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PHP - Help With Mysql Syntax
i have created a table named 'masteruseraccounts' in mysql.it has fields : key(auto increment,int);name(varchar(50));idNo(varchar(12));password(varchar(50));mobileNo(varchar(10));email(varchar(50)).
i entered one row as follows: key | name | idNo | password | mobileNo | email 1 | nikhil upadhyay | 2009AAPS092H | nikhil |9505042041 | aliasnikhil@gmail.com and the idNo is the login name and is stored in a session variable($_SESSION['MM_Username']).i confirmed that the idNo is properly getting stored in session variable by using echo statement.i am also using the session_start().depending on this idNo i was trying to display the rest of information.so i tried to use the WHERE clause but in vain here is my syntax Code: [Select] <?php session_start(); mysql_select_db("mydb"); $userId = mysql_real_escape_string($_SESSION['MM_Username']); $query = "SELECT * " . "FROM masteruseraccounts". "WHERE idNo = '$userId' "; $results = mysql_query($query) or die(mysql_error()); while ($row = mysql_fetch_array($results)) { extract($row); echo $idNo; echo " - "; echo $email; echo " - "; echo $mobileNo; echo "<br>"; } ?> it 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 '= '2009AAPS092H'' at line 1" instead if i do not use the where clause then it displays all info correctly.i even tried removing the single quotes around $userId.also tried the WHERE clause as( "WHERE idNo = " .$userId. ""; ) still it gives error. i even tried hardcoding it with the WHERE statement as WHERE idNo = 2009AAPS092H but it doesnt work.putting quotes around 2009AAPS092H also doesnt help. please i am pulling my hair out.someone help me.i thought there might be some different syntax for varchar types but i tried hardcoding the WHERE clause as (WHERE key = 1) this also gives 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 'key = 1' at line 1" now this key field being int type i know it doesnt require quotes around it. any help is very much appreciated. Similar TutorialsI 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.
Hi.. I create mysql syntax for query testing before i input to my php code here is my mysql code: Code: [Select] set @t = 0; set @rqty=31968; SELECT LOT_CODE as code, DATE_ENTRY, CASE WHEN @t+OUTPUT_QTY > @rqty THEN @rqty -@t ELSE OUTPUT_QTY END as qty, @t := @t + d.OUTPUT_QTY as cumulative FROM dipping d WHERE SUBSTR(LOT_CODE, 9,4) = 'P28' AND (@t < @rqty); and i attach the sample output of the above query. Now that query test is work i will input that code to my php codes. $sql = "SELECT SKUCode, Materials, Comp, Qty FROM bom WHERE SKUCode = '$SKUCode'"; $res = mysql_query($sql, $con); ($row = mysql_fetch_assoc($res)); $Materials = $row['Materials']; $Qty = $row['Qty']; $Comp = $row['Comp']; //P28 //-----Compute Req Qty and Save to table---// $ReqQty = $Qty * $POReq; // 31968 $sql = "UPDATE bom SET ReqQty = '$ReqQty' WHERE SKUCode = '$SKUCode' AND Materials = '$Materials'"; $resReqQty = mysql_query($sql, $con); $t = 0; $sql = "SELECT LOT_CODE as code, DATE_ENTRY, CASE WHEN $t+OUTPUT_QTY > $ReqQty THEN $ReqQty -$t ELSE OUTPUT_QTY END as qty, $t := $t + d.OUTPUT_QTY as cumulative FROM dipping d WHERE SUBSTR(LOT_CODE, 9,4) = '$Comp' AND ($t < $ReqQty)"; when I echo the query: I got this: SELECT LOT_CODE as code, DATE_ENTRY, CASE WHEN 0+OUTPUT_QTY > 31968 THEN 31968 -0 ELSE OUTPUT_QTY END as qty, 0 := 0 + d.OUTPUT_QTY as cumulative FROM dipping d WHERE SUBSTR(LOT_CODE, 9,4) = 'P28' AND (0 < 31968) then I run it to the sql and I got an error: Error Code : 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 ':= 0 + d.OUTPUT_QTY as cumulative FROM dipping d WHERE SUBSTR(LOT_CODE, 9,4) = '' at line 1 (0 ms taken) Any help is highly appreciated Thank you so much Howdy folks, I am creating a Facebook app for a bit of fun and practice and getting the following error in index.php: Code: [Select] Invalid query -- SELECT * FROM `results` WHERE `resultLow` <= AND `resultHigh`>= -- 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 'AND `resultHigh`>=' at line 1 Here is the area: Code: [Select] $res = query("SELECT * FROM `results` WHERE `resultLow` <=$user_score AND `resultHigh`>=$user_score"); Any help would be appreciated. Hi guys i have the following code which is misbehaving can anyone see where its wrong? Code: [Select] <?php include 'dbc.php'; page_protect(); company(); $Referrer = mysql_query("SELECT * FROM Referrer WHERE SentOut='0' "); if (isset($_POST['submit'])) { //Assign each array to a variable $StaffMember = $_POST['StaffMember']; $referrer = $_POST['referrer']; $referred = $_POST['referred']; $SentOut = $_POST['SentOut']; $today = date("y.m.d H:i:s"); $user_id = $_SESSION['user_id']; $IssueNum = $_POST['Referrerid']; $limit = count($StaffMember); for($k=0;$k<$limit;$k++){ $msg[] = "$limit New KPI's Added"; $values[$k] = array( $StaffMember[$k],$referrer[$k],$referred[$k],$SentOut[$k],$today,$user_id ); // build the array of values for the query string } foreach( $values as $key => $value ) { $query = "UPDATE `Referrer` (StaffMember, referer, referred, SentOut, SentOutDate, SentOutBy) VALUES ('" . implode( '\', \'', $value ) . "') WHERE IssueNum= '{$IssueNum[$key]}'"; mysql_query($query) or die(mysql_error()); } } if (checkAdmin()) { ?> <html> <head> <title>Book Off Holiday</title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"> <script src="php_calendar/scripts.js" type="text/javascript"></script> <link href="styles.css" rel="stylesheet" type="text/css"> </head> <body> <form name="form" action="SendReferrers.php" method="post"> <table width="100%" border="0" cellspacing="0" cellpadding="5" class="main"> <tr> <td colspan="3"> </td> </tr> <td width="160" valign="top"> <?php if (isset($_SESSION['user_id'])) { } ?> <a href="admin.php">Admin CP </a> </td> <td width="732" valign="top"> <p> <h3 class="titlehdr">New KPI</h3> <table width="300px" border="0" align="Centre" cellpadding="2" cellspacing="0"> <tr bgcolor="#000050"> <td width="20px"><h3 class="Text2">Referrer ID</h3></td> <td width="20px"><h3 class="Text2">Staff Member</h3></td> <td width="20px"><h3 class="Text2">referrer</h3></td> <td width="20px"><h3 class="Text2">referred</h3></td> <td width="40px"><h3 class="Text2">Sent Out</h3></td> </tr> <?php while ($rrows = mysql_fetch_array($Referrer)) {?> <tr> <td><h3 class="Text3"><input type="" name="Referrerid[]" id="Referrerid[]" size="4" value="<?php echo $rrows['IssueNum'];?>" /></h3></td> <td><h3 class="Text3"><input type="" name="StaffMember[]" id="StaffMember[]" size="4" value="<?php echo $rrows['StaffMember'];?>" /></h3></td> <td><h3 class="Text3"><input type="" name="referrer[]" id="referrer[]" value="<?php echo $rrows['referer'];?>" /></h3></td> <td><h3 class="Text3"><input type="" name="referred[]" id="referred[]" value="<?php echo $rrows['referred'];?>" /></h3></td> <td><h3 class="Text3"><input name="SentOut[]" type="checkbox" value="1" id="SentOut[]"></h3></td> </tr> <?php } ?> </table> <input name="submit" type="submit" id="submit" value="Create"> </table> </form> </body> </html> <?php } ?> the error i get when the submit button is clicked is 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 '(StaffMember, referer, referred, SentOut, SentOutDate, SentOutBy) VALUES' at line 1 any help would be appriciative Ok I solved the problem, this was my initial query which didn't work: $query3 = "INSERT INTO user (avatar) VALUES ('$avatar') WHERE user_id = '$dbuser_id'"; but this one works: "UPDATE user set avatar = '$avatar' WHERE user_id = '$dbuser_id'"; Guys, I'm not at all able to insert characters like " ' ` and all sorts into my database or it will always return an error. I just created a textarea field in my site of which I just want to istore all those collected datas into my database for later retrieval and all sorts. Please help! Hey all, I keep getting this 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 2 When I use the script below. I'm finding it a bit confusing because everything about it continues to work, it's just it gives me an error. When the script runs, the outcome is "Success! Your dog now looks far more energetic! Looks like this food is all used up." Followed by the error, which cuts the remainder of the page off. Does anybody know why it's doing this? $dogyay = $_POST['dogid']; $checkenergy = "SELECT energy FROM dogs WHERE id=$dogyay"; $energylevel = mysql_query($checkenergy) or die(mysql_error()); $row = mysql_fetch_array($energylevel) or die(mysql_error()); if($row['energy'] >= 100) { echo "<b>Oops!</b> Looks like your dog is full right now...";} else{ echo "<b>Success! Your dog now looks far more energetic!</b><br><br>"; $sql11="UPDATE dogs SET energy=energy + 50 WHERE id=$dogyay"; $result11=mysql_query($sql11); $sql12="UPDATE items SET uses = uses - 1 WHERE itemid=$id"; $result12=mysql_query($sql12); $checkuses = "SELECT uses FROM items WHERE itemid=$id"; $useslevel = mysql_query($checkuses) or die(mysql_error()); $row = mysql_fetch_array($useslevel) or die(mysql_error()); if($row['uses'] == 0) { echo "Looks like this food is all used up.<bR><br>"; mysql_query("DELETE FROM items WHERE itemid='$id'") or die(mysql_error());} Thanks ! This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=352460.0 Hi good people, i'm vary new in this and i'm having trouble with PHP while writing some project for school and because i find many answers on this forum till now i decide to post this.. so here is my problem: I'm trying to make a web page for students and profesors where students (when they are loged in) will be able to sign a date for their exam so i made a form like this : Code: [Select] <form method="POST" action=""> choose exam: <p><select name="exams"> <option value="k1D">exam 1</option> <option value="k2D">exam 2</option> <option value="k3D">exam 3</option> </select></p> choose date: <p><select name="dates"> <optgroup label="Zimski rokovi"> <option value="2011-02-01">01.02.2011.</option> <option value="2011-02-07">07.02.2011.</option> <option value="2011-02-15">15.02.2011 </option> <optgroup label="Ljetni rokovi"> <option value="2011-05-21">21.05.2011.</option> <option value="2011-05-28">28.05.2011.</option> <option value="2011-06-04">04.06.2011.</option> </select></p> <input type="submit" value="Prijavi ispit" name="prijavi"> </form> table for students in mysql has columns for every exam (k1D , k2D..) but how can i make so that student can pick wich exam he wants to sign on some of dates (wich column he wants to fill with wich of dates) ? i tryed some variations of : Code: [Select] $k1D = $_POST['dates']; $kol = $_POST['exams']; mysql_query(" UPDATE studenti SET '$kol' = '$k1D' WHERE ID = '3'"); but i'm just getting different errors.. 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 This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=342913.0 I have been pulling my hair out for the lasy 3 hours i am trying to update a MySql table but i cant get it too work, i just keep getting MySql error #1064 - You have an error in your SQL syntax; if i just update 1 field it works fine but if i try to update more than 1 field it dosent work, Help Please! <?php $root = $_SERVER['DOCUMENT_ROOT']; require("$root/include/mysqldb.php"); require("$root/include/incpost.php"); $con = mysql_connect("$dbhost","$dbuser","$dbpass"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("$dbame", $con); mysql_query("UPDATE Reg_Profile_p SET build='$build' col='$col' size='$size' WHERE uin = '$uinco'"); ?> Which of the below syntax is the better to be using? $bal = 100; if($bal > 100) { echo "you have over 100 pounds; } else { echo "you have less than 100 pounds"; } if ($bal > 100); echo "you have over than 100"; else ; echo "you have less than 100"; endif; 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.
create table mimi (mimiId int(11) not null, mimiBody varchar(255) );
<?php
//connecting to database
include_once ('conn.php');
$sql ="SELECT mimiId, mimiBody FROM mimi";
$result = mysqli_query($conn, $sql );
$mimi = mysqli_fetch_assoc($result);
$mimiId ='<span>No: '.$mimi['mimiId'].'</span>';
$mimiBody ='<p class="leading text-justify">'.$mimi['mimiBody'].'</p>';
?>
//what is next?
i want to download pdf or text document after clicking button or link how to do that hey all, here is the xml code Code: [Select] <title>Dogged determination</title> <link rel="alternate" href="http://video.news.com.au/1959670201/Dogged-determination" /> <author> <name>mr mittens</name> </author> now to parse this to get the title i would do this but what if i want the link? Code: [Select] <?php$feed = simplexml_load_file('http://video.news.com.au/feed.atom');$link = (string) $feed->entry->title;echo $link;?> but what do i do to get the 'link'? i tried link an that obviously didnt work. cheers nick I need help with some syntax for this line of code. I know what I have surrently isn't right. echo "<h3>Welcome $_SESSION['FirstName'] </h3>"; just trying to echo out a session var in a string. |
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