PHP - Syntax Error In Defining Variable

Ok this is puzzleing. I am geting "Could not delete data: 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". but its is deleting the entry that needs to be removed. The "1" is the entry. Just not sure what is causing the error. I do have another delete php but I have put that on the back burning for the time being.     <?php

         $con = mysqli_connect("localhost","user","password","part_inventory");
    // Check connection
if (mysqli_connect_errno()) {
  printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
} else {
    
    $result = mysqli_query($con, "SELECT * FROM amp20 ");
    
    $amp20ptid = $_POST['amp20ptid'];                // escape variables for security
    $amp20ptid = mysqli_real_escape_string($con, $_POST['amp20ptid']);
    
    mysqli_query($con, "DELETE FROM amp20 WHERE amp20ptid = '$amp20ptid'");
      
    
if (!mysqli_query($con, $amp20ptid));

{
 die('Could not delete data: ' . mysqli_error($con));    
}
 echo "Part has been deleted to the database!!!\n";
    mysqli_close($con);
}
?>

Hi guys   I have this code below and all works fine when submitting this online application apart from when someone types either ' # & into one of the comment fields in which it throws up the error. Have tried various fixes from across the internet but no joy. Can anyone offer suggestions?   <?php $con = mysql_connect("localhost:3306","root","password"); if (!$con)   {   die('Could not connect: ' . mysql_error());   } mysql_select_db('sfapp', $con); $sql="INSERT INTO 'sfapp' ('surname_add','forename_add','dob_add','hometele_add','mobiletele_add','homeadd_add','siblings_add','schoolname_add','headname_add','schooladd_add','schooltele_add','schoolem_add','alevel_add','personstate_add','nameprovided_add','pe_add','se_add','PredGrade_Art','PredGrade_AScience','PredGrade_BusStudies','PredGrade_Electronics','PredGrade_EnglishLang','PredGrade_EnglishLit','PredGrade_French','PredGrade_German','PredGrade_Geog','PredGrade_Graphics','PredGrade_History','PredGrade_Maths','PredGrade_SepScience','PredGrade_ProductDesign','PredGrade_Spanish','PredGrade_Other','Gender_Male','Gender_Female','Sub_EnglishLit','Sub_Maths','Sub_FurtherMaths','Sub_Biology','Sub_Chemistry','Sub_Physics','Sub_French','Sub_German','Sub_Spanish','Sub_Geography','Sub_History','Sub_RE','Sub_FineArt','Sub_Business','Sub_Computing','Sub_GlobPersp','Sub_DramaAndTheatre','Sub_PE','Sub_Dance','Sub_Politics','Sub_Psychology','Sub_Sociology','readprospect_chk','Sib_Yes','Sib_No','Current_Student_Yes','Current_Student_No','I_Understand_chk','Current_Education_chk','Local_Care_chk','Staff_Cwhls_chk','Sub_Film')   VALUES ('$_POST[surname_add]','$_POST[forename_add]','$_POST[dob_add]','$_POST[hometele_add]','$_POST[mobiletele_add]','$_POST[homeadd_add]','$_POST[siblings_add]','$_POST[schoolname_add]','$_POST[headname_add]','$_POST[schooladd_add]','$_POST[schooltele_add]','$_POST[schoolem_add]','$_POST[alevel_add]','$_POST[personstate_add]','$_POST[nameprovided_add]','$_POST[pe_add]','$_POST[se_add]','$_POST[PredGrade_Art]','$_POST[PredGrade_AScience]','$_POST[PredGrade_BusStudies]','$_POST[PredGrade_Electronics]','$_POST[PredGrade_EnglishLang]','$_POST[PredGrade_EnglishLit]','$_POST[PredGrade_French]','$_POST[PredGrade_German]','$_POST[PredGrade_Geog]','$_POST[PredGrade_Graphics]','$_POST[PredGrade_History]','$_POST[PredGrade_Maths]','$_POST[PredGrade_SepScience]','$_POST[PredGrade_ProductDesign]','$_POST[PredGrade_Spanish]','$_POST[PredGrade_Other]','$_POST[Gender_Male]','$_POST[Gender_Female]','$_POST[Sub_EnglishLit]','$_POST[Sub_Maths]','$_POST[Sub_FurtherMaths]','$_POST[Sub_Biology]','$_POST[Sub_Chemistry]','$_POST[Sub_Physics]','$_POST[Sub_French]','$_POST[Sub_German]','$_POST[Sub_Spanish]','$_POST[Sub_Geography]','$_POST[Sub_History]','$_POST[Sub_RE]','$_POST[Sub_FineArt]','$_POST[Sub_Business]','$_POST[Sub_Computing]','$_POST[Sub_GlobPersp]','$_POST[Sub_DramaAndTheatre]','$_POST[Sub_PE]','$_POST[Sub_Dance]','$_POST[Sub_Politics]','$_POST[Sub_Psychology]','$_POST[Sub_Sociology]','$_POST[readprospect_chk]','$_POST[Sib_Yes]','$_POST[Sib_No]','$_POST[Current_Student_Yes]','$_POST[Current_Student_No]','$_POST[I_Understand_chk]','$_POST[Current_Education_chk]','$_POST[Local_Care_chk]','$_POST[Staff_Cwhls_chk]','$_POST[Sub_Film]')"; if (!mysql_query($sql,$con))   {   die('Error: ' . mysql_error());   }   ?> <?php //if "email" variable is filled out, send email   if (isset($_REQUEST['pe_add']))  {     //Email information     $admin_email = $_REQUEST['pe_add'];   $forename = $_REQUEST['forename_add'];   $email = "autoreply@testing.com";   $subject = "Application";   $desc =   "Dear $forename   Thank you for submitting your online application, we will be in touch shortly. " ;       //send email   mail($admin_email, "$subject", "$desc", "From:" . $email);     //Email response   echo "Thank you for contacting us!";   }     //if "email" variable is not filled out, display the form   else  { ?> If you are seeing this, you need to go back and fill out the Personal Email section! <?php   }   header("location:complete.php"); mysql_close($con) ?>     Thanks in advance.

Hello all,   Appreciate if you folks could pls. help me understand (and more importantly resolve) this very weird error:     Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[42000]: Syntax error or access violation: 1064 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 'ASC, purchase_later_flag ASC, shopper1_buy_flag AS' at line 3' in /var/www/index.php:67 Stack trace: #0 /var/www/index.php(67): PDO->query('SELECT shoplist...') #1 {main} thrown in /var/www/index.php on line 67   Everything seems to work fine when/if I use the following SQL query (which can also be seen commented out in my code towards the end of this post) :  

$sql = "SELECT shoplist.*, store_master.store_name, item_master.item_name
           FROM shoplist, store_master, item_master
           WHERE     shoplist.store_id = store_master.store_id
                   AND shoplist.item_id = item_master.item_id";

  However, the moment I change my query to the following, which essentially just includes/adds the ORDER BY clause, I receive the error quoted above:

$sql = "SELECT shoplist.*, store_master.store_name, item_master.item_name 
		FROM shoplist, store_master, item_master 
		ORDER BY purchased_flag ASC, 
				purchase_later_flag ASC, 
				shopper1_buy_flag ASC, 
				shopper2_buy_flag ASC, 
				store_name ASC) 
		WHERE 	shoplist.store_id = store_master.store_id 
				AND shoplist.item_id = item_master.item_id";

In googling for this error I came across posts that suggested using "ORDER BY FIND_IN_SET()" and "ORDER BY FIELD()"...both of which I tried with no success.   Here's the portion of my code which seems to have a problem, and line # 67 is the 3rd from bottom (third last) statement in the code below:  

                    <?php
                        /*
                        $sql = "SELECT shoplist.*, store_master.store_name, item_master.item_name
                                FROM shoplist, store_master, item_master
                                WHERE     shoplist.store_id = store_master.store_id
                                        AND shoplist.item_id = item_master.item_id";
                        */
                    
                        $sql = "SELECT shoplist.*, store_master.store_name, item_master.item_name
                                FROM shoplist, store_master, item_master
                                ORDER BY FIND_IN_SET(purchased_flag ASC,
                                                 purchase_later_flag ASC,
                                                 shopper1_buy_flag ASC,
                                                 shopper2_buy_flag ASC,
                                                 store_name ASC)
                                WHERE     shoplist.store_id = store_master.store_id
                                        AND shoplist.item_id = item_master.item_id";
                        $result = $pdo->query($sql);                        
                        
                        // foreach ($pdo->query($sql) as $row) {
                        foreach ($result as $row) {
                            echo '<tr>';
                                print '<td><span class="filler-checkbox"><input type="checkbox" name="IDnumber[]" value="' . $row["idnumber"] . '" /></span></td>';

    Thanks  

Parse error: syntax error, unexpected T_STRING in C:\xampp\htdocs\mywork\unique.php on line 15


<html>
<head>
<title>

</title>
</head>
<body bgproperties="fixed">
<?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$con = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql');
$dbname = 'mywork';
mysql_select_db($dbname, $con);
$sql=mysql_query(insert into users (regno,name,gender,date,month,year,emailid,cell,paddress,caddress,incometype,incomeamt,dad,fyes,dadocup,mom,myes,momocup,password) VALUES ('$_POST[regno]','$_POST[name]','$_POST[gender]','$_POST[date]','$_POST[month]','$_POST[year]','$_POST[emailid]','$_POST[cell]','$_POST[paddress]','$_POST[caddress]','$_POST[incometype]','$_POST[incomeamt]','$_POST[dad]','$_POST[fyes]','$_POST[dadocup]','$_POST[mom]','$_POST[myes]','$_POST[momocup]','$_POST[password]')");
$sql1=mysql_fetch_array($sql);
$result = @mysql_query($SQl1);
$result="SELECT * FROM users WHERE regno='$regno'";
while($row = mysql_fetch_array($result))
{
    //echo $row['regno']."regno<br>";
//echo $row['name']."name<br>";
//echo $row['gender']."gender<br>";
//echo $row['date']."date<br>";
//echo $row['month']."month<br>";
//echo $row['year']."year<br>";
//echo $row['emailid']."emailid<br>";
//echo $row['cell']."cell<br>";
//echo $row['paddress']."paddress<br>";
//echo $row['caddress']."caddress<br>";
//echo $row['incometype']."incometype<br>";
//echo $row['incomeamt']."incomeamt<br>";
//echo $row['dad']."dad<br>";
//echo $row['fyes']."fyes<br>";
//echo $row['dadocup']."dadocup<br>";
//echo $row['mom']."mom<br>";
//echo $row['myes']."myes<br>";
//echo $row['momocup']."momocup<br>";
//echo $row['password']."password<br>";

}
echo "Thanks for Register!";
if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";
mysql_close($con);
?>
<form name="security" action="index.php" method="post">

<input type="submit" value="click here to login">
</form>
</body>
</html>

The issue is there seems to be NO syntax error.

1. There is no relevant code before or after this line. 2. Yes, one would think a ! should be there as did I (I didn't write the code), however, even with the ! it still gives the same error.

if (function_exists('gzcompress')) die(FUNCTION_NOT_FOUND); is giving me a

syntax error, unexpected 'if', expecting 'function' or 'const'

I am updating this code from PHP5.3 to PHP7.4 and I can't figure out what the syntax problem is since PHP allows this. I am using Eclipse PHP to do the conversion.

Code: [Select]
<?php
mysql_connect ("-","-","-") or die ('Error');
mysql_select_db ("-");

$out = mysql_query("SELECT * FROM guestbook ORDER BY id DESC");

while($row = mysql_fetch_assoc($out); --and this one if that braces is deleted
{ ----this is where im getting the error
$name = $row['name'];
$email = $row['email'];
$txt = $row['comment'];

        $msg = "Are you sure you want to delete";

/* @var $_REQUEST <type> */
if (isset($_REQUEST ["action"]) && $_REQUEST["action"] == "del") 
    {
                    $id = intval($_REQUEST['id']);
                    mysql_query("DELETE FROM guestbook WHERE id=$id;");
                   echo "<action=index.php>";
                }

echo "<font face='verdana' size='1'>";
echo "<table border='0'>
<tr><td>Name: ".$name."</td></tr>"."
<tr><td>Email: ".$email."</td></tr>
<tr><td colspan='2'>Comment:</td></tr>
<tr><td colspan='2' width='500'><b>".$txt."</b></td></tr>
                     <tr><td><a onclick=\"return confirm('.$msg.');\" href='index.php?action=del&amp;id=".$row['id']."'><span class='red'>["."Delete"."]</span></a>
                        </td></tr>
  </table><br />";


echo "<hr size='1' width='500' align='left'></font>";
}
?>

Kindly help me please. When i delete ({) the error will become the ( i dont know what to do already. Thanks.

Hello everyone,
I am a total newbie in PHP and so do here. I am trying to build a table consisting foreign key and primary key and it is showing error like "Parse error: syntax error, unexpected 'cities' (T_STRING) in line number 29". I have marked up my lines according to sublime text 3 editor. I am trying to create a database. 

 

1 <?php
2 $servername = "localhost";
3 $username = "username";
4 $password = "password";
5 $dbname = "newDB";

7 // Create connection
8 $conn = new mysqli($servername, $username, $password, $dbname);

10 // Check connection
11 if ($conn->connect_error) {
12   die("Connection failed: " . $conn->connect_error);}

14 $firstname = $conn->real_escape_string($_REQUEST['firstname']);
15 $lastname = $conn->real_escape_string($_REQUEST['lastname']);
16 $address = $conn->real_escape_string($_REQUEST['address']);
17 $city = $conn->real_escape_string($_REQUEST['city']);
18 $country = $conn->real_escape_string($_REQUEST['country']);
19 $phone_number = $conn->real_escape_string($_REQUEST['phone_number']);
20 $email = $conn->real_escape_string($_REQUEST['email']);
 
22 // Attempt insert query execution
23 $sql1 = "INSERT INTO cities VALUES ('$city')";

25 $sql2 = "INSERT INTO countries VALUES ('$country')";

27 $sql3 = "INSERT INTO Contacts (firstname, lastname, address, city, country, phone, email) VALUES ('$firstname', '$lastname', '$address', $city, $country, '$phone_number','$email')";

29 SELECT * FROM cities;
SELECT * FROM countries;
SELECT * FROM Contacts;

if($conn->query($sql1) === true){
    echo "City added successfully.";
} else{
    echo "ERROR: Could not able to execute " . $conn->error;
}

if($conn->query($sql2) === true){
    echo "Çountry added successfully.";
} else{
    echo "ERROR: Could not able to execute " . $conn->error;
}

if($conn->query($sql3) === true){
    echo "Çontact added successfully.";
} else{
    echo "ERROR: Could not able to execute " . $conn->error;
}
 
// Close connection
$conn->close();
?>

I have been trying to get my files to upload onto a computer and I receive this message: Parse error: syntax error, unexpected T_STRING in /home/content/19/6550319/html/listing.php on line 27. Line 27 is how the php logs into my SQL. The problem is that I was able to log in before. I just made changes to the form by adding a dropdown menu and price and now it says it doesnt parse.

Can anyone figure this out. I will include the code without the login information because the forum is public but I did put the words left out for you to see where I took out the passcodes.

Code: [Select]
<?php




//This is the directory where images will be saved
$target = "potofiles/";
$target = $target . basename( $_FILES['photo']['name']);

//This gets all the other information from the form
$price=$_POST['price'];
$gig=$_POST['giga'];
$yesg=$_POST['yesg'];
$pic=($_FILES['photo']['name']);
$pic2=($_FILES['phototwo']['name']);
$pic3=($_FILES['photothree']['name']);
$pic4=($_FILES['photofour']['name']);
$description=$_POST['iPadDescription'];
$condition=$_POST['condition'];
$fname=$_POST['firstName'];
$lname=$_POST['lastName'];
$email=$_POST['email']



// Connects to your Database
mysql_connect ("left out", "left out", "left out") or die(mysql_error()) ;
mysql_select_db("left out") or die(mysql_error()) ;

//Writes the information to the database
mysql_query("INSERT INTO listing (price,giga,yesg,photo,phototwo,photothree,photofour,iPadDescription,condition,firstName,lastName,email)
VALUES ('$price', '$gig', '$yesg', '$pic', '$pic2', '$pic3', '$pic4', '$description', '$condition', '$fname', '$lname', '$email')") ;

//Writes the photo to the server
if(move_uploaded_file($_FILES['photo']['tmp_name'], $target))
{

//Tells you if its all ok
echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded, and your information has been added to the directory";
}
else {

//Gives and error if its not
echo "Sorry, there was a problem uploading your file.";
}

echo date("m/d/y : H:i:s", time()) 


?>

I just enabled error reporting and I am not that familiar with it.  I know I have an error some where around line 33.  I know I am missing a bracket or a comma or some other syntax error I just cannot find where the error is.  Below is my script.
Thanks for any help.

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">
<head>
<title>Airline Survey</title>
<meta http-equiv="content-type" content="text/html; charset=iso-8859-1" />
<meta name="author" content="Revised by abc1234"/>
</head>
<body>
<?php
$WaitTime = addslashes($_POST["wait_time"]);
$Friendliness = addslashes($_POST["friendliness"]);
$Space = addslashes($_POST["space"]);
$Comfort = addslashes($_POST["comfort"]);
$Cleanliness = addslashes($_POST["cleanliness"]);
$Noise = addslashes($_POST["noise"]);
if (empty($WaitTime) ||
    empty($Friendliness) ||
    empty($Space) ||
empty($Comfort) ||
    empty($Cleanliness) ||
    empty($Noise))
    echo "<hr /><p>You must enter a value in each field. Click
     your browser's Back button to return to the form.</p><hr />";
else {
        $Entry = $WaitTime . "\n";
        $Entry .= $Friendliness . "\n";
        $Entry .= $Space . "\n";
        $Entry .= $Comfort . "\n";
        $Entry .= $Cleanliness . "\n";
        $Entry .= $Noise . "\n";
        $SurveyFile = fopen("survey.txt", "w")
}
        if (flock($SurveyFile, LOCK_EX)) {
                if (fwrite($SurveyFile, $Entry) > 0) {
                        echo "<p>The entry has been successfully added.</p>";
                        flock($SurveyFile, LOCK_UN;
                        fclose($SurveyFile);
  else
                        echo "<p>The entry could not be saved!</p>";
        }
        else
                echo "<p>The entry could not be saved!</p>";
}
?d>
<p><a href="AirlineSurvey.html">Return to Airline Survey</a></p>
</body>
</html>

I don`t get it, waht is wrong?!
Code: [Select]
<?php
require_once 'auth.php';
if (!isset($_SESSION['SESS_VERIFY'])) {
header("location: access-denied.php");
exit();
}
if ($_SESSION['lang'] == 'Ro') {

// setare data romania
date_default_timezone_set('Europe/Bucharest');

$today = getdate();


$zi = $today['mday'];
$luna = $today['mon'];
$lunastring = $today['month'];
$an = $today['year'];
$data = $zi.$luna.$an;
$data = (string)$data;
$ora = date('H:i:s'); 

$msg = array();
$err = array();

$luni = array (
1=>'Ianuarie',
2=>'Februarie',
3=>'Martie',
4=>'Aprilie',
5=>'Mai',
6=>'Iunie',
7=>'Iulie',
8=>'August',
9=>'Septembrie',
10=>'Octobrie',
11=>'Noiembrie',
12=>'Decembrie');

// comun
const SQL_ERR = 'SQL statement failed with error: ';
const ADD_MODEL = 'ADAUGA UN MODEL NOU';
.
.many constants..
.
}
elseif ($_SESSION['lang'] == 'It') {...

Thank you!

Hi folks,

I am a complete n00b at php and mysql. I am teaching myself from books and the WWW, but alas I am stuck...

the error I get is:
Parse error: syntax error, unexpected T_STRING in X:\xampp\htdocs\search.php on line 7

here is the code:




<?php
  
  mysql_connect ("localhost", "user", "password") or die (mysql_error());
  mysql_select_db ("it_homehelp_test") or die (mysql_error());
  
  $term = $_POST['term'];
  $sql = $mysql_query(select * from it_homehelp_test where ClientName1 like '%term%');   <<<------this is line 7
  while ($row = mysql_fetch_array($sql)){
   echo 'Client Name:' .$row['ClientName1'];
echo 'Address:' .$row['Address1'];
echo 'Phone:' .$row['Tel1'];
}
?>



Any help you can offer would be great. I can also post the ".html" file that creates the search bar if it is needed.

Thanks

I have been getting that error and I cannot figure out why it is happening

Here is the error: Parse error: syntax error, unexpected T_ENDWHILE, expecting ',' or ';' in /home/scswc188/public_html/index.php on line 23

Here is my Code (Database Credentials removed for obvious reasons)
<?PHP

// Conect to the Mysql Server
$connect = mysql_connect("IP","USER","PASS");

//connect to the database
mysql_select_db("TABLE");

//query the database
$query = mysql_query("SELECT * FROM users_online WHERE online = 1");

// fetch the results / convert into an array

        WHILE($rows = mysql_fetch_array($query)):

 

 

        $users = $rows['name'];

echo "'<font color='black'>Online:<font color='green'>$users, </font></font>;"

endwhile;

?>

or here http://pastebin.com/ZYh4t2pD

Thanks

Edit: Found the php tag