PHP - Just Spent An Hour Typing This Code, Getting A Db Query Error For No Reason!

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?

This topic has been remastered in brilliant HD quality in JavaScript Help.

http://www.phpfreaks.com/forums/index.php?topic=356870.0

Hi, suppose the user requests a page www.foo2.php. Now if it is a link embedded in another page, say www.foo1.php, then Javascript can send the server browser info before furnishing the page. But is it possible to let the server know which browser is being used when the user requests the page by just typing the url http://www.foo2.php and hitting enter?

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.

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?

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 />";}

 ?>
 

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.

&#65279;&#65279;
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')

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!

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);
?>

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 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();
}
?>

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'
Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean given in /home/csi/public_html/resources/preview1920/sectional.php on line 19

 

Passing variable via the URL:

https://www.courtsideindiana.com/season-preview/19-20/sectional1920/?sectional=8&amp;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>';

}