I am trying to submit my query into mySQL database but it keeps stating that it fails. It connects to the database fine but will not integrate the query into the table called 'ticket'.
<?php
if (isset($_POST['submit'])){
include 'mysqli_connect.php';
$query = "INSERT INTO ticket (Ticket_ID, Submit_Date, F_Name,
L_Name, Email, Ph_Num, Subject, Priority, Description)
VALUES ('$_POST[Ticket_ID]', '$_POST[Submit_Date]',
'$_POST[F_Name]', '$_POST[L_Name]', '$_POST[Email]',
'$_POST[Ph_Num]', '$_POST[Subject]', '$_POST[Priority]',
'$_POST[Description]')";
$result = mysqli_query($query) or die ('Query Failed:' .
mysqli_error());
mysql_close($conn);
} else {
echo "No submit";
}
?>
Outputs:
Connected Database SuccessfullyQuery Failed:
mysqli_query() has 2 parameters, first is the variable connection, and second is variable of query...
so I think it could be
$result = mysqli_query($conn, $query) or die ('Query Failed:' .mysqli_error($conn));
Try this code:
$query = "INSERT INTO ticket (Ticket_ID, Submit_Date, F_Name,
L_Name, Email, Ph_Num, Subject, Priority, Description)
VALUES ('{$_POST['Ticket_ID']}', '{$_POST['Submit_Date']}',
'{$_POST['F_Name']}', '{$_POST['L_Name']}', '{$_POST['Email']}',
'{$_POST['Ph_Num']}', '{$_POST['Subject']}', '{$_POST['Priority']}',
'{$_POST['Description']}')";
and replace mysqli_query($query), mysqli_error(), mysql_close($conn) with
mysqli_query($conn, $query), mysqli_error($conn), mysqli_close($conn)
it should be like this
<?php
if (isset($_POST['submit'])){
include 'mysqli_connect.php';
$query = "INSERT INTO ticket (Ticket_ID, Submit_Date, F_Name,
L_Name, Email, Ph_Num, Subject, Priority, Description)
VALUES ('".$_POST['Ticket_ID']."', '".$_POST['Submit_Date']."',
'".$_POST['F_Name']."', '".$_POST['L_Name']."', '".$_POST['Email']."',
'".$_POST['Ph_Num']."', '".$_POST['Subject']."', '".$_POST['Priority']."',
'".$_POST['Description']."')";
$result = mysqli_query($conn,$query) or die ('Query Failed:' .
mysqli_error($conn));
mysqli_close($conn);
} else {
echo "No submit";
}
?>
Related
I was trying to insert data into multiple data tables. It's only working for single data tables, I'm just wondering how I would be able to insert data into two data tables. I've been struggling with this issue for the past few hours and can't seem to get to the bottom of it. If anyone has any advice please let me know. :)
<?php
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("localhost","ivodatat","","");
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
// Inputs for security
$fname = mysqli_real_escape_string($link, $_REQUEST['fname']);
$sname = mysqli_real_escape_string($link, $_REQUEST['sname']);
$address = mysqli_real_escape_string($link, $_REQUEST['address']);
$email = mysqli_real_escape_string($link, $_REQUEST['email']);
$phone = mysqli_real_escape_string($link, $_REQUEST['phone']);
$mac = mysqli_real_escape_string($link, $_REQUEST['mac']);
$installer = mysqli_real_escape_string($link, $_REQUEST['installer']);
$status = mysqli_real_escape_string($link, $_REQUEST['status']);
// Insert Query
$sql1 = "INSERT INTO leadlist (fname, sname, address, email, phone, mac, installer, status) VALUES ('$fname', '$sname', '$address', '$email', '$phone', '$mac', '$installer', '$status')";
$sql2 = "INSERT INTO $installer (fname, sname, address, email, phone, mac, installer, status) VALUES ('$fname', '$sname', '$address', '$email', '$phone', '$mac', '$installer', '$status')";
if (mysqli_multi_query($link, $sql1, $sql2)){
mysqli_close($conn);
header("Location: installercontrol.php");
exit;
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
// Close The Connection
mysqli_close($link);
?>
To use mysqli_multi_query you need to append the queries to each other as it only takes one query argument. From the manual:
Executes one or multiple queries which are concatenated by a semicolon.
Try this instead:
mysqli_multi_query($link, $sql1 . ';' . $sql2)
You should probably also update your error message:
echo "ERROR: Could not able to execute $sql1;$sql2. " . mysqli_error($link);
I need to insert two pieces of data into two different tables. It successfully does it with one of the tables but not the second. I have used or die mysqli_error to see if it will tell me the error, but it does not show anything. See the code below:
$sql = "INSERT INTO ticketUsers
(name, emailAddress, password)
SELECT * FROM (SELECT '$name', '$emailAddress', '$dbPassword') AS tmp
WHERE NOT EXISTS (
SELECT name
FROM ticketUsers
WHERE emailAddress = '$emailAddress'
)
LIMIT 1";
$query = mysqli_query($connection, $sql);
if($query)
{
echo "Success entering ticket Users";
}
else if(!$result)
{
echo "Cant enter information";
}
$sql = "INSERT INTO tickets
(id, emailAddress, urgency, subject,
description, relevantURL, status)
VALUES ('$id', '$emailAddress', '$username', '$urgency',
'$subject', '$description2', '$relevantURL', 'Open')";
$query = mysqli_query($connection, $sql);
if($query)
{
echo "Success entering tickts";
}
else if(!$result)
{
echo "Cant enter information";
}
if (!sql)
{
echo "There has been an error creating your ticket.";
}
In your second query, you try to insert in a table with 7 fields 8 values.
I think you don't want to insert '$username' in the query.
I'm on ubuntu running on apache2.
<?php
$con = mysqli_connect("localhost", "root", "password", "loops");
if (!mysqli_connect_errno()) {
echo "Error to connect: ".mysqli_connect_error();
}
$firstName = $_POST['firstname'];
$lastName = $_POST['lastname'];
$age = $_POST['age'];
echo "First: ".$firstName."<br />Age: ".$age;
$sql = "INSERT INTO persons (FirstName, LastName, Age)
values (".$firstName.",".$lastName.",".$age.")";
if (!mysqli_query($con)) {
die("Error :".mysqli_error($con);
}
echo "1 added";
mysqli_close($con);
?>
I don't know what to do now, but the error seems to be on the '$sql', 'cause when I put that
under a comment, the rest works...
You forgot the infamous mistake of wrapping the varchar variables with single quote '
"INSERT INTO persons (FirstName, LastName, Age)
values ('".$firstName."','".$lastName."','".$age."')";
Additionally, It's highly risky for you, not be using protections against SQL Injections
Try this, mysqli_query($con, $sql) You need to execute the query so that it will store into the database
$sql = "INSERT INTO persons (FirstName, LastName, Age)
values ('".$firstName."','".$lastName."','".$age."')";
if (!mysqli_query($con, $sql)) {
...................^
die("Error :".mysqli_error($con);
}
Try this:
$sql = "INSERT INTO persons (`FirstName`, `LastName`, `Age`) values (".$firstName.",".$lastName.",".$age.")";
If this doesn't work, what error do you get.
When my form is submitted (via Ajax), I'm getting the following error message:
[17-Oct-2012 11:46:29] PHP Warning: mysqli_query() [<a href='function.mysqli-query'>function.mysqli-query</a>]: Empty query in /home1/xenongro/public_html/testing/enrolment/thanks.php on line 32
I have a suspicion that it's something to do with the if/else statements, but not sure what the actual problem is.
Can anyone help?
<?php
$firstname = htmlspecialchars(trim($_POST['fname']));
$lastname = htmlspecialchars(trim($_POST['lname']));
$worktel = htmlspecialchars(trim($_POST['worktel']));
$dbc = mysqli_connect('localhost', 'xxxxx', '<xxxx>', 'xxxx')
or die ('Could not connect to MySQL server.');
if ($level != "IOSH Managing Safely"){
if ($funding == "Self Funding"){
$query = "INSERT INTO enrolments (fname, lname, worktel)" .
"VALUES ('$firstname', '$lastname', '$worktel')";
}
else if ($funding == "Employer Funding"){
$query = "INSERT INTO enrolments (fname, lname, worktel)" .
"VALUES ('$firstname', '$lastname', '$worktel')";
}
}
else if ($level == "IOSH Managing Safely"){
if ($funding == "Self Funding"){
$query = "INSERT INTO enrolments (fname, lname, worktel)" .
"VALUES ('$firstname', '$lastname', '$worktel')";
}
else if ($funding == "Employer Funding"){
$query = "INSERT INTO enrolments (fname, lname, worktel)" .
"VALUES ('$firstname', '$lastname', '$worktel')";
}
}
$result = mysqli_query($dbc, $query)
or die ('error querying database');
mysqli_close($dbc);
?>
try
var_dump($query);
var_dump($funding);
just before
$result = mysqli_query($dbc, $query);
it'll give you more information
I suspect that $funding might have slight variation to your constant strings
might be typo / extra space / cap case
There are two situation where no query is being set:
the level does not match the string, or the funding does not match the string.
It might be a problem with the spaces.
Worse, you don't use mysql_real_escape_string and unless magic_quotes_gpc is on, this allows an attacker to inject his SQL.
$funding doesn't appear to be defined in the code example provided, so none of your if's will match.
<?php
$ip = $_SERVER['REMOTE_ADDR'];
$date = date("d/m/y : H:i:s", time());
$dbc = mysqli_connect('localhost', 'root', 'derp', 'derpdb')
or die("Database connection fried.");
$query = "INSERT INTO ipstore (tstamp, ip), " .
"VALUES ('$date', '$ip')";
mysqli_query($dbc, $query);
mysqli_close($dbc);
?>
Can anyone tell me what's wrong with this code? It's meant to store the users IP/date they requested the page in the database. I've tried replacing localhost with 127.0.0.1, no luck. It doesn't bring a message, so it must be connected, however when it comes to querying it just doesn't do it. And it doesn't give a warning. I've checked the DB, nothings there.
Also don't worry, nothing sensitive is there ;)
Thanks
$query = "INSERT INTO ipstore (tstamp, ip), " . "VALUES ('$date', '$ip')";
You are not supposed to use a comma after specifying columns - try
$query = "INSERT INTO ipstore (tstamp, ip) VALUES ('$date', '$ip')";
try it this way
$query = mysql_query("INSERT INTO ipstore (tstamp,ip) VALUES ('$date', '$ip')") or die(mysql_error()); if($query) {echo 'Success'; esle { echo 'Failed'; }
And you will get success for sure