is sent like www.yoursite.com/reset_password.php?userid=0&code= and not http://your.url/set_new_password.php?userid=564979&code=54c4a2767c2f485185ab72cdcf03ab59 so, the problem is that i can get the value of hash and the userid. If i do an echo "$hash" in the function code(); it shows the value like 564979&code=54c4a2767c2f485185ab72cdcf03ab59 my question is, how i can do the same in the last if? above two functions. Same occur in the last select select userid from password_reset where code=? none userid is showed, it is always zero. Code: [Select] <? function check($sql, $db, $email) { if(!empty($_POST['email'])) { $email = $_POST["email"]; if ($sql = $db->prepare("select email from users where email=?")) { $sql->bind_param('s', $email); $sql->execute(); $sql->bind_result($email); if ($sql->fetch()) { return true; } else { return false; } } } } function code($sql, $db, $hash, $pwdHasher, $userExists, $sendPass) { if (check($sql, $db, $email)) { $pwdHasher = new PasswordHash(8, FALSE); $hash = $pwdHasher->HashPassword($userExists["email"]); $sendPass=$hash; ($sql = $db->prepare('insert into password_reset (code) values (?)')); $sql->bind_param('s', $hash); $sql->execute(); $sql->fetch(); return true; } } if (code($sql, $db, $hash, $pwdHasher, $userExists, $sendPass)) { ($sql = $db->prepare("select userid from password_reset where code=?")); $sql->bind_param('s', $hash); $sql->execute(); $sql->bind_result($hash); $sql->fetch(); echo $hash; $pwrurl = "www.yoursite.com/reset_password.php?userid=" .$hash . "&code=" . $sendPass; $mailbody = "Dear user,<br><br>If this e-mail does not apply to you please ignore it. It appears that you have requested a password reset at our website www.yoursitehere.com<br> To reset your password, please click the link below. If you cannot click it, please paste it into your web browser's address bar.<br> <a href='$pwrurl'>$pwrurl</a> <br> <br> Thanks,\nThe Administration"; $mail->MsgHTML($mailbody); $mail->AddAddress($email,"Membro"); $mail->IsHTML(true); if(!$mail->Send()) { echo "Deu erro: " . $mail->ErrorInfo; } else { echo "Enviado com sucesso"; } $sql->close(); $db->close(); } ?> any help? thanks Ok I am trying to use mysqli instead of the usual mysql. Mysql would be outdated. With mysqli, sgl-injection is impossible if you use the "?" in those codes. I would normally use a function but I've made a simple script to find the error. I use $parameters and $sql because these are the data I need to give as parameters to the function, so I used it here too but without the function actually. Code: [Select] ini_set('display_errors',1); // 1 == aan , 0 == uit error_reporting(E_ALL | E_STRICT); # sql debug define('DEBUG_MODE',true); // true == aan, false == uit $userid = 11; $lang = 1; $newLink = "testing123"; $db_host = "localhost"; $db_gebruiker = "root"; $db_wachtwoord = ''; $db_naam = "projecteasywebsite"; $sql= "INSERT tbl_link(userid,linkcat,linksubid,linklang,linkactive,linktitle) VALUES(?, ?, ?, ?, ?, ?)"; $parameters = '"iiisis", $userid, 1, 0, $lang, 1, $newLink'; echo $parameters; $mysqli = new mysqli($db_host, $db_gebruiker, $db_wachtwoord, $db_naam); $stmt = $mysqli->prepare($sql); $stmt->bind_param($parameters); $stmt->execute(); echo "<br><br>". mysqli_connect_errno(); echo "<br><br>". mysqli_report(MYSQLI_REPORT_ERROR); $stmt->close(); $mysqli->close(); I got Wrong parameter count for mysqli_stmt::bind_param() So naturally a problem when we execute : Warning: mysqli_stmt::execute() [mysqli-stmt.execute]: (HY000/2031): No data supplied for parameters in prepared statement ($stmt->execute() Is someone using mysqli too ? Hello everyone, For two weeks now, I'm trying to get this database connection in my query. Can someone give me a solution and tell me what I've done wrong? Am I overlooking something? <?php class Mysql{ public function connect(){ $mysqli = new mysqli('localhost','root','','login'); } } class Query extends Mysql{ public function runQuery(){ $this->result = parent::connect()->query("select bla bla from bla bla"); } } $query = new Query; $query->runQuery(); ?> hello , I'm starting to use mysqli and i have few questions. is there a guide for mysqli? and how do i use this functions at mysqli ? mysql_num_rows mysql_query mysql_fetch_assoc mysql_fetch_array thanks , Mor. Hi, The following code is what I want in that it creates a menu and I can select and display a table row.
I still need to use that selection to update the "lastused". I really appreciate your help. <!DOCTYPE><html><head><title>email menu</title></head> <body><center> <form name="form" method="post" action=""> <?php $con=mysqli_connect("localhost","root","cookie","homedb"); //============== check connection if(mysqli_errno($con)) {echo "Can't Connect to mySQL:".mysqli_connect_error();} else {echo "Connected to mySQL</br>";} //This creates the drop down box echo "<select name= 'target'>"; echo '<option value="">'.'--- Select email account ---'.'</option>'; $query = mysqli_query($con,"SELECT target FROM emailtbl"); $query_display = mysqli_query($con,"SELECT * FROM emailtbl"); while($row=mysqli_fetch_array($query)) {echo "<option value='". $row['target']."'>".$row['target'] .'</option>';} echo '</select>'; ?> <input type="submit" name="submit" value="Submit"/><!-- update "lastused" using selected "target"--> </form></body></html> <!DOCTYPE><html><head><title>email menu</title></head> <body><center> <?php $con=mysqli_connect("localhost","root","cookie","homedb"); if(mysqli_errno($con)) {echo "Can't Connect to mySQL:".mysqli_connect_error();} if(isset($_POST['target'])) { $name = $_POST['target']; $fetch="SELECT target,username,password,emailused,lastused, purpose, saved FROM emailtbl WHERE target = '".$name."'"; $result = mysqli_query($con,$fetch); if(!$result) {echo "Error:".(mysqli_error($con));} $lastused = "CURDATE()"; // update "lastused" using selected "target" //display the table echo '<table border="1">'.'<tr>'.'<td bgcolor="#ccffff align="center">'. 'Email menu'. '</td>'.'</tr>'; echo '<tr>'.'<td>'.'<table border="1">'.'<tr>'.'<td bgcolor="#ccffff align="center">'.'target'.'</td>'.'<td bgcolor="#ccffff align="center">'.'username'.'</td>'.'<td bgcolor="#ccffff align="center">'.'password'.'</td>'.'<td bgcolor="#ccffff align="center">'.'emailused'.'</td>'.'<td bgcolor="#ccffff align="center">'.'lastused'.'</td>'.'<td bgcolor="#ccffff align="center">'.'purpose'. '</td>'.'<td bgcolor="#ccffff align="center">'. 'saved' .'</td>'.'</tr>'; while($data=mysqli_fetch_row($result)) {echo ("<tr><td>$data[0]</td><td>$data[1]</td><td>$data[2]</td><td>$data[3]</td><td>$data[4]</td><td>$data[5]</td><td>$data[6]</td></tr>");} echo '</table>'.'</td>'.'</tr>'.'</table>'; } ?> </body></html> I have just started using MySQLi and am clueless it is giving me the follow errors in which i do not understand
Warning: mysqli_select_db() expects exactly 2 parameters, 1 given in C:\xampp\htdocs\Login\connect.php on line 23 Notice: Trying to get property of non-object in C:\xampp\htdocs\Login\connect.php on line 25 Notice: Use of undefined constant mysqli - assumed 'mysqli' in C:\xampp\htdocs\Login\connect.php on line 32 Warning: mysqli_query() expects parameter 1 to be mysqli, string given in C:\xampp\htdocs\Login\connect.php on line 32 Warning: mysql_fetch_assoc() expects parameter 1 to be resource, null given in C:\xampp\htdocs\Login\connect.php on line 33 can someone please explain to me why i am getting these? and my code is
$mysqli_db = mysqli_select_db("$db_name");
if($mysqli_db->connect_errno) {
printf("Database not found: %s\n", $mysql->connect_error);
exit();
}
$sql = "SELECT * FROM $tbl_name WHERE username='$username' AND password='$password'";
$result = mysqli_query($sql);
$row = mysqli_fetch_assoc($result);
I just got rid off most the errors the only ones left are
Warning: mysqli_query() expects at least 2 parameters, 1 given in C:\xampp\htdocs\Login\connect.php on line 32Fatal error: Call to undefined function mysqli_result() in C:\xampp\htdocs\Login\connect.php on line 33 Code Updated:
$mysqli_db = mysqli_select_db($mysqli_connect, $db_name);
if(!$mysqli_db) {
printf("Database not found: %s\n", $mysqli->connect_error);
exit();
}
$sql = "SELECT * FROM $tbl_name WHERE username='$username' AND password='$password'";
$query = mysqli_query($sql);
$result = mysqli_result($query);
$row = mysqli_fetch_assoc($result);
Edited by Tom8001, 30 November 2014 - 12:43 PM. When running the following code i get the error: Call to undefined method mysqli::errno() the code: $conn = new mysqli(HOST, USER, PASSWORD, DATABASE); if ($conn->errno() !== 0) { $msg = $conn->error(); throw new connErrorException($msg, 'Connect'); } I am fairly new to classes but as i understand it this should be correct. I am using mysql 5.1 so mysqli is on by default. I have even checked the php ini and everything looks fine there in respect to this. Any advice? I am using mysqli, OO, to connect to MySQL. I have only today started looking at this and am used to: Code: [Select] <?php $con = mysql_connect();//etc mysql_close($connection); ?> Am I right that with mysqli (OO) that I don't need to set a connection variable wither when connecting or closing?? Code: [Select] <?php mysqli::connect();//etc mysqli::close(); ?> What about with multiple databases, does mysqli keep track for me, as I am used to this: Code: [Select] <?php $con1 = mysql_connect();//db1 $con2 = mysql_connect();//db2 ?> //etc
The below code produces a dropdown and when a selection is made and submitted produces ---------------------------------------------------------------------------
<!DOCTYPE><html><head>
<title>lookup menu</title>
</head>
<body><center><b>
<form name="form" method="post" action="">
<?php
// error_reporting(0);
error_reporting(E_ALL ^ E_NOTICE);
include 'homedb-connect.php';
//This creates the drop down box
echo "<select name= 'target'>";
echo '<option value="">'.'--- Select account ---'.'</option>';
$query = mysqli_query($con,"SELECT target FROM lookuptbl");
$query_display = mysqli_query($con,"SELECT * FROM lookuptbl");
while($row=mysqli_fetch_array($query))
{echo "<option value='". $row['target']."'>".$row['target']
.'</option>';}
echo '</select>';
?>
<input type="submit" name="submit" value="Submit"/>
</form><center>
<?php
// error_reporting(0);
error_reporting(E_ALL ^ E_NOTICE);
include 'homedb-connect.php';
if(isset($_POST['target']))
{
$name = $_POST['target'];
$fetch="SELECT target, purpose, user, password, email, visits, date,
saved FROM lookuptbl WHERE target = '".$name."'";
$result = mysqli_query($con,$fetch);
if(!$result)
{echo "Error:".(mysqli_error($con));}
//display the table
echo '<table border="1"><tr><td bgcolor="#ccffff" align="center">lookup menu</td></tr>
<tr><td>
<table border="1">
<tr>
<td> Target </td>
<td> Purpose </td>
<td> User </td>
<td> Password </td>
<td> Email </td>
<td> Visits </td>
<td> Date </td>
<td> Saved </td>
</tr>';
while($data=mysqli_fetch_row($result))
{
$url= "http://localhost/home/crud-link.php?target=". $data[0];
$link= '<a href="'.$url.'">'. $data[0]. '</a>';
echo ("<tr><td> $link </td><td>$data[1]</td><td>$data[2]</td><td>$data[3]</td>
<td>$data[4]</td><td>$data[5]</td><td>$data[6]</td><td>$data[7]</td></tr>");
}
echo '</table>
</td></tr></table>';
}
?>
</body></html>
I dont know whether the statement is correct.....i just tried it.....and it didn't work. $stmt->bind_param('ssiiiss',$_POST['name'],$_POST['email'],$_POST['d'],$_POST['m'],$_POST['y'],$_POST['add'],$_POST['phone']); here my first two values are strings and next 2 tiny int's next is int and last 2 again strings. The following code is supposed to submit everything in the database on a table, Currently its doing the first row in the database. Not sure whats wrong.. function getNews(){ $query = $this->con->query("SELECT * FROM `". $this->prefix ."news`"); while ($result = $query->fetch_assoc()) { if($result['serveradded'] == 1){ $queryy = $this->con->query("SELECT * FROM `". $this->prefix ."news` WHERE serveradded=1"); while ($resultt = $queryy->fetch_assoc()) { return "<tr><td>The Advertisement ". ucfirst($resultt['name']) ." was created</td><td>". $resultt['date'] ."</td></tr>"; } } elseif($result['servupdate'] == 1){ $queryyy = $this->con->query("SELECT * FROM `". $this->prefix ."news` WHERE servupdate=1"); while ($resulttt = $queryyy->fetch_assoc()) { return "<tr><td>". $result['type'] ." ". ucfirst($resulttt['name']) ." was Updated by ". ucfirst($resulttt['updatedby']) ." </td><td>". $resulttt['date'] ." </td></tr>"; } } elseif($result['newuser'] == 1){ $queryyyy = $this->con->query("SELECT * FROM `". $this->prefix ."news` WHERE newuser=1"); while ($resultttt = $queryyyy->fetch_assoc()) { return "<tr><td>Account Created: ". ucfirst($resultttt['name']) ." was created</td><td>". $resultttt['date'] ."</td></tr>"; } } } } I am trying to extend my knowledge from using basic mysql_connect etc. into something better. I am searching on Google, and looking up as much info as possible but find that I learn best with a brief overview that allows me to channel down my searches. (1) I know PDO v mysql_connect v mysqli isn;t right - I think PDO is a class, mysql_connect is a command and mysqli is something else. Am I on the right track? (2) What is the best solution? I am using a full OOP coding style and want to learn the best - I am a novice at this but would consider myself an 8/10 programmer at other parts of PHP. An very brief overview to get me started and channel my Google searches would be much appreciated. - I have heard PDO is slower. - I don't need support for different databases (only MySQL). Thanks in advance. Greeting all. I am brand new to the forum and I would like to ask a question regarding mysqli. I am not certain why the following code works: Code: [Select] ... $sqlCmd = "select * from usrdata"; $rs = $mysqli->query($sqlCmd); ... and the following code gives me a 500 internal error: Code: [Select] ... $sqlCmd = "select * from usrdata"; $result= dbQuery($sqlCmd); ... function dbQuery($sqlCmd){ $rt = array(); $rs = $mysqli->query($sqlCmd); while($row = $rs->fetch_assoc ()){ $rt[] = $row; } return 'records: ' . json_encode($rt); } The error occurs on the mysqli->query. Thanks for any help. Cheers! Hey. I was writing my class object for a database connection and while I was writing the query part, I came to wonder whether I should use mysql_real_escape_string or mysqli_real_escape_string to make the query mysql friendly, what's the difference? Dear Sir/Madame I am making a website where user can insert data and wait for the admin to approve/reject the form. Now i am stuck with the update status where an admin can submit with a click pending to approval or reject with comments. I am new to PHP programming. Can somebody help me with the issue. Part 1 is inserting the data and part two is fetching the data but i am unable to solve the status approve/reject and comment at the same time on the view.php? page. Kindly help. Thank you.
<?php
$host="localhost";
$username="root";
$pass="";
$db="ems1";
$conn=mysqli_connect($host,$username,$pass,$db);
if(!$conn){
die("Database connection error");
}
// insert query for register page
if(isset($_REQUEST['proposal']))
{
$details=$_POST['details'];
$location=$_POST['location'];
$date=$_POST['date'];
$time=$_POST['time'];
$status="Pending";
$comment=$_POST['comment'];
$query="INSERT INTO `proposal` (`details`,`location`,`date`,`time`,`status`,`comment`) VALUES ('$details','$location','$date','$time','$status','$comment')";
$res=mysqli_query($conn,$query);
if($res){
$_SESSION['success']="Not Inserted successfully!";
header('Location:');
}else{
echo "Leave not Applied, please try again!";
}
}
?>
<div class="col-xs-6 col-xs-push-3 well">
<form class="form-horizontal" method="post" action="" >
<input type="hidden" name="proposal" value="">
<fieldset>
<legend>New Proposals </legend>
<!----left box----------->
<!----right box----------->
<div class="col-xs-9">
<div class="form-group">
<label for="inputEmail" class="col-lg-3"><b>Details:</b></label>
<div class="col-lg-9">
<input type="text" name="details" class="form-control">
</div>
</div>
<div class="form-group">
<label for="inputEmail" class="col-lg-3"><b>Location:</b></label>
<div class="col-lg-9">
<input type="text" name="location" class="form-control" >
</div>
</div>
<div class="form-group">
<label for="inputEmail" class="col-lg-3"><b>Date:</b></label>
<div class="col-lg-9">
<input type="date" name="date" class="form-control">
</div>
</div>
<div class="form-group">
<label for="inputEmail" class="col-lg-3"><b>Time:</b></label>
<div class="col-lg-9">
<input type="time" name="time" class="form-control" >
</div>
</div>
<div class="col-lg-9">
<input type="hidden" name="status" class="form-control" >
</div>
</div>
<div class="form-group">
<label for="inputEmail" class="col-lg-3"><b></b></label>
<div class="col-lg-9">
<input type="hidden" name="comment" class="form-control">
</div>
</div>
</div>
<div class="form-group">
<div class="col-lg-12">
<button type="reset" class="btn btn-default">Cancel</button>
<button type="submit" class="btn btn-primary">Submit</button>
</div>
</div>
</fieldset>
</form>
</div>
</div>
<body>
<h2 style="text-align:center; color:orangered;">
DASHBOARD
</h2>
<table>
<h3>
<tr style="background-color:#E4EBC5; color:orangered;">
<th>ID</th>
<th>Details</th>
<th>Location</th>
<th>Status</th>
<th>Comment</th>
</tr>
</h3>
</table>
<?Php
////////////////////////////////////////////
require "dbconfig.php"; // MySQL connection string
$count="SELECT id,details,location,time,status,comment FROM proposal";
if($stmt = $connection->query($count)){
while ($nt = $stmt->fetch_assoc()) {
echo "
<body>
<table>
<tr>
<td><a href=view.php?id=$nt[id]>$nt[id]</a></td>
<td>$nt[details]</td>
<td>$nt[location]</td>
<td>$nt[status]</td>
<td>$nt[comment]</td>
</tr>
</table>
</body>
";
}
}else{
echo $connection->error;
}
?>
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