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I am trying to insert some data in the database but it gives me a really confusing error
Now the error is:
Parse error: syntax error, unexpected ')'
Code:
$query = mysql_query("INSERT INTO `members` VALUES (''," .$username ."," . $password . "," . $date . "," .$email . ",1'"));
You forgot some dots over there.
$query = mysql_query("INSERT INTO `members` VALUES (" .$username ."," . $password . "," . $date . "," .$email . ",1)");
you have many mistakes , you forgot points . you forget ) .
you should also specify the columns names
this should work for you
$query = mysql_query("INSERT INTO `members` (firstcolumn ,username , password ,date,email , lastcolumn) VALUES ('' ,'$username','$password','$date','$email',1 ) ");
^----------^--------^-----^----^-----^^---your columns
HERE genaeral rule how to use insert :
INSERT into table (column1 , column2 , column3) VALUES (value1 , value2 , value3)
Try this (replace x,y,z,a,b with column name):
$query = mysql_query("INSERT INTO `members` (`x`,`y`,`z`,`a`,`b`) VALUES ('','.$username.','.$password.','.$date .','.$email.','1'"));
You haven't used ' and " in correct place.
write your query and add at the end
or die(mysql_error());
what it says?
I've worked it out now.
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Closed 6 years ago.
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I am currently using PHP and SQL to throw back some paramters I enter in a form.
I can search numbers perfectly fine and it gives me the correct results but anytime I use a search like "443265dsa44dd" it displays nothing even though it's in the database.
$searchedID = $_POST['uuid'];
$sql = "SELECT name, contact, phone, address FROM test WHERE id = '.$searchedID.'";
if ($result->num_rows > 0) {
while ($row = $result->fetch_assoc()) {
echo "Name: " . $row["name"] . "<br>" . "Contact: " . $row["contact"] . "<br>" . "Phone: " . $row["phone"] . "<br>" . "Address: " . $row["address"] . " ";
}
}
The id is a primary key and set to VARCHAR, any ideas what is happening here?
You have an error when trying to include the searchedID into the sql-string.
Either concat like this:
$sql = "SELECT name, contact, phone, address FROM test WHERE id = '" . $searchedID . "'"
// note, the additional quotes
OR
let php parse that var for you (possible only inside double-quotes):
$sql = "SELECT name, contact, phone, address FROM test WHERE id = '$searchedID'"
BUT
You are vulnerable to sql-injection. So use prepared statements!
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Closed 7 years ago.
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I am new to php. I am trying to connect android with phpmyadmin using webservice .
php Code
<?php
include_once('configuration.php');
$UserId = $_POST['UserId'];
$ProductId = $_POST['ProductId'];
$DesiredQuantity = $_POST['DesiredQuantity'];
$cartstable=mysql_query("SELECT `UserId`, `ProductId`, `DesiredQuantity` FROM `carts` WHERE UId='".$UserId. "' AND ProductId='".$ProductId. "'");
$num_rows = mysql_num_rows($cartstable);
if($num_rows>0){
$updateqry=mysql_query("Update `carts` set `DesiredQuantity`= `DesiredQuantity` + $DesiredQuantity) WHERE UId='".$UserId. "' AND ProductId='".$ProductId. "');
}
else
{
$insertqry=mysql_query ("Insert into `carts` (`UId`, `ProductId`, `DesiredQuantity`) VALUES ('".$UserId. "','".$ProductId. "',$DesiredQuantity)");
}
$carts_ful=mysql_query("SELECT `UserId`, `ProductId`, `DesiredQuantity` FROM `CARTS` WHERE UId='".$UserId. "'");
while($carts = mysql_fetch_array($carts_ful)){
extract($carts);
$result[] = array("UserId" => $UserId,"ProductId" => $ProductId,"DesiredQuantity" => $DesiredQuantity);
}
$json = array("Updated Cart Details" => $result);
#mysql_close($conn);
header('Content-type: application/json');
// echo "Selected Product is added to the Cart !";
echo json_encode($json);
?>
When I tried running,I see the following error
<b>Parse error</b>: syntax error, unexpected 'insert' .
If I Cut and paste,
$insertqry=mysql_query ("Insert into `carts` (`UId`, `ProductId`, `DesiredQuantity`) VALUES ('".$UserId. "','".$ProductId. "',$DesiredQuantity)");
line above the if statement ,It works fine.
I could not understand where is the problem .Please help me finding the solution .
Stack Overflow's syntax highlighting should have been enough to spot the error.
You have missed a closing quote from one of your SQL queries. Find the amendment below.
$updateqry=mysql_query("Update `carts` set `DesiredQuantity`= `DesiredQuantity` + $DesiredQuantity) WHERE UId='".$UserId. "' AND ProductId='".$ProductId."'");
}
else
{
$insertqry=mysql_query ("Insert into `carts` (`UId`, `ProductId`, `DesiredQuantity`) VALUES ('".$UserId. "','".$ProductId. "',$DesiredQuantity)");
}
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Closed 7 years ago.
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I get this two queries with the same variables :
$query = 'UPDATE payee SET payee="oui", datePaiement=\'' . $datePaiement . '\',paiement="'.$paiement.'", typePaiement="' . utf8_decode($moyenPaiement) . '" WHERE id_commande=' . $commande->getNum() . '';
$connexion->exec($query);
$query2 = 'UPDATE commande SET mpaiement="' . utf8_decode($moyenPaiement) . '",pxttc="'.$paiement.'" WHERE noCommande=" . $commande->getNum() . "';
$connexion->exec($query2);
For the first one, my $paiement isn't save in my DB. I get a $paiement = 0 while in my second one $paiement is save as I want.
I have the same pattern in my second query $moyenPaiement, moyenPaiement is save as I want but he is not save in my first query.
Sorry for my explanation, it's maybe confused.
You should not combine both single and double quotes, it will be confusing. Try this:
$query = 'UPDATE payee SET payee="oui", datePaiement="' . $datePaiement . '", paiement="'.$paiement.'", typePaiement="' . utf8_decode($moyenPaiement) . '" WHERE id_commande="' . $commande->getNum() . '"';
$connexion->exec($query);
$query2 = 'UPDATE commande SET mpaiement="' . utf8_decode($moyenPaiement) . '", pxttc="'.$paiement.'" WHERE noCommande="' . $commande->getNum() . '"';
$connexion->exec($query2);
Later edit:
Don't forget to print your queries if something weird happens. These queries can be tested in phpMyAdmin too.
print_r($query);
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these are my query codes. Please help me.
PDO Error: Array
PDO Eror Code: 00000
<?php if ($_POST){
$title = trim($_POST['title']);
$content = trim($_POST['content']);
$id = $_GET['id'];
$save = $PDO->prepare("UPDATE `news` SET `title` = :title WHERE `id` = :id");
$save->execute(array(
"title" => $title,
"id" => $id
));
print_r("Error: ".$save->errorInfo());
print $save->errorCode();
}
?>
It the OK status code.
You always print error but you should to print that only when the query failed.
$sql = $save->execute(...)
if ($sql === FALSE) {
print ('Error: ' . $save->errorCode());
}
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Closed 9 years ago.
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I have the following code, but for some reason I am getting an unexpected T_Variable. I can't seem to figure out where I am getting the error at. Any assistance will be greatly appreciated. Thanks
<? php
$status = mysql_query('SELECT count(*) FROM AHG WHERE `Survey Tech Initials` = 'Jeff' AND completed = 'yes');
if (!$result) {
die('Invalid query: ' . mysql_error());
}
?>
See below
<? php
$status = mysql_query("SELECT count(*) FROM AHG WHERE `Survey Tech Initials` = 'Jeff' AND completed = 'yes'");
if (!$result) {
die('Invalid query: ' . mysql_error());
}
?>