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PHP - Sql Query Error, Not Sure What's Best In This Situation
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? 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 On one of my pages, I have 1 form, which just a few text boxes / check boxes. Currently, I have the form submit a new record into the DB and then refreshes the entire page to show the new result. My question is, should I display all records using JQuery or PHP. PHP refreshed the whole page, thus more queries are run per page. Does it matter much? Thanks. 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? I spent several hours trying to figure this thing out. Thought I had it nailed, but still getting duplicate record entries into the MySQL DB when I do NOT want them.
Here's the plot:
People filling out the possible attendance form for a Ham Radio event *sometimes* bring a 2nd person (either a spouse or a friend). The 2nd person may, or may not, also has a Callsign which I need to put INSERT the same MySQL Callsign column. In any event, to also identify the 2nd person as coming 'with' the 1st person.
MOST of the attendees are individuals with NO 2nd person.
My entry form has these primary fields:
callsign
fullname
AND...
callsign2
fullname2
What I came up with was to process the MySQL INSERT for the primary callsign & fullname into their respective MySQL DB Columns (which works fine), and then........... immediately following the main Query INSERT, to do a substitution type thing depending on whether or not a form entry was made in the callsign2 field, AND/OR, the fullname2 field.
This partially works, but if there is ONLY a primary callsign and fullname in the form, I'm still getting a duplicate record entry which includes the callsign in the `with` column (which should ONLY take place IF there is a 2nd person indicated).
Confusing?
Here is what I have been wrestling with to try and accomplish the objective, and now my eyes are glazed over ;-(
// TRICKY PART HERE
// If a 2nd Callsign AND a Fullname
if ($callsign2 != ' ' && $fullname2 != ' ') {
// Still make reference to the primary Callsign in the MySQL DB `with` column
$with = $callsign;
// Assignment to allow 2nd Callsign to be entered in the MySQL `callsign` column
$callsign=$callsign2;
$fullname=$fullname2;
$sql="INSERT INTO `mytable` (`callsign`, `fullname`, `with`)
VALUES ('$callsign', '$fullname', '$with')";
// If NO 2nd Callsign BUT a Fullname
} elseif ($callsign2 = ' ' && $fullname2 != ' ') {
// Make reference to the primary Callsign in the MySQL DB `with` column
$with = $callsign;
$callsign=$callsign2;
$fullname=$fullname2;
$sql="INSERT INTO `mytable` (`callsign`, `fullname`, `with`)
VALUES ('$callsign', '$fullname', '$with')";
} else {
// The only thing I could thing of to (hopefully) NOT make a 2nd entry
// record in the MySQL DB IF there is NO 2nd person referenced
$with = $callsign;
}
if (!mysqli_query($con,$sql)) {
die('Error: ' . mysqli_error($con));
}
This mostly works EXCEPT if only a single (primary) person entry.
The recap the objetives:
1. If ONLY a primary/single person entry on the form:
* callsign & fullname get INSERTed into the `callsign` and `fullname` columns in the DB as ONLY one record entry
2. If BOTH a primary and 2nd person on the form:
A. IF the 2nd person has a Callsign, then the 2nd record entry would be:
* callsign2 & fullname2 get INSERTed into the `callsign` & `fullname` columns in the 2nd DB as a separate record entry
* callsign of the primary person also gets INSERTED into the `with` column in the same 2nd DB record entry
B. IF the 2nd person does NOT have a callsign, then the 2nd record entry would be:
* fullname2 gets INSERTed into the `fullname` column in the DB as a separate 2nd DB record entry
* callsign of the primary person also gets INSERTED into the `with` column in the same 2nd DB record entry
I obvioiusly have overlooked something, but just can't seem to figure it out at this point {SIGH}.
Thanks for any enlightenment.
-FreakingOUT
Hello I am busy trying to setup a free to use budgeting and cashflow tool. I was recently going through my finances and whilst I found lots of spreadsheets on the subject, I didn't find a website that would allow me to store my data anonymously and access it when and where I felt fit. Bare in mind that I am a bit of a novice so go easy on me... So firstly, I've setup a page to collect 'account' information...that is, income items and expense items. I ask them to enter the amount, the frequency and when the next due date is. The DB looks like this: Code: [Select] CREATE TABLE IF NOT EXISTS `income_accounts` ( `inc_acc_ID` int(11) NOT NULL AUTO_INCREMENT, `user_ID` int(11) NOT NULL, `freq_ID` int(11) NOT NULL, `amount` decimal(7,2) NOT NULL, `next_due` date NOT NULL, `income_typeID` int(11) NOT NULL, PRIMARY KEY (`inc_acc_ID`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=105 ; and the same for expense 'accounts': Code: [Select] CREATE TABLE IF NOT EXISTS `expense_accounts` ( `exp_acc_ID` int(11) NOT NULL AUTO_INCREMENT, `user_ID` int(11) NOT NULL, `freq_ID` int(11) NOT NULL, `amount` decimal(7,2) NOT NULL, `next_due` date NOT NULL, `ex_typeID` int(11) NOT NULL, PRIMARY KEY (`exp_acc_ID`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=13 ; So once I have these entered by the user I should be able to then generate a cashflow forecast. For example, I give 5 options for the user: '4 weeks' , '8 weeks', '3 months', '6 months' '12 months'. Selecting one of these will then generate a table listing all of the items that apply for each period with a total at the bottom. Lets take '4 weeks' for an example. First I need to determine if the 'next_due' value is within the time period selected (4 weeks). If it is, what frequency is it? If it is every week then each column of the 4 column table created (i.e. 4 weeks/4 columns) will have it in it. If it is monthly frequency it will be only in the column for which it is due (i.e. column 1 would be todays date to todays date +6, column 2 = todays date + 7 to todays date +13 etc...) This will go through both the income table and the expense table and once all the items have been checked and added to the page table a further row would be added to total all the items in each column, and give a sum (i.e. sum of incomes less sum of expenses). For the '6 month' and '12 month' option, I want to display 1 month per column, rather than 1 week...just to not squeeze things up to much... I do not have a great deal of experience with arrays but my instinct tells me that they are the answer....otherwise I guess I could create a temporary DB table? I'm happy to share the code I've written so far if it helps...otherwise any advice on the best way to achieve this would be great. 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. 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 />";} ?>  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, 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. 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. 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.
<?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); ?> 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 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(); } ?> 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>';
}
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); Can anyone tell me whats wrong with this code? The mysql error I get is generic and says
"Could not enter 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 'Tickets (Name,Tech,Node,Address,Tap,Ped,Lash,Hardline,Other,Comments) VALUES ('C' at line 1"
<?php
$name=$_POST['name'];
$id=$_POST['id'];
$node=$_POST['node'];
$ped=$_POST['ped'];
$tap=$_POST['tap'];
$lash=$_POST['lash'];
$hardline=$_POST['hardline'];
$other=$_POST['other'];
$address=$_POST['address'];
$city=$_POST['city'];
$comments=$_POST['comments'];
$dbhost = 'xxxx';
$dbuser = 'xxxx';
$dbpass = 'xxxx';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
die('Could not connect: ' . mysql_error());
}
$sql = "INSERT INTO Damage Tickets (Name,Tech,Node,Address,Tap,Ped,Lash,Hardline,Other,Comments) VALUES ('$name','$id','$node','$address','$city','$tap','$ped','$lash','$hardline','$other','$comments')";
mysql_select_db('xxxxx');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Could not enter data: ' . mysql_error());
}
echo "Your Damage has been Submitted! <a href=\"damage.php\">Go Back</a>";
mysql_close($conn);
?>
No idea whats wrong with it? Thanks for any help
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