You have an error in your SQL syntax; check the manual that corresponds to your MySQL !! !! $ !!



I have 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
and I don't know how does this happen because whenever I check the SQL that I have made including the inserted data, it shows no error..

How does this happen..

Need Help..

Thanks


Related to : You have an error in your SQL syntax; check the manual that corresponds to your MySQL !! !! $ !!
You have an error in your SQL syntax; check the manual that corresponds to your MySQL !! $ !!
VB & VBnet
I have this Table, forum_posts (forum I'm building).
<?php
require_once "connect_to_mysql.php";// Tell ourselves on screen if we have connected
print "Success in database CONNECTION.....<br />";
$result = "create table IF NOT EXISTS `forum_posts` (
id int(11) NOT NULL AUTO_INCREMENT,
post_author varchar(24) NOT NULL,
post_author_id int(11) NOT NULL,
otid int(11) NOT NULL,
date_time datetime NOT NULL default '0000-00-00 00:00:00',
replyButton int(11) NOT NULL,
type enum('a','b') NOT NULL default,
view_count int(11) NOT NULL,
sect
You have an error in your SQL syntax; check the manual that corresponds to your MySQL !
VB & VBnet

hai i am new to php. i am getting this as error

---You have an error in your SQL syntax; check the manual that corresponds to your MySQL---
can anyone tel wher is the error. Thank u..!!

<html><body><form action="login.php" method="post"><div><table width="100%"> <tr> <td><img src="Logofinalcopy.gif"></td> </tr> <tr> <td bgcolor="aqua"><h2>Login</h2></td> </tr></table><table align="right" style="width:40%"> <br> <tr> <td>Username:</td> <td><input type="text" name="userna
You have an error in your SQL syntax; check the manual that corresponds to your MySQL !! !! $ !!
VB & VBnet

I have 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
and I don't know how does this happen because whenever I check the SQL that I have made including the inserted data, it shows no error..

How does this happen..

Need Help..

Thanks


You have an error in your SQL syntax; check the manual that corresponds to your MySQL !!!
VB & VBnet

i got this error when i trying to submit a form..

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 'order(firstname,lastname,email,address,city,postalcode,country,totalprice,paymen' at line 1

appreciate if you could assist me..

$sql2= mysql_query("INSERT INTO order(firstname,lastname,email,address,city,postalcode,country,totalprice,payment,status) VALUES( '$_POST[firstname]', '$_POST[lastname]', '$_POST[email]', '$_POST[address]', '$_POST[city]', '$_POST[postalcode]', '$_POST[country]', '0', '$_POST[payment]', 'Successful')")or die(mysql_error());
You have an error in your SQL syntax; check the manual that corresponds to your MySQL -
VB & VBnet

Hi, i am try to display and insert info on a person into my database but i keep getting this error whenever i enter the info

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 'join (FName, LName) VALUES ('flsdjflsjfl','jsdlfjslfjvlj')' at line 1

<?phpinclude ("include.php");if(!$_POST){ $display_block = " <form method="post" action="".$_SERVER["PHP_SELF"].""> <p><strong>First/Last Names:</strong><br/> <input type="text" name="FName" size="30" maxlength="75"> <input type="text" name="LName" size="35" maxlength="50"></p> <p>
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
VB & VBnet

I really cant figure out what i did wrong. I´m passing values to the database. It works, as the values are stored. They are checkboxes from previous page, only with a value of 1, or automatic 0 in phpMyadmin.

This is my code:

/* SQL COMMANDS */
$mysqli = new mysqli("localhost", "root", "****" , "buffetkeuze");
if ($mysqli->connect_errno) {
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ")
" . $mysqli->connect_error; } echo 'Connected... ' .
mysqli_get_host_info($mysqli) . "
";  

$insert = mysqli_query($mysqli, "INSERT INTO `buffetkeuze` (
`Gourmand` , `Party` , `Polder` ,       `Frites` ,
`Gebakken_aardappels` , `Rijst` , `Gegratineerde_aardappels` , `Pasta`
, `Ardennerham` , `Kipsate` , `Varkenssate` , `Spareribs` ,
`KipPiriPiri` , `Gehaktballetjes` , `Beenham` , `Minischnitzels` ,
`Gerooktezalm` , `Haringplateau` , `Zalmfilet` , `Kibbeling` ,
`Quiche` ,  `Tempuragroenten` , `Rundercarpaccio` , `Wildpate` ,
`Ossenhaaspuntjes` , `Ribeye` , `Blacktiger` , `ganzenborst` ,
`Wildragout`  , `Fazantenfilet`  , `kaasplateau`  , `geitenkaas` ,
`paling` , `wildezalm` , `Oesters` , `Kreeft` , `Tonijn` ,
`Zeetongfilet` , `Reebiefstuk` , `Hertencarpaccio` ,
`gerookteeendenborst` , `garnalensalade` ) 
VALUES ( 
'$_POST[Gourmand]','$_POST[Party]','$_POST[Polder]','$_POST[Frites]','$_POST[Gebakken_aardappels]','$
  _POST[Rijst]'
,'$_POST[Gegratineerde_aardappels]','$_POST[Pasta]','$_POST[Ardennerham]','$_POST[Kipsate]','$_POST[Varkenssate]','$_POST[Spareribs]','$_POST[KipPiriPiri]','$_POST[Gehaktballetjes]','$_POST[Beenham]','$_POST[Minischnitzels]','$_POST[garnalensalade]','$_POST[Gerooktezalm]','$_POST[Haringplateau]','$_POST[Zalmfilet]','$_POST[Kibbeling]','$_POST[Quiche]','$_POST[Tempuragroenten]','$_POST[Rundercarpaccio]','$_POST[Wildpate]','$_POST[Ossenhaaspuntjes]','$_POST[Ribeye]','$_POST[Blacktiger]','$_POST[ganzenborst]','$_POST[Wildragout]','$_POST[Fazantenfilet]','$_POST[kaasplateau]','$_POST[geitenkaas]','$_POST[paling]','$_POST[wildezalm]','$_POST[Oesters]','$_POST[Kreeft]','$_POST[Tonijn]','$_POST[Zeetongfilet]','$_POST[Reebiefstuk]','$_POST[Hertencarpaccio]','$_POST[gerookteeendenborst]'
) " );

if (!mysqli_query($mysqli, $insert)) { die('Error: ' .
mysqli_error($mysqli)); } echo "values inserted " . "
"; 

$sql =  mysqli_query($mysqli,"SELECT * FROM `buffetkeuze` ") ;

if (!mysqli_query($mysqli, $sql)) { die('Error: ' .
mysqli_error($mysqli)); } echo "values selected " . "
"; while ($row = mysqli_fetch_assoc($mysqli,$sql)) {
$Gourmand = $row['Gourmand'];
$Party = $row['Party'];
$Polder = $row['Polder'];
$Frites = $row['Frites'];
$Gebakkenaardappels = $row['Gebakken_aardappels'];
$Rijst = $row['Rijst'];
$Gegratineerdeaardappels = $row['Gegratineerde_aardappels'];
$Pasta = $row['Pasta'];

etc.    

What is my mistake in this one?



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