|
PHP - Mysqli_query Returning Null
Okay so I created a theme for my website that will also utilize a custom CMS script I developed and its not cooperating with me nicely.
Here's my head of my header file for my template: Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" <?php language_attributes(); ?>> <head> <title><?php bloginfo('name') ?>: <?php bloginfo('description') ?></title> <link rel="stylesheet" type="text/css" media="all" href="<?php bloginfo( 'stylesheet_url' ); ?>" /> <meta http-equiv="content-type" content='<?php bloginfo("html_type"); ?> charset=<?php bloginfo('charset'); ?>' /> <?php if(is_singular()){ wp_enqueue_script('comment-reply');} ?> <?php wp_head(); ?> <?php require ('efedmanager/inc/dbconfig.php'); ?> </head> Now on my index page that calls the header file I DO NOT get a error saying that the file could not be located. So that's good news. Here's my dbconfig.php file. Code: [Select] <?php /** * @author Jeff Davidson * @copyright 2010 */ // This file contains the database access information and establishes a connection to MySQL and selects the database // Set the database access information as contstants DEFINE ('DB_USER', '?'); DEFINE ('DB_PASSWORD', '?'); DEFINE ('DB_HOST', '?'); DEFINE ('DB_NAME', '?'); // Make the database connection $dbc = mysqli_connect (DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); if (!$dbc) { die('Connect Error (' . mysqli_connect_errno() . ') ' . mysqli_connect_error()); } ?> Now when I go to one of my custom pages like this one it brings up the message: Warning: mysqli_query() expects parameter 1 to be mysqli, null given in /home/xtremer/public_html/wp-content/plugins/exec-php/includes/runtime.php(42) : eval()'d code on line 38 Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in /home/xtremer/public_html/wp-content/plugins/exec-php/includes/runtime.php(42) : eval()'d code on line 39 Code: [Select] <div id="champions" class="content"> <h1 class="pageheading">KOW Champions and Contenders</h1> <?php $championsQuery = " SELECT titles.titleName, titles.shortName, champions.champID, champions.con1ID, champions.con2ID, champions.con3ID, biochamp.shortName AS championshortName, biochamp.characterName AS champion, biocon1.shortName AS con1shortName, biocon1.characterName AS con1, biocon2.shortName AS con2shortName, biocon2.characterName AS con2, biocon3.shortName AS con3shortName, biocon3.characterName AS con3 FROM champions LEFT JOIN titles AS titles ON titles.ID = champions.titleID LEFT JOIN characters AS biochamp ON champions.champID = biochamp.ID LEFT JOIN characters AS biocon1 ON champions.con1ID = biocon1.ID LEFT JOIN characters AS biocon2 ON champions.con2ID = biocon2.ID LEFT JOIN characters AS biocon3 ON champions.con3ID = biocon3.ID WHERE titles.statusID = '1' ORDER BY titles.ID"; $championsResult = mysqli_query($dbc, $championsQuery); while ( $row = mysqli_fetch_array ( $championsResult, MYSQLI_ASSOC ) ) { $fieldarray=array('titleName','shortName','championID','championshortName','champion','con1ID','con1','con1shortName','con2ID','con2','con2shortName','con3ID','con3','con3shortName'); foreach ($fieldarray as $fieldlabel) { ${$fieldlabel} = $row[$fieldlabel]; } ?> <div id="title"><span class="large">< ?php echo $titleName ?></span> < ?php if (file_exists('images/championshots/'.$titleshortName.'/'.$championshortame.'.png')) { echo "<img class=champion src=images/championshots/".$shortName."/".$champion.".png alt= />\n"; } else { echo "<img class=champion src=images/championshots/".$shortName."/".$shortName.".png alt= />\n"; } if (strlen ($champion) < 1) { echo "<span class=medium>Vacant"; } else { echo "<span class=medium><a href=/bio?shortName=".$championshortName.">".$champion."</a></span>\n"; echo "<span class=medium>(Since TBD)</span>"; } ?> <span class="contender">Contenders</span> <ul> < ?php if ( strlen ($con1) < 1) { echo "<li><span class=medium>TBD</span>"; } else { echo "<li><a href=/bio?shortName=".$con1shortName.">".$con1."</a></li>\n"; } if (strlen ($con2) < 1) { echo "<li><span class=medium>TBD</span>"; } else { echo "<li><a href=/bio?shortName=".$con2shortName.">".$con2."</a></li>\n"; } if (strlen ($con3) < 1) { echo "<li><span class=medium>TBD</span>"; } else { echo "<li><a href=/bio?shortName=".$con3shortName.">".$con3."</a></li>\n"; } ?> </ul> </div> < ?php } ?> </div> Similar TutorialsHey guys,
Earlier I mentioned making search criteria in my database. I also had to put in something to make it add to the database. I made some code like this:
<form id="form" name="form" method="post" action="Website2.php">
<p>
<label for="Genre">Genre</label>
<input type="text" name="Genre" id="Genre" />
</p>
<p>
<label for="Naam">Naam</label>
<input type="text" name="Naam" id="Naam" />
</p>
<p>
<label for="Jaar">Jaar</label>
<input type="text" name="Jaar" id="Jaar" />
</p>
<p>
<label for="Regisseur">Regisseur</label>
<input type="text" name="Regisseur" id="Regisseur" />
</p>
<p>
<input type="submit" name="Verzenden" id="Verzenden" value="Verzenden" />
</p>
</form>
<?php
$Genre = $_POST["Genre"];
$Naam = $_POST["Naam"];
$Jaar = $_POST["Jaar"];
$Regisseur = $_POST["Regisseur"];
/*Hier hoeft geen verbinden meer gemaakt te worden met de database*/
if (isset($_POST['Verzenden']) && trim($_POST['Verzenden'])!=''){
$sql = "INSERT INTO
Movies (Genre, Naam, Jaar, Regisseur)
VALUES ('$Genre', '$Naam', '$Jaar', '$Regisseur')";
$resultaat = mysqli_query($db, $sql);
$verbreken = mysqli_close($db);
echo "De gegevens van $Naam zijn opgeslagen in de database.";}
else echo 'Hier kunt u iets toevoegen';
?>
However it didn't work.
I got the problem "Warning: mysqli_query() expects parameter 1 to be mysqli, null given in... on line 119"
Can you guys help please?
Hello
I am having an issue with encoding an array to JSON, my array seems to be formed correctly but when I try and echo the json_encode version I get "null". I echo the JSON error code and it is "0" indicating no issue..
I have read that non UTF-8 characters can be an issue, but this is simple data, no strange characters... but I added the utf8_encode anyway....
Here is my code:
$dataArray = array(
'title' => 'Test Item',
'icon_url' => 'http://url/api/testimage.jpg',
'item_id' => '12345',
'expiration' => 600
);
utf8_encode($dataArray);
$responsArray = array(
'success' => '1',
'data' => $dataArray
);
utf8_encode($responsArray);
header('Content-type: application/json');
echo print_r($responsArray);
echo json_encode ($responseArray);
echo json_last_error();
Hi all,
I'm posting JSON to an API call. If I use a var_dump within the API it will return correctly, although if I try to var_dump within the function it will return NULL.
API:
$json = file_get_contents('php://input');
if(isset($_GET['form_id']) && intval($_GET['form_id'])>0)
{
if(isset($json))
{
$records = $questionnaire->logFormAnswers($_POST['answers'],$_GET['form_id']);
Then using the logFormAnswers function...
Function:
public function logFormAnswers($answers = "", $form_id = 0)
{
$answers = json_decode($json,true);
var_dump($answers);
The JSON:
[{"question_id":"3","answer":"dsf"},{"question_id":"3","answer":"sdfsdfdsfsd"}]
Eventually, once the JSON has been decoded, it will be inserted into a database within the function.
Trying to figure out how to better write this chunk of code. I'm wanting to get the list of the roster members and then create an array of options for the view dropdown to display inside the select dropdown and also have it have an option to display "Please Select An Option". However what if what is returned from the getAllRoster function is NULL which is what I have returned if no results are returned from a query. How should I handle that which I just want the empty option displayed. Also I need to think about is do a function to retrieve all the allies for that specific matter and then display that ally as the default ally in the dropdown for each dropdown. Controller: Code: [Select] $rosterList = $this->bios->getAllRoster(); $allies = array(); $allies[''] = 'Please Select An Opion'; foreach ($rosterList AS $ally) { $allies[$ally->id] = $ally->rosterName; } View: Code: [Select] <?php echo form_label( 'Ally 1', 'ally1'); ?> <div> <?php echo form_dropdown( 'ally1', $allies, ''); ?> </div> <?php echo form_label( 'Ally 2', 'ally2'); ?> <div> <?php echo form_dropdown( 'ally2', $allies, ''); ?> </div> <?php echo form_label( 'Ally 3', 'ally3'); ?> <div> <?php echo form_dropdown( 'ally3', $allies, ''); ?> </div> Ok, this may be just because I have been programming all day and my mind has gone blank (happens alot), but this is my PHP script: Code: [Select] <?php $query_distinct_item_types = mysql_query("SELECT DISTINCT name FROM item_types"); while($item_types = mysql_fetch_array($query_distinct_item_types)){ $distinct_item_types[] = $item_types['name']; } foreach($distinct_item_types as $item){ $query_item_total = mysql_query("SELECT item_type, SUM(price) WHERE item_type='$item' FROM costs GROUP BY item_type"); while($item_total = mysql_fetch_array($query_total_price)){ $item_totals[] = $item_total['SUM(price)']; } } $item_summery = $item_totals; ?> $item_summery which is = to $item_totals is returning null, any idea's? Hello, I'm trying to take the value from an HTML form and insert it into a database on a button click. It inserts a null value into the database. The script is called submitColumnDetails.php. This is where I create the text field that I want to take the information from. This is in a separate file. Code: [Select] echo <<<END <form action = "submitColumnDetails.php" method = "POST"> <input type = "text" name = "columnField"/> </form> END; This is the submitColumnDetails.php file Code: [Select] <?php //Submit Column Data //-----------------------------------------------------// //Connect to localhost server $connector = mysql_connect("localhost", "root", "root"); if(!$connector){ //If user can't connect to database die('Could not connect: ' . mysql_error()); //Throw an error } //-----------------------------------------------------// mysql_select_db("colin_db", $connector); $newValue = $_POST["columnField"]; //Data from column field. THIS IS WHAT RETURNS NULL $newColumnQuery = "INSERT INTO `colin_db`.`allColumnFields` (`DATA`) VALUES ('$newValue')"; mysql_query($newColumnQuery); //Query to add form to main table $newColumnIntoMainTableQuery = "ALTER TABLE colin_table ADD ('$newValue' varchar(50))"; mysql_query($newColumnIntoMainTableQuery); //Query to add column to main table mysql_close($connector); //Close database connection echo "<script type = 'text/javascript'> window.location = 'index.php'</script>"; //Go back to original page ?> Even when I print out the $newValue, it does not print anything. What am I doing incorrectly? I wrote a simple mysql record update.
mysqli_query($link, "UPDATE photo01 SET PhotoText='abcd' WHERE id= '1' ");
Work fine until I use a variable:
mysqli_query($link, "UPDATE photo01 SET PhotoText= $ZZ WHERE id= $x ");
If $ZZ contains only number it updated the cell.
ANY letters does not.
The cell is text and standard US ASCII
I tried:
mysqli_query($link, "UPDATE photo01 SET PhotoText={$ZZ} WHERE id={$x}");
same problem
Here is my loop:
<?php for ($x = 0; $x < 131 ;$x++) { $ZZ = $_POST["z".$x]; if ($ZZ != '') { mysqli_query($link, "UPDATE photo01 SET PhotoText={$ZZ} WHERE id={$x}"); } } ?> After overcoming this-> error "Deprecated: mysql_connect(): The mysql extension is deprecated and will be removed in the futu use mysqli or PDO instead in this-> $myconnection = mysql_connect($server, $user, $pass);" by changing mysql_query to mysqli_query($con, $query), I started getting this specific error on line 106= $result = mysqli_query($myconnection, $query) or $this->debugAndDie($query);
Warning: mysqli_query() expects parameter 1 to be mysqli, null given on line 106=="$result = mysqli_query($myconnection, $query) or $this->debugAndDie($query);"
This header wont go, I tried $myconnection = mysql_connect($query); and it says expecting 2 parameters.
Please I'm open to ideas to solve this,
<snip - code is posted in next post>
Edited by mac_gyver, 05 June 2014 - 03:28 PM. removed code not in code tags as the op posted it later Hi guys, I have coded this, however I have received an error msg, can someone advice me in this? Thank you Warning: mysqli_query() expects at least 2 parameters, 1 given in D:\inetpub\vhosts\abc.com\httpdocs\report.php on line 166 Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in D:\inetpub\vhosts\abc.com\httpdocs\report.php on line 167 <?php $sql = mysqli_query("SELECT * FROM tutor_preferred_district ORDER BY district_id ASC"); while($data = mysqli_fetch_array($sql)) { echo '<input name="district" type="checkbox" id="'.$data['district_id'].'" value="'.$data['district_id'].' class="required" title="Please check at least 1 location."> <label for="'.$data['district_id'].'">'.$data['district_name'].'</label>'; } ?> Hi, I'm by no means an expert in php but I use and continually try and figure out how to update some very old soccer stats scripts that break from time to time when newer versions of php are released. Anyway, I would appreciate any pointers with this error please: Warning: mysqli_query() expects parameter 3 to be integer, object given in /blah/blah/blah.php on line 81 The code is: 79 - mysqli_query($connection,"INSERT INTO seasons SET 80 - SeasonID = '$seasonid', 81 - SeasonPlayerID = '$player_id'",$connection) 82 - or die(mysqli_error($connection)); Server is running php 7.2.28 Thanks in advance. Hi I am new to php, I am trying to capture the url and place into a variable but I only get the 1st digit to show, I just cant see what I am doing wrong. Sorry to ask such a basic question but I just can't work it out, I have attached a screen shot of all me code, your help would be very very much appreciated. I am trying to update the database with isset to set the value of the variable to either null or the value entered by the user. Currently, if no value has been entered null is being set as a string causing the date field to populate with zeros. How do I code it so that it populates or sets the value as null and not as string null?
Here's my code: (yes I know sql injections it's still in development )
<?php I'm trying to get this to either show "Final" or "Final Overtime" but it just continues to show "Final" even when the XML key value says Overtime...
What am I missing here?
Thank you!
Ok I have this code retrieving information from mysql: if (isset($memid)) { $query_meminfo = "SELECT * FROM ".MEMBER_PROFILE_TABLE." WHERE `user_id`=".$memid; $query_result = mysql_query($query_meminfo, $db); $MemName = mysql_result($query_result, 0, 'display_name'); $MemGroup = mysql_result($query_result, 0, 'Group'); $LevelQuery = "SELECT group_level FROM ".MEMBER_GROUPS." WHERE group_id='".$MemGroup."'"; $LevelResult = mysql_result(mysql_query($LevelQuery, $db), 0, 'group_level'); $MemLevel = intval($LevelResult); } It is definitely retrieving the correct info as it is correctly checking member levels to display certain information. Part 2 of code: <table width="100%" id="userlinks" cellspacing="6"> <tr> <td>Welcome <?php if (isset($MemName)) { Echo $MemName + " "; ?> (<a href="index.php?act=logout">Logout</a> <?php if ($MemLevel >= 9000) { echo " | <a href='admin.php'>Admin CP</a>"; } ?> ) <?php } Else { ?> Guest (<a href="index.php?act=login">Login</a> | <a href="index.php?act=register">Register</a>) <?php } ?> </td> </tr> </table> For some reason, instead of displaying the person's display name it is just displaying the number 0. i am having issues returning all sent messages. it will only return one for some reason. Code: [Select] //If there are sent messages, display them if ($row = mysql_num_rows($result) > 0) { $row = mysql_fetch_array($result) or die(mysql_error()); //Open table and create headers echo "<table border=\"1\">\n"; echo " <tr>\n"; echo " <th>Recipient</th>\n"; echo " <th>Subject</th>\n"; echo " </tr>\n"; while(mysql_fetch_array($result)) { //Show messages $userIDTo = $row['userIDTo']; // Get the recipient's ID number $recipient = checkRecipient($userIDTo); // Get the sender's Username $messageID = $row['ID']; echo " <tr>\n"; echo " <td>{$recipient}</td>\n"; echo " <td><a href='messageDetails.php?messageID=$messageID' target='_blank'>{$row['subject']}</a></td>\n"; echo " <tr>\n"; } thanks in advance. SELECT i.item_id, i.title, i.price, i.p_and_p, SUM(i.price + i.p_and_p) AS `total_price`, i.listing, i.condition, i.start_date_time, i.listing_duration, CONVERT_TZ(DATE_ADD(i.start_date_time, INTERVAL concat(i.listing_duration) DAY), '+00:00', u.time_zone) AS `end_date_time` FROM items i LEFT JOIN sub_categories sc ON sc.sub_category_id = i.sub_category_id LEFT JOIN categories c ON c.name = 'test' JOIN users u WHERE u.username = 'Destramic' AND i.start_date_time < NOW() AND DATE_ADD(i.start_date_time, INTERVAL concat(i.listing_duration) DAY) >= NOW()I'm having a problem with my query returning more than 1 rows...I've even copied the row which is returning to see if that'll return 2 rows but it doesn't can anyone explain why this is happening please? Hi All Please take a look at the code below. I am sucessfully accessing a database and pulling the details through into a HTML form with a drop down. However I then want to use a variable (test) in another page. However I cannot get the variable userrow['userid'] to send a value to my getquiz.php. What am I doing wrong? <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML> <HEAD> <TITLE> New Document </TITLE> <META NAME="Generator" CONTENT="EditPlus"> <META NAME="Author" CONTENT=""> <META NAME="Keywords" CONTENT=""> <META NAME="Description" CONTENT=""> </HEAD> <?php include_once "db.inc.php"; ?> <body> <form action="getquiz.php" method="GET"> <tr><td class="title">Please Select User</td> <td><select name="test"> <?php $courseid = mysqli_query($link, 'SELECT id FROM mdl_quiz WHERE course = "225"; '); if (!courseid) { $error = ' Error fetching course id: ' . mysqli_error($link); include 'error.html.php'; exit(); } else { while($row = mysqli_fetch_array($courseid)) { $user = mysqli_query($link, 'SELECT userid FROM mdl_quiz_attempts WHERE quiz =' . $row['id']); while ($userrow = mysqli_fetch_array($user)) { $names = mysqli_query($link, 'SELECT firstname,lastname FROM mdl_user WHERE id=' . $userrow['userid']); while ($usernames = mysqli_fetch_array($names)) {?> <option value="<?php $userrow['userid']?>"> <?php echo $usernames ['firstname'] . " " . $usernames['lastname'] . $userrow['userid']; ?> </option> <?php } } } } ?> </select> </td></tr> <center><input type="submit" value="Submit"></center> </table> </form> </body> </html> My aim is for this what if the table is empty which means no results will be retrieved in the first query then I want it to repeat the $testArray as "TBD" as well. Code: [Select] function getTop5() { $this->db->select('character1_id, character2_id, character3_id, character4_id, character5_id'); $this->db->from('site_top5'); $this->db->where('status_id', '1'); $this->db->order_by('id', 'desc'); $query = $this->db->get(); $row = $query->row(); $ids = array( $row->character1_id, $row->character2_id, $row->character3_id, $row->character4_id, $row->character5_id ); $testArray = array(); foreach ($ids as $id) { if($id !== "0") { $this->db->select('character_name'); $this->db->from('characters'); $this->db->where('id', $id); $query = $this->db->get(); $row = $query->row(); $testArray[] = $row->character_name; } else { $testArray[] = "TBA"; } } return $testArray; } hello, anybody able to help me with why this is only returning the first staff member's hours? Code: [Select] <?php if(isset($_POST['view'])) { $y3=$_POST['y']; $m3=$_POST['m']; $d3=$_POST['d']; $pdate=$y3."-".$m3. "-".$d3; $pdate1 = date( 'D M j', strtotime($pdate) ); } else { $pdate = date('Y-m-d', strtotime("-1 day") ); $pdate1 = date( 'D M j', strtotime($pdate) ); } echo "<table border='1' style='border-collapse: collapse' bordercolorlight='#000000' bordercolordark='#000000' width='98%' align='center'>"; echo "<tr><td width='100%' colspan='9' align='center'><b>Timesheets For $pdate1</b></td></tr>"; $result = mysql_query("SELECT * FROM staff ORDER BY name"); while($row = mysql_fetch_array($result)) { $eid=$row['eid']; $name=$row['name']; echo "<tr>"; echo "<td align='left' colspan='9'><b>" . $name . "</b></td>"; echo "</tr>"; echo "<tr> <th align='center'>Date</th> <th align='center'>Job Number</th> <th align='center' width='30%'>Service Report</th> <th align='center'>Sign In Time</th> <th align='center'>Sign Out Time</th> <th align='center'>Lunch</th> <th align='center'>Time Billed</th> <th align='center'>Estimated</th> </tr>"; $result3 = mysql_query("SELECT * FROM timesheet WHERE date = '$pdate' AND eid = '$eid'"); $talltime = 0; while($row3 = mysql_fetch_array($result3)) { $tid=$row3['id']; $tdate=$row3['date']; $tjobnumber=$row3['jobnumber']; $teid=$row3['eid']; $tdescription=$row3['description']; $tsignin=$row3['start']; $tfinish=$row3['finish']; $tlunch=$row3['lunch']; $tkms=$row3['kms']; $tschednum=$row3['schednum']; $tdate1 = date( 'M j, Y', strtotime($tdate) ); $tsignin1 = date( 'g:i a', strtotime($tsignin) ); $tfinish1 = date( 'g:i a', strtotime($tfinish) ); if( empty($tfinish) ) { $tfinish2="<i>In Progress"; } else { $tfinish2="$tfinish1"; } $shortid=substr($tjobnumber, 5, -1); //remove the first 5 characters and minus the last character $result = mysql_query("SELECT * FROM jobs WHERE id = '$shortid'"); while($row = mysql_fetch_array($result)) { $jstatus=$row['status']; } $result4 = mysql_query("SELECT * FROM schedule WHERE id = '$tschednum'"); while($row4 = mysql_fetch_array($result4)) { $sid=$row4['id']; $sdate=$row4['date']; $seid=$row4['eid']; $sjobnumber=$row4['jobnumber']; $sstarttime=$row4['starttime']; $sstatus=$row4['status']; $setime=$row4['etime']; } $log_in_time_string = strtotime($tsignin); $log_out_time_string = strtotime($tfinish); $difference_in_seconds = ($log_out_time_string - $log_in_time_string); $tbilled = ($difference_in_seconds / 3600); if($tbilled < 0) { $tbilled1 = $tbilled + 24; } else { $tbilled1=$tbilled; } $tbilled2 = number_format(round($tbilled*4)/4,2); $tbilled3 = $tbilled2 - $tlunch ; $talltime += $tbilled3; echo "<tr>"; echo "<td align='center'>" . $tdate1 . "</td>"; echo "<td align='center'>" . $tjobnumber . "</td>"; echo "<td align='center'>" . $tdescription . "</td>"; echo "<td align='center'>" . $tsignin1 . "</td>"; echo "<td align='center'>" . $tfinish2 . "</td>"; echo "<td align='center'>" . $tlunch . " hour</td>"; echo "<td align='center'>"; if ($tbilled3 > $setime ) { echo "<font color='red'><b>*** " . $tbilled3 . " hours ***</b></font>"; } else { echo "" . $tbilled3 . " hours"; } echo "</td>"; echo "<td align='center'>" . $setime . " hours</td>"; echo "</tr>"; } echo "<tr>"; echo "<td colspan='6' align='right'><b>Totals :</td>"; echo "<td align='center' colspan='2'>"; if ($talltime > "8" ) { echo "<font color='red'><b>*** " . $talltime . " hours ***</b></font>"; } elseif ($talltime < "0" ) { echo "<font color='red'><b>Not Signed Out</b></font>"; } else { echo "" . $talltime . " hours"; } echo "</td>"; echo "</tr>"; echo "<tr>"; echo "<td align='center' colspan='8' bgcolor='#D9FFD9'><hr></td>"; echo "</tr>"; } echo "</table>"; include 'close.php'; ?> Hey there, Now I was curious if there was a way I could return an array in my function like the following, I currently get a error though: Code: [Select] Catchable fatal error: Object of class panel_accounts could not be converted to string in C:\wamp\www\X-HostLTD - Panel\PanelFrontend\index.php on line 21 But can you only return strings?. Here is the function that it errors on: function returnAccountInformation($AccountID) { return mysql_fetch_array(mysql_query("SELECT * FROM ****_***** WHERE *****_***** = '".mysql_escape_string($AccountID)."'")); } Could somebody help me on the matter?, Thank you. P.S sorry for staring out the table names but I don't really want people knowing my database structure hehe. |
|||||||