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PHP - Mysql_query Syntax Error Database Connection
Similar TutorialsHoping someone can help me, ive spent the last hour looking up examples of mysql_query(Insert) and changing my query but im still gettin a syntax error and I cant understand where. Ive tried removing apostrophies and putting the longest variable string outside of the speach marks without any joy. mysql_query("INSERT INTO $QA_table (SUB_CATEGORY, QUESTION, FROM, FROM_EMAIL, DISPLAY) VALUES ('$sub_cat','$message','$from','$from_email','2')") or die(mysql_error()); I get the following 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 'FROM, FROM_EMAIL, DISPLAY) VALUES ('Range','test question bla bla bla','Phil','phi' at line 1 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> I get the following error when trying to connect Warning: mysql_connect() [function.mysql-connect]: Can't connect to MySQL server on '72.18.129.104' (10061) in C:\Domains\crysvis.com\wwwroot\include\dbConnectAndSelect.php on line 8 Here's the code $host = "72.18.129.104"; $user = "deltron"; $pass = "masterconn"; $db = "customers"; $conn = mysql_connect("$host","$user","$pass"); if(!$conn) errorHandler("Msg 10:\nCould not connect to database with $host,$user,$pass in ". $_SERVER["PHP_SELF"]."\nCustId = $CustId\nmysql_error() = ".mysql_error()); if(!mysql_select_db("$db",$conn)) errorHandler("Msg 11:\nCould not select db=$db in ". $_SERVER["PHP_SELF"]."\nCustId=$CustId\nmysql_error() = ".mysql_error()); ?> What am I doing wrong? Any help will be appreciated Im building a list of offers and adding them to a table in a database. Pretty much all it is is HTML. Im inserting an ahref link that has a php echo in it. So it looks like this: <div class="offerlinks"><a href="http://website.com/offer/blahblah&blah=blah&sid=<?php echo $_SESSION['uid'];?>">Offer name</a><br><b>Info:</b> Signup<br><b>Value</b> 1 pt</div> When I insert this (through my form) I get mysql error 1064 which is syntax error. I tested it without the php & it gives me 0, which worked fine. I need the php code so I can append userid to the SID var. Am I doing something wrong? Well I guess I obviously am so the real question is what am I doing wrong & how could I do it the right way? Thanks guys hey I am still little new in php can you help me in this one func.php function confirm_query($result_set){ if(!$result_set){ die(mysql_error()); } function get_subjects_all() { global $connection ; $sub_query = "SELECT * FROM subjects ORDER BY position ASC"; $subject_query = mysql_query($sub_query,$connection); confirm_query($subject_query) ; return $subject_query; } function get_pages_all($subject_id) { global $connection; $query = "SELECT * FROM pages WHERE subject_id = '$subject_id' ORDER BY position ASC"; $page_set = mysql_query($query, $connection); confirm_query($page_set); return $page_set; } test.php require_once("config.php"); require_once("func.php"); $subject_query = get_subjects_all(); while($subject = mysql_fetch_array($subject_query)) { echo "<a href= \"test.php?subj=" . urlencode($subject["id"]) . "\">{$subject['menu_name']} <br />"; } $an = $subject["id"] ; $page_set = get_pages_all($an); while($page = mysql_fetch_array($page_set)) { echo "<a href = \"test.php?page=" . urldecode($page["id"]) . "\">{$page['menu_name']} </a><br />" ; // echo "<b> <a href= '1.php?page={$page['id']}' >{$page['menu_name']} </a></b> <br />"; } in this line WHERE subject_id = '$subject_id' the query don't work and gives me sql error i tried with this one ' ".$subject_id." ' but not worked but if i changed $subject_id to a number it work and read the data from the table thanks in advanced . Is there something wrong this? If so, I'm not seeing it...I keep getting this error: Quote ( ! ) Parse error: syntax error, unexpected T_STRING in C:\Program Files\wamp\www\register.php on line 1059 Here's my code for registering a member: Code: [Select] <?php mysql_query (INSERT INTO fans (id, username, email, password, country, region, gender, status, account, subscription, time_registered, date_registered, birthdate, name, website, age, activation_key) VALUES ('', '{$_SESSION["user"]["username"]}', '{$_SESSION["user"]["email"]}', '{$_SESSION["user"]["password"]}', '{$_SESSION["user"]["country"]}', '{$_SESSION["user"]["region"]}', '{$_SESSION["user"]["gender"]}', 'Offline', 'Inactive', 'Unsubscribed', '{$_SESSION["user"]["time"]}', '{$_SESSION["user"]["date"]}', '{$_SESSION["user"]["birthdate"]}', '{$_SESSION["user"]["name"]}', '{$_SESSION["user"]["website"]}', '{$_SESSION["user"]["age"]}', '{$_SESSION["user"]["activation_key"]}')) or die ("Could not register member"); ?> Help, please and thank you! after cloasing connection of database i still got the values form database. Code: [Select] <?php session_start(); /* * To change this template, choose Tools | Templates * and open the template in the editor. */ require_once '../database/db_connecting.php'; $dbname="sahansevena";//set database name $con= setConnections();//make connections use implemented methode in db_connectiong.php mysql_select_db($dbname, $con); //update the time and date of the admin table $update_time="update admin set last_logged_date =CURDATE(), last_log_time=CURTIME() where username='$uname'limit 3,4"; //my admin table contain 5 colums they are id, username,password, last_logged_date, last_log_time $link= mysql_query($update_time); // mysql_select_db($dbname, $link); //$con=mysql_connect('localhost', 'root','ijts'); $result="select * from admin where username='a'"; $result=mysql_query($result); mysql_close($con); //here i just check after closing data baseconnection whether i do get reselts but i do, why? echo "after the cnnection was closed"; if(!$result){ echo "cont fetch data"; }else{ $row= mysql_fetch_array($result); echo "id".$row[0]."usrname".$row[1]."passwped".$row[2]."date".$row[3]."time".$row[4]; } // echo "<html>"; //echo "<table border='1' cellspacing='1' cellpadding='2' align='center'>"; // echo "<thead>"; // echo"<tr>"; // echo "<th>"; // echo ID; // echo"</th>"; // echo" <th>";echo Username; echo"</th>"; // echo"<th>";echo Password; echo"</th>"; // echo"<th>";echo Last_logged_date; echo "</th>"; // echo "<th>";echo Last_logged_time; echo "</th>"; // echo" </tr>"; // echo" </thead>"; // echo" <tbody>"; //while($row= mysql_fetch_array($result,MYSQL_BOTH)){ // echo "<tr>"; // echo "<td>"; // echo $row[0]; // echo "</td>"; // echo "<td>"; // echo $row[1]; // echo "</td>"; // echo "<td>"; // echo $row[2]; // echo "</td>"; // echo "<td>"; // echo $row[3]; // echo "</td>"; // echo "<td>"; // echo $row[4]; // echo "</td>"; // echo "</tr>"; // } // echo" </tbody>"; // echo "</table>"; // echo "</html>"; session_destroy(); session_commit(); echo "session and database are closed but i still get values from doatabase session is destroyed".$_SESSION['admin']; ?> session is destroyed but database connection is not closed. thanks Hey, I found some code online that references a bunch of sql commands in there own functions that can be called from a require_once(), All the ones I have tried work expect mysql_query when its called it returns this error: Quote Warning: mysql_result() expects parameter 1 to be resource, boolean given in But once i paste the mysql_query into the main script it works fines. code below is mine and the online script. Webcode: Code: [Select] $sql = "SELECT cat_id, cat_parent_id, cat_name, cat_description, cat_image FROM tbl_category WHERE cat_parent_id = $catId ORDER BY cat_name"; /* Don't know what getPagingQuery does and couldnt find any reference to it so i removed it and replaced with simple query in my own code */ $result = dbQuery(getPagingQuery($sql, $rowsPerPage)); My Code: Code: [Select] $sql = "SELECT cat_id, cat_parent_id, cat_name, cat_description, cat_image FROM tbl_category WHERE cat_parent_id = $catId ORDER BY cat_name"; $result = dbQuery($sql); Include file: Code: [Select] $dbHost = "localhost"; //SQL Server $dbUser = "root"; // Database username $dbPass = ""; // Database password $dbName = "lh_shop"; // Database name $sqlCon = mysql_connect($dbHost, $dbUser, $dbPass) or die ('MYSQL connection Failed. ' . mysql_error()); mysql_select_db($dbName) or die('Cannot select database. ' . mysql_error()); // Database query functions function dbQuery($sql) { return mysql_query($sql) or die('Query failed. ' . mysql_error()); } Thanks, I really hope someone has had this issue I have been looking at this code most of the morning and do not have a clue what is wrong with the code. I am hoping its not a stupid mistake, can someone please help me out? thank you
<title>Inputing Travel Detials</title>
<header>
<h1 align="center"> Adding Travel Detials </h1>
<body>
<p> <center><img src="cyberwarfareimage1.png" alt="Squadron logo" style="width:200px;height:200px" style="middle"></center>
<table border="1">
<tr>
<td><a href="index.php"> Home Page </a></td>
<td><a href="administratorhomepage.html">Administrator Home Page </a></td>
<td><a href="viewhomepage.html">View Home Page </a></td>
<td><a href="Inputhomepage.html">Input Home Page </a></td>
<td><a href="traveldetials.html">Enter More Travel Detials </a></td>
</table>
</p>
<?php
include "connection.php";
$Applicant_ID = $_POST["Applicant_ID"];
$Method_Of_Travel = $_POST["Method_Of_Travel"];
$Cost = $_POST["Cost"];
$ETA = $_POST["ETA"];
$Main_Gate_Advised = $_POST["Main_Gate_Advised"];
$query = ("UPDATE `int_board_applicant` SET `Method_Of_Travel`=`$Method_Of_Travel', `Cost`=`$Cost', `ETA`='$ETA', `Main_Gate_Advised`='$Main_Gate_Advised' WHERE `Applicant_ID`='$Applicant_ID'");
$result = mysqli_query($dbhandle, $query)
or die(mysqli_error($dbhandle));
if($result){ echo "Success!"; }
else{ echo "Error."; }
// successfully insert data into database, displays message "Successful".
if($query){
echo "Successful";
}
else {
echo "Data not Submitted";
}
//closing the connection
mysqli_close($dbhandle)
?>
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
On some occasions I need to connect to a second and third database in the same script (maybe 5% of scripts have at least a second connection). Usually I would just select the new database. However, my host requires different users to be created for each database. What is the best way to do this? Close current connection (say db1) and open new (say db2) OR keep all open, creating 2nd and 3rd connections. I am happy with the design of my database, and don't want to merge all these tables into one db. Overall I am still happy with my host, so I'd rather not change. hey i need help im tryig to get information from my user and then process it in my database so i can use it to log them to a different web site im trying to use this method to get the information from the user but need help to get it please help me. Code: [Select] //make the database connection. $conn = mysql_connect("localhost", "Black Jack"); mysql_select_db("chaper7", $conn); //create a query $sql = "SELECT * FROM hero"; $result = mysql_query($sql, $conn); <?php class UserQuery { public function Adduser($id,$username,$email,$password) { $conn = new Config(); $sql =("INSERT INTO test.user (id, username, email, password) VALUES ('$id', '$username', '$email',$password)"); $conn->exec($sql); } }
getting an "exec doesnt exist " error, saying exec doesnt exist in my db file. it doesnt need to exist does it ? anyone any idea why ?
Through out my program I have used global $mysqli; as a connection to my database - this has worked fine so far. Now I have called $sql_statement = "SELECT * FROM items WHERE name='$itemName'"; $itemStats = mysqli_fetch_array(mysqli_query($mysqli, $sql_statement), MYSQLI_ASSOC); via a function and I get the following warnings - Quote Warning: mysqli_query() expects parameter 1 to be mysqli, null given in /functions/getdata.php on line 27 Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in /functions/getdata.php on line 27 it still works though and returns the correct information. If I take the code out of the function and use it normally then I get no Warnings - but this defeats the whole purpose of having functions! How do I get rid of the warnings? (and no, I don't mean turn the warnings off ) I'm connecting to my database using the following... @ $db = new mysqli('host', 'username', 'password', 'database') The .php file that is connecting to the database is in my root (htdocs) folder on the server. I know that I am not supposed to put my actual 'host', 'username', 'password', 'database' inside the mysqli function for security purposes. I know that I am supposed to put variables in instead. But here is where I am confused. Where do I set those variables? Do I set them in another file and include that file? If so, where do I store the file that holds the passwords, and what prevents a hacker from simply navigating to that file? Thanks for the help Hi guys, Hi, I am currently trying to build an Artist's website, the artist wanted a CMS that was completely customized to the site (so no Wordpress, Joomla, Drupel, etc) - because of this I am having to create a CMS completely from scratch.
The problem I am having is with the database connection (hence the topic title), other sites that I have built with this same code work fine - however this particular site does not seem to want to play ball. It keeps giving me this error:
SQLSTATE[HY000] [1045] Access denied for user 'web113-janesart'@'10.0.44.113' (using password: YES)I have tried obvious things such as spelling mistakes, wrong password/username/db name, nothing seems to get rid of this error. Any help on what else this could be would be appreciated. Dear friends i,m a php beginner and i got a problem with connecting to my database i created a database called (koora) with one table called (admins) and when i tried to connect to it (database ) ; it did not connect here is the code i used for that <?php $connectdb = mysql_connect('localhost','','') or die("not connected"); $selectdb = mysql_select_db("koora", $connectdb); if(!$selectdb) { die("error connecting table" .mysql_error()); } then when refreshing my phpmyadmin page i got that message error connecting tableAccess denied for user ''@'localhost' to database 'koora' koora is the name of the database so i need your help with this problem and what is the reason not to connect to the data base Thank you |
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