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PHP - Php Displaying Mysqli Array
Hey All.
I'm currently using the following code to source some information from a Mysql Database.
I've limited the results to 6 Values. I'm using the WHILE function to display a table with the results populated in.
As you'll see from the attached picture, the WHILE function is causing 5 tables to be created, each with 5 of the same image displayed. How do I go about displaying only 1 table with the 5 different pictures displayed in?
<?php
$result = mysqli_query($con, "SELECT * FROM photo ORDER BY id DESC LIMIT 6");
while($row = mysqli_fetch_array($result)){
echo
'<table>
<tr><td class="left">
<a href="photo.php?id='.$row['id'].'"><img src="images/'.$row['id'].'.jpg" height="232" alt=""/></a>
</td>
<td class="center">
<a href="photo.php?id='.$row['id'].'"><img src="images/'.$row['id'].'.jpg" height="112" alt=""/></a>
<a href="photo.php?id='.$row['id'].'"><img src="images/'.$row['id'].'.jpg" height="112" alt=""/></a>
</td>
<td class="right">
<a href="photo.php?id='.$row['id'].'"><img src="images/'.$row['id'].'.jpg" height="72" alt=""/></a>
<a href="photo.php?id='.$row['id'].'"><img src="images/'.$row['id'].'.jpg" height="72" alt=""/></a>
<a href="photo.php?id='.$row['id'].'"><img src="images/'.$row['id'].'.jpg" height="72" alt=""/></a>
</td></tr></table>';
}
?>
Screen Shot 2014-08-20 at 15.46.31 (2).jpg 35.86KB
0 downloads
Thanks
Edited by JimmyLeaman, 20 August 2014 - 09:49 AM. Similar TutorialsThanks to the great help from some great people on this message board I'm right at the finish line on this page. I'm trying to mesh all the coding together at this point, and make it display the information. I keep getting 'mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource....etc..(filename/location) on line 51. Am I way off here??? Code: [Select] <?php $db_host = 'localhost'; $db_user = 'root'; $db_pass = ''; $db_in = new MySQLi ( $db_host, $db_user, $db_pass, 'sign_in' ); $db_out = new MySQLi( $db_host, $db_user, $db_pass, 'sign_out' ); if( $db_in->connect_error || $db_out->connect_error ) trigger_error( 'Unable to initiate database connections', E_USER_ERROR ); $q_in = 'SELECT * FROM `customer sign-in`'; $q_out = 'SELECT * FROM `customer sign-out`'; if( ($r_in = $db_in->query($q_in)) === FALSE || ($r_out = $db_out->query($q_out)) === FALSE ) trigger_error( 'Unable to grab ticket information from databases', E_USER_ERROR ); $data_in = array(); $data_out = array(); while( $row = $r_in->fetch_assoc() ) $data_in[ $row['Ticket #'] ] = $row; $r_in->free(); while( $row = $r_out->fetch_assoc() ) $data_out[ $row['Ticket #'] ] = $row; $r_out->free(); $result = array_diff_key( $data_in, $data_out ); echo "<table border='1'> <tr> <th>ID</th> <th>Sign In Date</th> <th>RANK/CIV</th> <th>First Name</th> <th>Last Name</th> <th>Unit</th> <th>DSN/Roshan</th> <th>Classifications</th> <th>Services Requested</th> <th>Service Tag/ Serial Number</th> <th>Ticket #</th> <th>Make/ Model</th> </tr>"; while($row = mysql_fetch_assoc($result)) { // Retrieve data until no more { echo "<tr>"; echo "<td>" . $row['ID'] . "</td>"; echo "<td>" . $row['Sign in Date'] . "</td>"; echo "<td>" . $row['Rank/CIV'] . "</td>"; echo "<td>" . $row['First Name'] . "</td>"; echo "<td>" . $row['Last Name'] . "</td>"; echo "<td>" . $row['Unit'] . "</td>"; echo "<td>" . $row['DSN/Roshan'] . "</td>"; echo "<td>" . $row['Classifications'] . "</td>"; echo "<td>" . $row['Services Requested'] . "</td>"; echo "<td>" . $row['Service Tag/ Serial Number'] . "</td>"; echo "<td>" . $row['Ticket #'] . "</td>"; echo "<td>" . $row['Make/ Model'] . "</td>"; echo "</tr>"; } } ?> Consider this scenario: I'm using an ajax to make the form use only one page I got a multistep form with step3 being the result of step2, within step3 there is an "add more steps" button which duplicates step2 and step3, this can be done countless times. to complement this I added array to my input boxes something like this: Code: [Select] <input type="text" name="name[]" id="name"/> unfortunately, I'm not sure how to approach this in the php. In the step2 going to step3, I would call the brand id and use it in the database to get the results for step3, but now it's array I added the following lines $brand = $this->input->post('brand[]'); $this->load->model('MBest'); $q = $this->MBest->getbestmatching($brand[0]); But the $brand[0] doesn't works and triggers an error. And come to think of it, if I call a $brand[0] index, then everytime a person add a new step, it will always use the id of the first result. Are there any ideas on how to fix this problem? Thanks Hi, I have this array: $cats = array( "A"=>'Business Expansion Specialist', "B"=>'Affiliate Page', "C"=>'Personal: Hobbies', "D"=>'Personal: Pets', "E"=>'Personal: Holidays', "F"=>'Personal: Sport', "G"=>'Personal: C.V.', "H"=>'Personal: Other', "I"=>'For Sale: Antiques', "Z"=>'Other' );
And then from a table I get the variable $cat_cd. $cat_cd contains 'F' so it refers to 'Personal: Sport' If I want to put that in $category, how would I write that? I thought it might be .... $category = $cats['$cat_cd']; I tried that and it didn't work .
Thanks
Hi I appreciate everyone who helped me with my other problem trying to call my Array() I had issues with. Heres what that looks like http://crafted.horizon-host.com/ff/ Code from the above Code: [Select] <html> <head> <title>FINAL FANTASY XIV - Item Database</title> <style type="text/css"> a:link {text-decoration: none; color: #FFFFFF} a:visited {text-decoration: none; color: #FFFFFF} a:hover {text-decoration: none; color: #FFFFFF} a:active {text-decoration: none; color: #FFFFFF} p.title { font-size: medium; font-family: Verdana; color: #FFFFFF; } p.items { font-size: small; font-family: Verdana; color: #FFFFFF; } </style> <script src="http://static.yg.com/js/exsyndication.js" type="text/javascript"> </script> <script type="text/javascript"> YG.UserSettings.Syndication = { removeClasses: ['extern'], defaultClass: 'yg-iconsmall yg-name', clearTitle: true } </script> </head> <body background="../images/testbg.jpg" marginwidth=60 marginheight=60> <table align="center" id="Table_01" width=520 height=393 border=0 cellpadding=0 cellspacing=0> <tr> <td width=520 height=14 colspan=5 background="../images/test_01.gif"></td> </tr> <tr> <td width=20 height=23 background="../images/test_02.gif"></td> <td colspan=3 background="../images/test_03.gif" align="center" valign="middle"> <p class='title'>FINAL FANTASY XIV - Item Database</p> </td> <td width=14 height=23 background="../images/test_04.gif"></td> </tr> <tr> <td width=520 height=7 colspan=5 background="../images/test_05.gif"></td> </tr> <tr> <td background="../images/test_06.gif"></td> <td colspan=3 background="../images/test_07.gif" align="center" valign="top"> <p class='items'> <?php function grabdata($site){ $page_contents = file_get_contents($site); $pattern = '/\/item\/([a-z,-]+[^gil])\?id\=([0-9,]+)/'; preg_match_all($pattern, $page_contents, $matches); $arrRetList = array(); if(!empty($matches)){ foreach($matches as $item) { foreach($item as $pos => $row){ $arrRetList[$pos][] = $row; } } foreach($arrRetList as $row){ echo "<p class='items'><a href='http://ffxiv.yg.com" . $row[0] . "'>Unknown Item</a> - " . $row[1] . " - " . $row[2] . "<BR>"; } } } grabdata("http://ffxiv.yg.com/items?l=50:50;trd=0;cl=30"); ?> </p> </td> <td background="../images/test_08.gif"></td> </tr> <tr> <td colspan=2 width=23 height=17 background="../images/test_09.gif"></td> <td background="../images/test_10.gif"></td> <td colspan=2 width=17 height=17 background="../images/test_11.gif"></td> </tr> <tr> <td> <img src="../images/spacer.gif" width=20 height=1></td> <td> <img src="../images/spacer.gif" width=3 height=1></td> <td> <img src="../images/spacer.gif" width=480 height=1></td> <td> <img src="../images/spacer.gif" width=3 height=1></td> <td> <img src="../images/spacer.gif" width=14 height=1></td> </tr> </table> </body> </html> Heres the code im trying to run, has some added stuff to the above <?php ?> code but its not viewing at all, just shows a white page.. and would like to get some advice on what im doing wrong Code: [Select] <?php function grabdata($site){ $page_contents = file_get_contents($site); $pattern = '/\/item\/([a-z,-]+[^gil])\?id\=([0-9,]+)/'; preg_match_all($pattern, $page_contents, $matches); $arrRetList = array(); if(!empty($matches)){ foreach($matches as $item) { foreach($item as $pos => $row){ $arrRetList[$pos][] = $row; } } foreach($arrRetList as $row){ echo $row[1] . " - " . $row[2] . "<BR>"; echo "<table align='center' border='0' cellpadding='0' cellspacing='0' width='80%'><td align='left'><p class='items'>Item</td><td align='center'><p class='items'>NPC Price</td><tr><td align='left'><p class='items'><a href='http://ffxiv.yg.com" . $row[0] . "'>Unknown Item</a></td></td><td align='center'><p class='items'>"; $page_contents2 = file_get_contents("http://ffxiv.yg.com" . $row[0] . ""); $matches2 = array(); preg_match_all('/<td class=\"lefttd\">Normal<\/td><td align=\"right\">([0-9,]+)<\/td><\/tr>/', $page_contents2, $matches2); if ($matches2[1][0] == ""){ echo "0g - "; } else { print_r($matches2[1][0] . "g - "); } preg_match_all('/<td class=\"lefttd\">\+1<\/td><td align=\"right\">([0-9,]+)<\/td><\/tr>/', $page_contents2, $matches2); if ($matches2[1][0] == ""){ echo "0g - "; } else { print_r($matches2[1][0] . "g - "); } preg_match_all('/<td class=\"lefttd\">\+2<\/td><td align=\"right\">([0-9,]+)<\/td><\/tr>/', $page_contents2, $matches2); if ($matches2[1][0] == ""){ echo "0g - "; } else { print_r($matches2[1][0] . "g - "); } preg_match_all('/<td class=\"lefttd\">\+3<\/td><td align=\"right\">([0-9,]+)<\/td><\/tr>/', $page_contents2, $matches2); if ($matches2[1][0] == ""){ echo "0g"; } else { print_r($matches2[1][0] . "g"); } echo "</td></tr>"; } echo "</table><BR><BR><BR>"; } } grabdata("http://ffxiv.yg.com/items?l=50:50;trd=0;cl=30"); ?> To give you a idea of what im looking for, it already grabs each item and list it from grabdata(""); function, i wanna take that further by it grabs the item, and then uses that data to grab the item sell price (normal, +1, +2, +3 quality) for each display it out, something like.. Code: [Select] ITEMNAME1 132g - 145g - 165g - 198g ITEMNAME2 50g - 0g - 0g - 0g ITEMNAME3 63g - 78g - 90g - 110g etc etc from the first $page_contents from $site $row[0] will output: item url variables $row[1] will output: item name $row[2] will output: item # from the second $page_contents from "http://ffxiv.yg.com" . $row[0] . "" $matches2[1][0] will be numerical output of the prices for each preg_match_all These scripts work separately just need them to work together I have a function which finds a list of companies associated with a country from a custom taxonomy. It finds their logo and creates an output which is all logos associated with a country, and the logo has a href to link to a company page. The code for this is: function countryCompanies(){ global $post;
$echo = f_print(array(
The f_print function finds the logos, and their sizes. So the 'mainlogo' type is a pull down of the thumbnail at 200px size.
What I need to do is write CSS and PHP which will display these results in a grid, 4 colums across, and as many down as results from the array. At the moment all I can get is all the results in one column. What tutorials would I need to write appropriate CSS to create this, and then is it as simple as adding the div prior to the echo of the $echo variable? Thanks I am validating a form, and if there are errors in the input, the error array is filled with appropriate mssg and the "sticky" form with right values is shown back, and errors on top. Example: first name, last name, email, date of birth. each field is validated,and if for example first name does not pass,i will nullify the value and populate error array like below: $_SESSSION['error']['fname']="Please check the format of the name,can contain only alphabets!" $_SESSSION['error']['email']="Please check the format of email!" When I slap back the form, last name has passed, and so will have the previously entered value, but first name and email will be empty, and i need to know how to display the 2 errors on top of the form. It uses post method, and posts to itself. Hi, I've used this code before, but have had to make some modifications and am now getting a mysql fetch array error: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/tesolcla/public_html/test/results10k201010.php on line 134 If anyone can help, it would be much appreciated. Code: [Select] <?php $dbcnx = @mysql_connect('localhost', 'MYUSERNAME', 'MYPASSWORD'); //$dbcnx = @mysql_connect('localhost', 'root', 'mysql'); if (!$dbcnx) { exit('<p>Unable to connect to the ' . 'database server at this time.</p>'); } if (!@mysql_select_db('MYDATABASE')) { //if (!@mysql_select_db('rac')) { exit('<p>Unable to locate the results ' . 'database at this time.</p>'); } $asc_on = '<img src="images/results_sorting/Asc.gif" border="0" />'; $asc_off= '<img src="images/results_sorting/AscOff.gif" border="0" />'; $desc_on = '<img src="images/results_sorting/Desc.gif" border="0" />'; $desc_off= '<img src="images/results_sorting/DescOff.gif" border="0" />'; $sortfield = isset($_GET['sort']) ? $_GET['sort'] : '4'; $sorttype = isset($_GET['type']) ? $_GET['type'] : '1'; for($i=1; $i<5; $i++) { if($i==$sortfield) { if ($sorttype==1) $srt[$i] = $asc_on.'<a href="?sort='.$i.'&type=2">'.$desc_off.'</a>'; else $srt[$i] = '<a href="?sort='.$i.'&type=1">'.$asc_off.'</a>'.$desc_on; } else { $srt[$i] = '<a href="?sort='.$i.'&type=1">'.$asc_off.'</a><a href="?sort='.$i.'&type=2">'.$desc_off.'</a>'; } } $fields = array("firstname", "lastname", "time", "position"); $sorts = array("ASC", "DESC"); $field = $fields[$sortfield-1]; $sort = $sorts[$sorttype-1]; $field = $field=="" ? $fields[4] : $field; $sort = $sort=="" ? $sorts[0] : $sort; $sql = mysql_query("SELECT firstname, lastname, time, position FROM 10k_results ORDER BY $field $sort"); echo "<table border='1' align='center' bordercolor='#000000' CELLPADDING=5 cellspacing='0' STYLE='font-size:13px'>"; echo "<tr bgcolor='#008000' STYLE='color:white'> <td>*</td><td><H3>First name $srt[1]</h3></td> <td><H3>Lastname $srt[2]</H3></td> <td><H3>Time $srt[3]</H3></td> <td><H3>Position $srt[4]</H3></td></tr>"; // keeps getting the next row until there are no more to get $row_counter = 1; //create row counter with default value 0 // Print out the contents of each row into a table while ($row = mysql_fetch_array($sql)) { // Print out the contents of each row into a table echo "<tr>\n"; echo "</td><td>"; echo $row_counter++; echo "</td>"; echo "<td>{$row['firstname']}</td>\n"; echo "<td>{$row['lastname']}</td>\n"; echo "<td>{$row['time']}</td>\n"; echo "<td>{$row['position']}</td>\n"; echo "</tr>\n"; } echo "</table>"; ?> I am trying to give role based access here. I want to display the pages which user has access in checkbox format. Here is my code
$sql = "SELECT p.page_id AS pid,
p.page,
p.href,
ra.pages AS rpage
FROM pages p
INNER JOIN role_access ra
WHERE p.page_id IN (ra.page)
AND ra.role=1";
$query = mysqli_query($con, $sql) or die(mysqli_error($con));
$checked_arr = array();
while($row = mysqli_fetch_array($query)) {
$checked_arr = explode(",",$row['rpage']);
foreach ($checked_arr as $page) {
echo "<br/><input type='checkbox' name=\"pages[]\" value='$page' />$page<br>";
}
My tables are like this ROLE CREATE TABLE `role` ( `rid` int(5) NOT NULL, `role_name` varchar(50) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; INSERT INTO `role` (`rid`, `role_name`) VALUES (1, 'Admin'), (2, 'Others'); ALTER TABLE `role` ADD PRIMARY KEY (`rid`); ROLE-ACCESS CREATE TABLE `role_access` ( `id` int(10) NOT NULL, `page` varchar(160) NOT NULL, `role` int(7) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; INSERT INTO `role_access` (`id`, `page`, `role`) VALUES (1, '1,2,3,4,5', 1), (2, '2,4,5', 2); ALTER TABLE `role_access` ADD PRIMARY KEY (`id`); PAGES CREATE TABLE `pages` ( `page_id` int(11) NOT NULL, `code` varchar(10) NOT NULL, `page` varchar(100) NOT NULL, `href` varchar(255) NOT NULL ) ENGINE=InnoDB DEFAULT CHARSET=latin1; INSERT INTO `pages` (`page_id`, `code`, `page`, `href`) VALUES (1, 'Patient', 'Registration', '/patient_registration.php'), (2, 'Patient', 'List', '/patient_list.php'), (3, 'Patient', 'Edit', '/edit_patient.php'), (4, 'Patient', 'Export', '/export_patient.php'), (5, 'Patient', 'MRD', '/patient_MRD.php'); ALTER TABLE `pages` ADD PRIMARY KEY (`page_id`);
In above query i get result like this
But required result is
if i change checkbox display like
while($row = mysqli_fetch_array($query)) {
$checked_arr = explode(",",$row['rpage']);
$pname = $row['page'];
foreach ($checked_arr as $page) {
echo "<br/><input type='checkbox' name=\"pages[]\" value='$page' />$pname<br>";
}
}
i get result
How to get the names for that file.
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. 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. 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 ? 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 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>
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> 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 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(); ?> Here is my code so far... I can't login using this php, i am new to php and am trying my hardest to figure this out... i have been on this for 4 days and am about to pull all of my hair out... can anyone please help me... //******** start of login.php ******** <?php require_once('connectvars.php'); // Start the session session_start(); // Clear the error message // $error_msg = ""; // If the user isn't logged in, try to log them in if (!isset($_SESSION['user_id'])) { if (isset($_POST['submit'])) { // Connect to the database $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); if (!$dbc) { die('Could not connect: ' . mysqli_error()); } echo 'Connected successfully'; // Grab the user-entered log-in data $user_email = mysqli_real_escape_string($dbc, trim($_POST['email'])); $user_pass = mysqli_real_escape_string($dbc, trim($_POST['password'])); if (!empty($user_email) && !empty($user_password)) { // Look up the username and password in the database $query = "SELECT tb_user_id, tb_user_email FROM tb_users WHERE tb_user_email = '$user_email' AND tb_user_password = SHA('$user_pass')"; $data = mysqli_query($dbc, $query); if (mysqli_num_rows($data) == 1) { // The log-in is OK so set the user ID and username session vars (and cookies), and redirect to the home page $row = mysqli_fetch_array($data); $_SESSION['user_id'] = $row['tb_user_id']; $_SESSION['email'] = $row['tb_user_email']; setcookie('user_id', $row['tb_user_id'], time() + (60 * 60 * 24 * 30)); // expires in 30 days setcookie('email', $row['tb_user_email'], time() + (60 * 60 * 24 * 30)); // expires in 30 days $home_url = 'http://' . $_SERVER['HTTP_HOST'] . dirname($_SERVER['PHP_SELF']) . '/index.php'; header('Location: ' . $home_url); } else { // The username/password are incorrect so set an error message $error_msg = 'Sorry, you must enter a valid username and password to log in1.'; } } else { //*********** This is the error i keep getting // The username/password weren't entered so set an error message $error_msg = 'Sorry, you must enter your username and password to log in2.'; } } } // Insert the page header $page_title = 'Log In'; require_once('header.php'); // If the session var is empty, show any error message and the log-in form; otherwise confirm the log-in if (empty($_SESSION['user_id'])) { echo '<p class="error">' . $error_msg . '</p>'; ?> <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <fieldset> <legend>Log In</legend> <label for="email">Email Address:</label> <input type="text" name="email" value="<?php if (!empty($user_email)) echo $user_email; ?>" /><br /> <label for="password">Password:</label> <input type="password" name="password"/> </fieldset> <input type="submit" value="Log In" name="submit" /> </form> <?php } else { // Confirm the successful log-in echo('<p class="login">You are logged in as ' . $_SESSION['email'] . '.</p>'); } ?> <?php // Insert the page footer require_once('footer.php'); ?> Hi 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);
?>
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
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