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PHP - Error In Query.
Hi, i'm currently coding a new inbox for my website, but ive got an error which says:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'to='NoName' ORDER BY id DESC' at line 1 Im not sure why ive got that as everything seems to be fine :S My Code: $user=$_SESSION['username']; $get_messages = mysql_query("SELECT `id` FROM `inbox` WHERE to='$user' ORDER BY `id` DESC") or die("Error on line 9 - " . mysql_error()); Thanks for any help/advise given. Similar TutorialsHere is my code: // Start MySQL Query for Records $query = "SELECT codes_update_no_join_1b" . "SET orig_code_1 = new_code_1, orig_code_2 = new_code_2" . "WHERE concat(orig_code_1, orig_code_2) = concat(old_code_1, old_code_2)"; $results = mysql_query($query) or die(mysql_error()); // End MySQL Query for Records This query runs perfectly fine when run direct as SQL in phpMyAdmin, but throws this error when running in my script??? Why is this??? Code: [Select] You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '= new_code_1, orig_code_2 = new_code_2WHERE concat(orig_code_1, orig_c' at line 1 A friend of mine must of changed something on the site while I was asleep last night and now all the site says when you go to it is: Error Database query error Warning: mail() [function.mail]: SMTP server response: 530 SMTP authentication is required. in C:\xampp\htdocs\inc\utils.inc.php on line 449 I'm not exactly sure what he did since I can't contact him. Can anyone help me fix this? Code: [Select] mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean i get that error with this code: $online_query = $link->simple_query('u_username', 'users', 'u_online=1 AND u_hidden != 1', '0u_username'); while($online_info = $link->fetch_array($online_query)) //THIS LINE THROWS THE ERROR the simple_query function is: function simple_query($fields, $table, $clause, $order) { global $link, $config; if(!empty($clause)) { $clause = "WHERE $clause"; } else { $clause = ''; } if(!empty($order)) { $direction = $order[0]; switch($direction) { case '0': $direction = 'ASC"'; break; case '1': $direction = 'DESC'; break; } $order = substr($order, 1, strlen($order)); $order = "ORDER BY $order $direction"; } $query = mysqli_query($this->link, "SELECT $fields FROM ".TBL_PREFIX."$table $clause $order"); return $query; } when i use a normal query instead of my simple_query function it works fine. it also worked fine before i made the database class. Where am i going wrong? Hello, i have one little problem and can't pass it. Problem is i need to call new sql query inside another query. Am trying to make accordion which will put result of first query($sql) like title and result of second query($sql2) like list of current item. All time am getting error : Quote Warning: odbc_exec() [function.odbc-exec]: SQL error: [Microsoft][ODBC SQL Server Driver]Connection is busy with results for another hstmt, SQL state S1000 in SQLExecDirect in ..... I know reasone is because i use query inside query so am trying to figure is there any way to bypass it or make it work. Example: Code: [Select] $sql="EXECUTE _PROCEDURE1 '".$date."',''.$code."; $rs=odbc_exec($conn,$sql); if (!$rs){exit("Error in SQL");} while (odbc_fetch_row($rs)){ $id_number=odbc_result($rs,"ID"); $name=odbc_result($rs,"NAME"); echo $id_number.' - '.$name; $sql2="EXECUTE _PROCEDURE2 '".$id_number."',''.$name."; $name=odbc_exec($conn,$sql2); while(odbc_fetch_row($popust)){ $detail = odbc_result($sql2,"DETAILS"); $detail2 = odbc_result($sql2,"DETAILS2"); $detail3 = odbc_result($sql2,"DETAILS3"); echo $detail.' - '.$detail2.' - '.$detail3; } } I hope i explained it well. Thanks.  Warning: mysqli_query() expects parameter 1 to be mysqli, null given in /home/omerbsh/hallofblogs.com/autoTwit/Library/database.php on line 13 Warning: mysqli_error() expects exactly 1 parameter, 0 given in /home/omerbsh/hallofblogs.com/autoTwit/Library/database.php on line 13 Error: The Query:INSERT INTO twitter_profiles VALUES('dfggffg','dgfgfdg','1') Hi all, having a strange problem with my query Its only returning some of my data, and in the format User 1 <br /> <br /> User 2 <br /> <br /> <br /> <br /> Code: [Select] $newmembers = "SELECT * FROM users WHERE linked_user IS NOT NULL ORDER BY datejoined LIMIT 6"; $nmresult = mysql_query($newmembers); while($row = mysql_fetch_array($nmresult)){ echo $row['linked_user']; echo "<br />";} ?> Hello all, I am having this very frustrating issue, I am trying to print this query i got from my SQL table. However, If the query returns 10 [1,2,3,4,5,6,7,8,9] results it just prints 5 [2,4,6,8,10] . I have no idea what to do. Code: [Select] <!--Error Reporting Production Only[start]--> <?php error_reporting(E_ALL & ~E_NOTICE); ini_set('display_errors','1'); //Error Reporting Production Only[end]--> //Connect to Database [start]--> require_once 'sqllogin.php'; $db_server = mysql_connect($db_hostname, $db_username, $db_password); if (!$db_server) die("Unable to connect to MySQL: " . mysql_error()); mysql_select_db($db_database) or die("Unable to select database: " . mysql_error()); //Connect to Database [end]--> //Process the search [start]--> //Initialize the search output variable $search_output = ""; $sql_command = "SELECT * FROM `Alpha` WHERE 1"; //See if the posted search field is set and has a value if (isset($_POST['searchquery']) && $_POST['searchquery'] !="") {//IF START 1 //run code if condition is met //Filter the search query user input $searchquery = preg_replace('#[a-z 0-9 A-Z]#i', '', $_POST['searchquery']); // Search Query********************* if($_POST['filter1'] == "All") {//all filter code goes her $sql_command = "SELECT id, lastname, firstname, building, room FROM `Alpha` WHERE firstname LIKE '%$_POST[searchquery]%' OR lastname LIKE '%$_POST[searchquery]%' OR id LIKE '%$_POST[searchquery]%'"; } else if($_POST['filter1'] == "Last Name") {//Last Name filter code goes her $sql_command = "SELECT id, lastname, firstname, building, room FROM `Alpha` WHERE lastname LIKE '%$_POST[searchquery]%'"; } else if($_POST['filter1'] == "First Name") {//First Name filter code goes her $sql_command = "SELECT id, lastname, firstname, building, room FROM `Alpha` WHERE firstname LIKE '%$_POST[searchquery]%'"; } else if($_POST['filter1'] == "First Name") {//ID filter code goes her $sql_command = "SELECT id, lastname, firstname, building, room FROM `Alpha` WHERE id LIKE '%$_POST[searchquery]%'"; } }//IF END 1 $query = mysql_query($sql_command) or die(mysql_error()); $count = mysql_num_rows($query); if ($count>0) { $search_output.= "<hr /> $count results for <strong>$_POST[searchquery]</strong><hr /> $sql_command <hr />**IF**POST = $_POST[searchquery] Count=$count, Query= $query<hr />"; /*for ($j = 0 ; $j < $rows ; ++$j) { $id = $row["id"]; $lastname = $row["lastname"]; $firstname = $row["firstname"]; $building = $row["building"]; $room = $row["room"]; $search_output .= "$id \t $firstname \t\t $lastname \t\t $building \t$room<br />"; } *************** Can be used instead but not here try putting it where $search_output is echo 'ID: ' . $row[0] . '<br />'; echo 'Last Name: ' . $row[1] . '<br />'; echo 'First Name: ' . $row[2] . '<br />'; echo 'Building: ' . $row[3] . '<br />'; echo 'Room: ' . $row[4] . '<br /><br />'; *********************** */ while($row = mysql_fetch_array($query)) { $row = mysql_fetch_array($query); $id = $row[0]; $lastname=$row[1]; $firstname=$row[2]; $building=$row[3]; $room=$row[4]; $search_output .= "$id $firstname $lastname $building $room<br />"; } } else { $search_output ="<hr /> 0 results for <strong>$searchquery</strong><hr /> $sql_command <hr />****ELSE**POST = $_POST[searchquery] Count=$count, Query= $query<hr /" ; } ?> <!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"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>MC Reslife WEB APP Developed by; JB</title> <link href="styles.css" rel="stylesheet" type="text/css" /> </head> <body> <div id="Header">Monroe College Resident's Life Web App</div> <div id="Page"> <div id="Menu"> <h1>Search</h1> <form action="<?php echo $_SERVER['PHP_SELF'];?>" method="POST"> <input name="searchquery" type="text" maxlength="19" size="19"/> <input name="myBTN" type="submit" /> <br /> Search By: <select name="filter1"> <option value="All">All</option> <option value="Last Name">Last Name</option> <option value="First Name">First Name</option> <option value="ID Number">ID Number</option> </select> </form> <hr /> <h1>Menu</h1> <p><a href="#">Alpha List</a></p> <p><a href="#">Lock Out Log</a></p> <p><a href="#">Current Probations</a></p> </div> <div id="Content"> <h1>Alpha List</h1> <?php echo $search_output; ?> <p>*************************</p> <?php echo "this is what in post $_POST[searchquery]"?> </div> </div> <div id="Footer">Content for id "Footer" Goes Here</div> </body> </html> please help me Quote
Unknown column 'Carmel' in 'where clause'
Passing variable via the URL: https://www.courtsideindiana.com/season-preview/19-20/sectional1920/?sectional=8&school=Carmel Sectional = 8 School = Carmel Before I added the &school=Carmel, it was working, just echoing the total list of schools in the table.
$sectional = $_GET['sectional'];
$school = $_GET['school'];
echo $school;
// Query
$query = "SELECT * FROM a_schools WHERE sectional=".$sectional." AND school=" . $school ."";
$results = mysqli_query($con,$query);
echo mysqli_error($con);
while($row = mysqli_fetch_assoc($results)) {
echo $row['school'] . '<br>';
}
Hi all !,
I am stuck on the following piece of code which does not give an error nor does it give a result. ( i.e. it gives 0 num_rows which should be > 1).
If, however, I execute the query in phpmyadmin by simply substituting the values of $pp,$ll and $room_no in the query it gives the correct result.
Please can someone tell me what I may be doing wrong here. Thanks !
$fcon = mysqli_connect($db_host,$db_user,$db_pass,$db_database) or die('Unable to establish a DB connection');
$pp = "(ms.level = 'Beginner' || ms.level = 'Intermediate')";
$ll = 'ms.diff <= 7';
$room_no = 4;
$query = "SELECT md.Member_reg_id, md.fname, md.lname, md.email, md.cell,
ms.level, ms.diff, ms.score,
r.ID_Status
FROM register as r
JOIN member_detail as md ON r.ID = md.Member_reg_id
JOIN memstatus as ms On r.ID = ms.ID
WHERE r.CENTERCODE = ? AND r.ID_Status ='A' AND ? AND ?
ORDER by level, diff, score DESC";
$stmt=$fcon->prepare($query);
$stmt->bind_param('iss',$room_no,$pp,$ll);
if(!$stmt->execute()) die('Failed to execute the query'.$fcon->error);
else
{
echo "Executed";
$stmt->bind_result($Member_reg_id,$fname,$lname,$email,$cell,$level,$diff ,$score,$ID_Status);
$numrows = $stmt->num_rows;
$stmt->store_result();
// echo $numrows;
while($stmt->fetch())
{
echo "<br>".$fname.' '.$lname;
echo "<br>".$level;
echo "<br>".$diff;
echo "<br>".$score;
echo "<br>".$cell;
echo "<br>".$email;
}
}
Edited by ajoo, 03 January 2015 - 08:00 AM. I am getting an empty query error on this line Code: [Select] $mysql->query($insert); here is my code Code: [Select] if(isset($_POST['submit'])){ $firstName = $_POST['firstName']; $lastName = $_POST['lastName']; $companyName = $_POST['companyName']; $homePhone = $_POST['homePhone']; $cellPhone = $_POST['cellPhone']; $companyPhone = $_POST['companyPhone']; //checking the values are filled //echo $firstName. " " . $lastName . " " . $companyName . " " . $homePhone . " " . $cellPhone . " " . $companyPhone; $insert = $mysql->query("INSERT INTO names('firstName','lastName','companyName'), phone('home','cell','company') values('$firstName','$lastName','$companyName','$homePhone','$cellPhone','$companyPhone'"); $mysql->query($insert); if($insert){ echo "success"; } else { mysql_error(); } } I got a connection to the database, I tested for it. I can't find the problem. this is the error I get Warning: mysqli::query() [mysqli.query]: Empty query in C:\wamp\www\test\formData.php on line 23 which is this line Code: [Select] $mysql->query($insert); Here's my build a querty code... Code: [Select] $sql = "UPDATE $tablename SET DIST_PART_NUM = '$dist_part_num', DIST_PUB = '$dist_pub', MFR_PART_NUM = '$mfr_part_num', MFR_PUB = '$mfr_pub', ITEM_DESC = '$item_desc', ITEM_STD_DESC = '$item_std_desc', ITEM_COPY = '$item_copy', COST = '$cost', GP_MULT = '$gp_mult', ITEM_IMAGE = '$item_image', ITEM_URL = '$item_url', COUNTRY_OF_ORIG = '$country_of_orig', CATEGORY_ID = '$category_id', PICGROUP_KEY = '$picgroup_key', UPC_CODE = '$upc_code', LEADTIME = '$leadtime', UNITS = '$units', LENGTH = '$length', WIDTH = '$width', HEIGHT = '$height', WEIGHT = '$weight', FLAG_HAZARDOUS = '$flag_haz', FLAG_LTL = '$flag_ltl', FLAG_NON_RETURNABLE = '$flag_non', CAMPAIGN_KEY = '$campaign_key', "; foreach($atr as $key => $value) { $sql .= $value . "='" . addslashes($_POST[$value]) . "',"; } $sql2 = rtrim($sql,","); $sql2 .= " WHERE DIST_PART_NUM = '$partnumber' OR MFR_PART_NUM = '$partnumber'"; Now the query get's built 100% fine that's not the issue... my issue is that sine if the table fields are named something like #_of_shelves... and the # sign causes a MySQL error... You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 26 How can i allow for the # sign, or any other character that can cause issues like () . , * ~ etc.... i tried encompassing the field name with single quotes 'fieldname' and with those slanted ones `tablename` but that did not help. Any ideas? Thanks! Here is the error I get ERROR: Could not execute query: INSERT INTO items (quickti, longti, desc, maincat, subcat1, subcat2, colour, size, style, price, stock, pic1, pic2, pic3, pic4, pic5, pic6, sold, saledate) VALUES ('quick', 'main title', 'description', 'Jewelry', 'Gemstones', 'Cubic Zirconia', 'Black', '50m', 'red', '1.50', '100', 'quick1', 'quick2', 'quick3', 'quick4', 'quick5', 'quick6', '0', '0'). You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'desc, maincat, subcat1, subcat2, colour, size, style, price, stock, pic1, pic2, ' at line 1 Here is the relevant code $sql = "INSERT INTO items (quickti, longti, desc, maincat, subcat1, subcat2, colour, size, style, price, stock, pic1, pic2, pic3, pic4, pic5, pic6, sold, saledate) VALUES ('$quickheading', '$heading', '$content', '$cat1', '$cat2', '$cat3', '$colour', '$size', '$style', '$total', '$stock', '$quick1', '$quick2', '$quick3', '$quick4', '$quick5', '$quick6', '0', '0')"; if I remove desc and $content from the query, then it works. desc is a text field. So I wrote a question/answer script, and I am having issues with inputting symbols. It works great it you only have letters, number, dots, commas, and likely other symbols. However query fails every time you insert certain symbols. Here is an error: Quote You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '" ' ''' \ | / ) // ')' at line 3 My test input was following: Quote ' " ' ''' \ | / ) // I am not sure which symbol causes this, and I need help fixing this. How do I go about this without stripping any symbols? Maybe change table structure somehow?
<?php
<!DOCTYPE html>
I am having an error with this code You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near ''name','email','password','profile') SET ('Sasural','kill@1234.com','kill','ANDK' at line 1 I am stuck with this for last 5hrs Please deal with me for a moment as I try to explain what is occurring. I have created drop down menu that you choose a name and the php/MySQL will run multiple queries to display information regarding the chosen name. The query works great. There is no problem with the query. However when you leave the page, either by navigation or by X'ing out. When you return there are two errors that pop up: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in You can then use the drop down menu and choose a name and it all works great. It is my belief that these errors occur due to how the query was written, as the two queries that have these errors both have math in them. My question is, how do I block these errors from occurring when you return to the page? Here is the query: <?php //Worst Regular Season Record include_once('../other/functions.php'); $con = mysql_connect($hostname, $username, $password) OR DIE ('Unable to connect to database! Please try again later.'); $db = mysql_select_db($dbname, $con); $query = "SELECT win, loss, year, teamname FROM standings, owners WHERE owners.owner_id = standings.owner_id AND win = (SELECT MIN(win) FROM standings) AND standings.owner_id = $thing ORDER BY year"; $result = mysql_query($query); $row = mysql_fetch_array($result); @ mysql_data_seek($result, 0); if(mysql_num_rows($result)>0) { echo "<table CELLPADDING=5 border =1>"; echo "<tr>"; echo "<th align=center colspan=4 > Worst Regular Season Record </th>"; echo "</tr>"; while ($row = mysql_fetch_array($result)) { echo "<tr>"; echo '<td align=center> In '.$row['year'].', the '.$row['teamname'].' had '.$row['win'].' wins, and '.$row['loss'].' losses.</td>'; echo "</tr>"; } } else { } echo "</table>"; mysql_close($con); ?> I have following error when i try to enter my student info into the database. ! ) Warning: Header may not contain more than a single header, new line detected. in C:\wamp\www\Student registration\new student registration\newStudentRegistrationFormvalidation.php on line 43 newStudentRegistrationFormvalidation.php <?php $admission_no=$_POST['admission_no']; $admission_date=$_POST['admission_date']; $full_name=$_POST['full_name']; $name_with_initial=$_POST['name_with_initial']; $date_of_birth=$_POST['date_of_birth']; $religion=$_POST['religion']; $address=$_POST['address']; $telephone=$_POST['telephone']; $grade_on_admission=$_POST['grade_on_admission']; $grade_ID=$_POST['grade_ID']; $stream_ID=isset($_POST['stream_ID']) ? $_POST['stream_ID'] :''; $class_ID=$_POST['class_ID']; $student_house=$_POST['student_house']; $password=$_POST['password']; $description_about_st=$_POST['description_about_st']; $payment=$_POST["payment"]; $currentdate=getdate(time()); $year=$currentdate["year"]; //admission number validation $answer=''; $con=mysql_connect("localhost","root",""); mysql_select_db("student_management",$con); $query="SELECT admission_no FROM student_info WHERE student_info.admission_no='$admission_no'"; $result=mysql_query($query); while($row=mysql_fetch_array($result)){ $answer=$row['admission_no'];} if($answer==0) { //line 43 header("location:student registrationDatabase.php?admission_no=".$admission_no."&year=".$year."&admission_date=".$admission_date."&full_name=".$full_name."&name_with_initial=".$name_with_initial."&date_of_birth=".$date_of_birth."&religion=".$religion."&address=".$address."&telephone=".$telephone."&grade_on_admission=".$grade_on_admission."&grade_ID=".$grade_ID."&stream_ID=".$stream_ID."&class_ID=".$class_ID."&student_house=".$student_house."&password=".$password."&description_about_st=".$description_about_st."&payment=".$payment); exit(); }else{ ?> <body> <?php echo "Admission number".$admission_no."This student has been alredy entered to the system ."."<BR>"."<BR>"."<BR>"; echo "<a href='newStudentRegistrationForm.php'>GO to manage student details page</a> "; exit(); }?> studentregistrationDatabase.php <?PHP $admission_no=$_POST['admission_no']; $admission_date=$_POST['admission_date']; $full_name=$_POST['full_name']; $name_with_initials=$_POST['name_with_initial']; $dob=$_POST['date_of_birth']; $religion=$_POST['religion']; //$gender=$_GET['gender']; $address=$_POST['address']; $telephone=$_POST['telephone']; $grade_on_admission=$_POST['grade_on_admission']; $present_grade=$_POST['grade_ID']; $stream=$_POST['stream_ID']; $present_class=$_POST['class_ID']; $student_house=$_POST['student_house']; $password=$_POST['password']; $description_about_st=$_POST['description_about_st']; $payment=$_POST["payment"]; $current=getdate(time()); $year=$current["year"]; $today = date("Y-m-d"); $con=mysql_connect("localhost","root",""); mysql_select_db("student_management",$con); //insert to database $query="select count(*) from student_info where admission_no='$admission_no'"; $result=mysql_query($query); $row=mysql_fetch_array($result); if($row[0]==0) { $query="insert into student_info values(null,'$admission_no','$admission_date','$full_name','$name_with_initials','$dob','$religion','$address','$telephone','$grade_on_admission','$password','$student_house','$description_about_st')"; $result=mysql_query($query); if($stream!=null){ $query="select class_id from class where class.grade_id='$present_grade' and class.class_name='$present_class' and class.stream='$stream'"; }else{ $query="select class_id from class where class.grade_id='$present_grade' and class.class_name='$present_class'"; } $result=mysql_query($query); while($row=mysql_fetch_array($result)){ $class_id=$row['class_id']; } $query="insert into student_class values('$admission_no','$class_id','$year')"; $result=mysql_query($query); if($payment=="2000") { $query="insert into payment_amount values(null,'2000','$today')"; $result=mysql_query($query); $query="SELECT amount_id FROM payment_amount order by amount_id DESC limit 1"; $result=mysql_query($query); while($row = mysql_fetch_array($result)) { $amount_id=$row['amount_id']; } $query="insert into payment values('$admission_no','$amount_id')"; $result=mysql_query($query); }else{ $query="insert into payment_amount values(null,'0','$today')"; $result=mysql_query($query); $query="SELECT amount_id FROM payment_amount order by amount_id DESC limit 1"; $result=mysql_query($query); while($row = mysql_fetch_array($result)) { $amount_id=$row['amount_id']; } $query="insert into payment values('$admission_no','$amount_id')"; $result=mysql_query($query); } header("location:../new student registration-parent details/studentRegistrationParentDetailsForm.php?admission_no=".$admission_no); exit(); } else{ ?> <?php echo "Admission number".$admission_no."This student has been already entered to the system"."<BR>"."<BR>"."<BR>"; echo "<a href='student registration.php'>GO to registration page</a>"; exit(); } ?> Im sure this is simple, but I cannot see what my problem is! I am hitting an error on my insert query Code: [Select] Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1 I know the output of $user_id is 1, so my error is on $mysavepath Code: [Select] $mysavepath = $folder.'/'.$worldname.'_'.date("dMjY"); echo $mysavepath; $savepath = mysql_query("INSERT INTO saves (user_id,savepath) VALUES ('$user_id','$mysavepath')"); echo '<br>'.$savepath; if(!mysql_query($savepath)) { die('<br>Error: ' . mysql_error()); } however it all echos out ok? Code: [Select] 188ea678f0dcdc8252aeb15e3c910408/world_15Jan152012 1 Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1 Can anyone see the problem? Cheers Dave Hi i have created a query where i want some columns to be retrieved. i get these 2 errors and i dont understand why.
Notice: Undefined variable: db
Fatal error: Call to a member function query() on a non-object in
this is my query
function get_products_all() {
$connect = get_connected_db();
try {
$results = $db->query(" ***** error on this line****
SELECT id, product, description, price, picc
FROM products ORDER BY id DESC
");
} catch (Exception $e) {
echo "Data could not be retrieved from the database.";
exit;
}
$result = $results->fetchALL(PDO::FETCH_ASSOC);
return $result;
}
what am i doing wrong?
when i run the below code i get an error...
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in...
but when i take out the h>0 then there is no errors in the code. in table users, i need all columns in the array. How to get h>0 working?
$query = "SELECT * FROM users WHERE h>0 ORDER BY id";
$result = mysqli_query($link, $query);
$data = array();
while($row1 = mysqli_fetch_array($result))
{
$data[] = $row1;
}
print_r($data);
Edited by kalster, 27 October 2014 - 07:38 PM. This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=325938.0 |
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