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PHP - Display Content As Default From Mysqli
Hi folks, I am in the middle of creating a CMS as a project. It's going pretty well so far but I am stuck and hoping to get some guidance. When loading the main website, I want the contents from "Home" in the database to display unless a menu item is clicked. Here is what I have so far:
<?php
include_once('includes/header.php');
require_once('admin/includes/config.php');
?>
<div class="hero-wrap" style="background-image: url('images/uluru.jpg');" data-stellar-background-ratio="0.5">
<div class="overlay"></div>
<div class="container">
<div class="row no-gutters slider-text justify-content-start align-items-center">
<div class="col-lg-6 col-md-6 ftco-animate d-flex align-items-end">
<div class="text">
<h1 class="mb-4">Coaches For Hire <span>Book Now!</span></h1>
<p style="font-size: 18px;">The local Anangu, the Pitjantjatjara people, call the landmark Uluṟu (Pitjantjatjara [ʊlʊɻʊ]). This word is a proper noun, with no further particular meaning in the Pitjantjatjara dialect, although it is used as a local family name by the senior Traditional Owners of Uluru.</p>
<a href="https://www.youtube.com/watch?v=biuYA54nb7Y" class="icon-wrap popup-vimeo d-flex align-items-center mt-4">
<div class="icon d-flex align-items-center justify-content-center">
<span class="ion-ios-play"></span>
</div>
<div class="heading-title ml-5">
<span>Learn more about Uluru</span>
</div>
</a>
</div>
</div>
<div class="col-lg-2 col"></div>
<div class="col-lg-4 col-md-6 mt-0 mt-md-5 d-flex">
<form action="#" class="request-form ftco-animate">
<h2>Get A Quote</h2>
<div id="searchBoxContainer" class="form-group">
<label for="searchBox" class="label">Pick-Up Location</label>
<input class="form-control" type="text" id="searchBox" placeholder="Start Typing..." />
</div>
<div id="searchBoxContainerAlt" class="form-group">
<label for="searchBoxAlt" class="label">Drop-Off Location</label>
<input type="text" class="form-control" id="searchBoxAlt" placeholder="Start Typing..." />
</div>
<div class="d-flex">
<div class="form-group mr-2">
<label for="" class="label">Departure Date</label>
<input type="text" class="form-control" id="book_pick_date" placeholder="Date">
</div>
<div class="form-group ml-2">
<label for="" class="label">Return Date</label>
<input type="text" class="form-control" id="book_off_date" placeholder="Date">
</div>
</div>
<div class="d-flex">
<div class="form-group mr-2">
<label for="" class="label">Pick-Up Time</label>
<input type="text" class="form-control" id="time_pick" placeholder="Time">
</div>
<div class="form-group ml-2">
<label for"" class="label">Passenger Numbers</label>
<input type="number" class="form-control" placeholder="Amount" />
</div>
</div>
<div class="form-group">
<input type="submit" value="Request Quote" class="btn btn-primary py-3 px-4">
</div>
</form>
</div>
</div>
</div>
</div>
<script type="text/javascript" src="https://www.bing.com/api/maps/mapcontrol?key=AqIY0ivSCCdBIe3-EKGuox9cwBFw2wWRWIErZi1iy57EfD67PoiSra9wl_wu48de&callback=bingMapsReady" async defer></script>
<?php
$id = $_GET['id'];
$sql = "SELECT * FROM pages WHERE id = $id";
$result = $con->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_array()) {
?>
<!-- HOW IT WORKS -->
<section class="ftco-section ftco-no-pt ftco-no-pb">
<div class="container">
<div class="row no-gutters">
<div class="col-md-12 wrap-about py-md-5 ftco-animate">
<div class="heading-section mb-5 pl-md-5">
<span class="subheading"><?php echo $row['description']; ?>
</span>
<h2 class="heading"><?php echo $row['name']; ?></h2>
<?php echo $row['body']; ?>
</div>
</div>
</div>
</div>
</section>
<?php
}
}
?>
<!-- FOOTER -->
<?php
include_once('includes/footer.php');
?>
I hope you can help and that I am making sense. Cheers, Dan Edited January 9 by DanEthicalSimilar Tutorialshi trying to make a grid of two columns , data from db using mysqli and php, but it does not work...can you guys help me thanks in advance
<?php echo"<table border='0' align='center' cellpadding='2' cellspacing='2' width='70%'>";
$query = "SELECT * FROM posts";
while ($row = mysqli_fetch_assoc($select_all_categories_query)){
if($count ==0) { echo"<tr>"; }
echo"<div class='post-img'>"; }else { $count=0;
echo"<div class='post-img'>"; } $count++; } echo"</tr></table>"; ?>
this is the result
i want this kind of result
thank you Alright, I looked though the read me's, went over the FAQ's... Think it is time to post...
So, this lump is what I got from going though google and piecing together the bits that looked good.
<?php
$db_host = "****";
$db_user = "****";
$db_pwd = "****";
$database = "****";
$table = "****";
$query = "SELECT * FROM {$table}";
$conn = mysqli_connect($db_host, $db_user, $db_pwd, $database);
$result = mysqli_query($conn,$query) or trigger_error($query . ' - has encountered an error at:<br />' . mysqli_error($conn));
$fields = mysqli_fetch_fields($conn);
$field_count = mysqli_num_fields($conn);
echo '<table border="1" style="width:100%">' . "\n" . '<tr>';
$i = 0;
foreach($fields as $field)
{
if(++$i == 1) {
echo '<th colspan="' . $field_count . '">' . $field->table . '</th></tr><tr>';
}
echo "\n" . '<th>' . $field->name . '</th>';
}
echo '</tr>';
while($row = mysqli_fetch_row($result)) {
echo '<tr><td>' . implode('</td><td>' , $row) . '</td></tr>';
}
echo '</table>';
?>
The goal is to display the table and conditionally format the contents.
Let's forget about the conditional part (thinking if/than but don't know how I'm going to do that yet) and focus on the key problem: displaying the table.
I have tested this code. It works...to a point. it displays the table and its contents but no headers. I know my issue is in line 20 and 22 but I can not for the life of me figure it out. If it is stupidly easy (for you) please remember that I am not a coder. At best I am a scripter when it comes to linux. I am over my head on this stuff.
Thank you all for any help you can give me
Edited by TheAlmightyOS, 18 November 2014 - 03:30 PM. I've used a little PHP but I am not great. Is there any way to take information from a html page written in basic text and have it dynamically displayed. The content I want is found he http://www.kitco.com/texten/texten.html I want to take the information from the following chart - only Metal and Ask and create a ticker. New York Spot Price MARKET IS OPEN Will close in 2 hour 24 minutes ---------------------------------------------------------------------- Metals Bid Ask Change Low High ---------------------------------------------------------------------- Gold 1822.80 1823.80 -12.30 -0.67% 1810.70 1840.90 Silver 41.45 41.55 +0.10 +0.24% 41.08 42.15 Platinum 1844.00 1852.00 -7.00 -0.38% 1826.00 1861.00 Palladium 783.00 788.00 +10.00 +1.29% 775.00 794.00 ---------------------------------------------------------------------- So basically I would just take the word: Gold - $1823.80 Silver - $41.55 Platinum $1852.00 Palladium - $788.00 Can someone help me through this or point me in a direction where I can read more about accomplishing this? I've been searching google for articles but still haven't found what I'm looking for. nofx1728 Slideshow programs created with Slideshow Maker. The html version works but opens in a new window. I want it to open in the content window same as all other programs. menu2.php (partial) <li> <a href="#">Slideshows</a> <ul> <li> <a href="index.php?page=slideshow_shed">Shed construction</a> </li> <li> <a href="content/pages/slideshow_shed.html">Shed Construction</a> </li> </ul> slideshow_shed.html <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <title>Shed Pictures</title> </head> <body bgcolor="#FFFFFF"> <p> <p> <style type="text/css"> <!-- .style1 { color: #006633; font-weight: bold; } --> </style> <!-- saved from url=(0013)about:internet --> <table align="center" border="0" cellpadding=0 cellspacing=0> <tr> <td align="center">Shed Pictures</td> </tr> <tr><td align="center"><br /></td></tr> <tr> <td align="center"> <object classid="clsid:d27cdb6e-ae6d-11cf-96b8-444553540000" codebase="http://fpdownload.macromedia.com/pub/shockwave/cabs/flash/swflash.cab#version=6,0,0,0" width="768" height="576" id="tech" align="middle"> <param name="allowScriptAccess" value="sameDomain" /> <param name="movie" value="slideshow_shed.swf" /> <param name="quality" value="high" /> <embed src="slideshow_shed.swf" quality="high" width="768" height="576" name="tech" align="middle" allowScriptAccess="sameDomain" type="application/x-shockwave-flash" pluginspage="http://www.macromedia.com/go/getflashplayer" /> </object> </td> </tr> <tr> <td align="center" ><br /> Created by <a href="http://www.flash-slideshow-maker.com/index.php?srcid=glc" target="_blank">Flash Slideshow Maker</a> </td> </tr> <tr> <td align="center" valign="middle" bgcolor="#FFCC99"><br /> <a href="http://www.flash-slideshow-maker.com/buynow.php?srcid=glc" target="_blank"><strong>Order the full version now to remove the software logo at the end of slideshow</strong></a> </td> </tr> </table> </body> </html> slideshow_shed.php <?php ?> <style type="text/css"> <!-- .style1 { color: #006633; font-weight: bold; } --> </style> <table align="center" border="0" cellpadding=0 cellspacing=0> <tr> <td align="center">Shed Pictures</td> </tr> <tr><td align="center"><br /></td></tr> <tr> <td align="center"> <object classid="clsid:d27cdb6e-ae6d-11cf-96b8-444553540000" codebase="http://fpdownload.macromedia.com/pub/shockwave/cabs/flash/swflash.cab#version=6,0,0,0" width="768" height="576" id="tech" align="middle"> <param name="allowScriptAccess" value="sameDomain" /> <param name="movie" value="slideshow_shed.swf" /> <param name="quality" value="high" /> <embed src="slideshow_shed.swf" quality="high" width="768" height="576" name="tech" align="middle" allowScriptAccess="sameDomain" type="application/x-shockwave-flash" pluginspage="http://www.macromedia.com/go/getflashplayer" /> </object> </td> </tr> <tr> <td align="center" ><br /> Created by <a href="http://www.flash-slideshow-maker.com/index.php?srcid=glc" target="_blank">Flash Slideshow Maker</a> </td> </tr> <tr> <td align="center" valign="middle" bgcolor="#FFCC99"><br /> <a href="http://www.flash-slideshow-maker.com/buynow.php?srcid=glc" target="_blank"><strong>Order the full version now to remove the software logo at the end of slideshow</strong></a> </td> </tr> </table> I don't get any errors, just the title "Shed Pictures" displays. Most of the websites I have built have not been large and not dynamic in nature, but I need to make one now with a boatload of content on it. and I’m needing a little resource help where I can research something. I’m sure all developers do this, and I know I need a massive database or 2 and the PHP knowledge I already have, but I’m just looking for a little guidance about how to do this the easiest way. It may actually sound like a stupid question and I might be over-thinking a bit. So… Lets say I have 500 articles that I want to make available to a site visitor and give them the option to choose what they want to look at through nav menus and submenus. I want the same page to display to the user, regardless of what article is chosen. how does PHP play a part in that? Would I be right by saying the following? If the home page had this content on it: This website is a course in PHP coding. <a href="courseTemplate.php" target="_blank">Click here</a> to view course #1. and I wanted all the course content to be displayed through the use of the “courseTemplate.php” file, how simple is that? I would assume that these types of things would be the results and the techniques to accomplish the goal, right? => a resulting URL that looks like this: www.site.com/courseTemplate.php?id=1 => storing all the text of the course material in one single field of a DB record. NOT storing the layout-oriented code (HTML, etc.) and echoing it out with the rest of the text. => making use of the GET() function somehow to pull the course’s text content out of the database. Can someone show me a website that demonstrates this? I don’t think this is very difficult, and I’m sure there are web resources available that show how this is done. I’m almost sure that most news agency websites do this, and I know for a fact that most forum software has this template technique written into it, regardless if SEO is included or not. Sorry for the basic nature of this question. I know some of my previous posts have been such where I’ve asked much more difficult questions than this. I have seen youtube tutorials on issues similar to this, but nothing pulled up by google really shows this in a very simplistic nature. thanks. Name says it all. I have a form and 2 tables that are going to have to be linked with each other. Although what I'm trying to do now is very basic and I'm struggling haha Though this is my first php project Basicly I got a form with a drop down menu, which I want to display the contents of my table (course) which the user then selects from and submits to another table (student) Here is my code for this part <?php Code: [Select] <?php $q = "SELECT cname FROM course "; $result = mysql_query($q); $course = mysql_fetch_array($result); ?> <div id="apdiv3"> <FORM action = "demo.php" method ="post"> <p>Course name:</p> <select name="cname"> <OPTION> <?php echo $course['cname']; ?> </option> </SELECT>?> It only displays 1 item from my course table. What am I doing wrong? And also please explain so I can learn from my mistakes Hi there, I was wondering if there is a way to display content on only my index.php page? I believe there is a way and I have done it before, but some of my pages use index.php?category etc. Is there a way to show content just on index.php and not on any dynamic pages? Thanks What i want to is if content is more than one line but script should display only one line and add "...more" at the end of it. so that i can make it a link to the article. is this possible? Thanks in advance for the help and support. 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 ?
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. 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> 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(); ?> 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 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. 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 have some code I used to have in mysql and now im trying to convert to mysqli and I cant seem to find out what the problem is.
<?php
$username = $_SESSION['username'];
// Connect to server and select databse.
include "db_connect.php";
include "db_config.php";
// items tables selection
$sql = mysqli_query($my_database,"SELECT * FROM items_tbl WHERE level = '$account_info[player_level]' ORDER BY rand()");
//$result = mysqli_query($my_database,$sql);
// Put info into array (This Works)
while($item = mysqli_fetch_assoc($sql)){
//stats
$items_id['itemid'] = $item['itemid'];
$items_id['Level'] = $item['Level'];
$items_id['name'] = $item['name'];
$items_id['min_str'] = $item['min_str'];
$items_id['min_int'] = $item['min_int'];
$items_id['min_dex'] = $item['min_dex'];
$items_id['type'] = $item['type'];
$items_id['min_dmg'] = $item['min_dmg'];
$items_id['max_dmg'] = $item['max_dmg'];
$items_id['phys_defense'] = $item['phys_defense'];
$items_id['mag_defense'] = $item['mag_defense'];
}
?>
here is the error im getting:
Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean given
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. |
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